Since the last installment of Probability in Games, a reader requested an analysis and strategy suggestion for a dice game that is simple to explain and play, but complicated to analyze and strategize about. In fact, there is a thread discussing this hypothetical game on the Board Game Geek (BGG) Forums. When I began to work through analyzing this game, I found one very surprising insight, and one very expected complication. So I’ll split this analysis into two parts: the surprising insight will be discussed below, and the expected complication will be addressed in part two.
The game in question involves rolling five customs six-sided dice with sides labelled: FAIL, FAIL, FAIL, PASS, PASS, ACE. The player’s goal is to minimize the number of dice showing FAIL, although each die showing ACE negates a single FAIL when counting this score. After rolling all five dice, the player must choose to set aside one or more dice (to not be rolled again), and then reroll their remaing dice. Play continues in this way until all five dice are set aside, at which point their score is assessed.
One of the strategies proposed on BGG for this game is to always set aside any dice showing PASS or ACE, and to reroll as many dice showing FAIL as possible. But this strategy does not appear to be optimal. There are a few cases cited in that thread where rolling a die that shows PASS leads to better expected results. Between this article and its successor, we’ll investigate the optimal strategy for this game.
How can risking an extra FAIL by rerolliong a PASS be part of an optimal strategy? Since you must lock at least one die with every roll, choosing to roll more dice will give you the option of more future rolls before you must stop. And each of these extra rolls could give you additional chances to improve your score. In this specific game, the fact that a PASS has a chance of being improved to an ACE makes the value of these extra rolls more apparent. But what if that were not the case. Is there be a game where the optimal strategy involves rerolling dice showing their best possible face so that you have the option of additional rolls in the future?
To help understand the value of these extra rolls without the confound of ACEs negating FAILs, let’s consider a slight modification to the game described aboive. Let’s replace the ACE side on each custom die with a side labelled PASS. This gives each die a 50/50 chance to roll PASS/FAIL. Which for our purposes, functions similar to a fair coin toss. If we assign FAIL to the tail sides of our coins, then our score that we are trying to minimize will be the number of tails on these coins. I’ll stick with this coin flipping terminology for this variant to help distinguish it from the original game, but keep in mind how similar the two games truly are.
Let’s come up with an optimal strategy for this new game by computing the expected value for every possible set of coin flips. We’ll start with a single coin and then work our way up to five, so that we can can make use of our previous expected value computations when deciding between reflipping 0, 1, 2, 3, or 4 of coins.
The case of a single coin flip is a great one to quickly review the idea of expected value. We weight each of our possible scores by the chance of flipping them, and this is known as the expected value. In our game we have a 1/2 chance of flipping heads which leads to a score of 0, and a 1/2 chance of flipping tails which leads to a score of 1. When we need to reference the expected value of flipping a single coin in the future, I’ll refer to this value as E1 (E for expected value, and 1 to indicate the consideration of a single coin being flipped).E1 = (1/2 * 0) + (1/2 * 1) = 1/2
When flipping two coins, there are four possible outcomes. I’ve listed the probabilities of each below, with coins showing heads = H and tails = T. Since HT and TH are different permutations of the same outcome, they are combined into a single line below and this explains why the probability of this outcome is higher than the other two.| faces | chance |
| HH | 1/4 |
| HT | 1/2 |
| TT | 1/4 |
When flipping two coins, we get the option of re-flipping a single coin. So let’s extend this table to include the expected score both with (flip1) and without (flip0) this extra coin flip. Whenever possible, we’ll prefer to flip coins showing tails over coins showing heads. Since heads are listed to the left of tails in each row, you can think of this as flipping the right-most coins from this order.| faces | chance | flip0 | flip1 | min(flip 0,1) |
| HH | 1/4 | 0 | E1 | 0 |
| HT | 1/2 | 1 | E1 | E1 |
| TT | 1/4 | 2 | 1+E1 | 1+E1 |
To help ensure this table is clear, let’s step through the middle row. The chances of flipping HT on two coins is 1/2. The score of this flip without any reflips is 1, because it includes one coin showing tails. After reflipping the worst (right-most) coin showing tails here, our expected score is 0 (from the heads showing coin that we set aside) plus E1 (the expected score from a single coin flip, as calculated above).
Our best strategy will involve choosing between flip0 and flip1 to minimize the resulting expected score. This is the purpose of the final column in the table above. And we’ll always consider use this best strategy when computing the expected value for our coin flips.E2 = (1/4 * 0) + (1/2 * 1/2) + (1/4 * 3/2) = 5/8
When building a similar table for three coins, we’ll need an extra column to account for reflipping two of those three coins (labelled flip2). This will help us compute E3.| faces | chance | flip0 | flip1 | flip2 | min(flip 0,1,2) |
| HHH | 1/8 | 0 | E1 | E2 | 0 |
| HHT | 3/8 | 1 | E1 | E2 | E1 |
| HTT | 3/8 | 2 | 1+E1 | E2 | E2 |
| TTT | 1/8 | 3 | 2+E1 | 1+E2 | 1+E2 |E3 = (1/8 * 0) + (3/8 * 1/2) + (3/8 * 5/8) + (1/8 * 13/8) = 5/8
If you were surprised to see the trend of increasing expected values from E1=1/2 to E2=5/8 end with E3=E2, then brace yourself before looking at the table for four coins.| faces | chance | flip0 | flip1 | flip2 | flip3 | min(flip 0,1,2,3) |
| HHHH | 1/16 | 0 | E1 | E2 | E3 | 0 |
| HHHT | 4/16 | 1 | E1 | E2 | E3 | E1 |
| HHTT | 6/16 | 2 | 1+E1 | E2 | E3 | E2 = E3 |
| HTTT | 4/16 | 3 | 2+E1 | 1+E2 | E3 | E3 |
| TTTT | 1/16 | 4 | 3+E1 | 2+E2 | 1+E3 | 1+E3 |E4 = (1/16 * 0) + (4/16 * 1/2) + (6/16 * 5/8) + (4/16 * 5/8) + (1/16 * 13/8) = 79/128
The table above gives us our first chance to flip a coin showing heads without lowering our expected value. When we flip HHTT, we find the same expected value for reflipping TT as TTH. But the more surprising result above is that E4=79/128 (~0.6171), which is smaller than E2=E3=5/8 (0.625). So whenever we find two or three tails that are worth flipping, it will actually be BETTER to flip them as part of a group of 4 coins… even when that means reflipping coins that show heads.
Enough foreshadowing… let’s get to the final table and expected score when flipping all five coins.| faces | chance | flip0 | flip1 | flip2 | flip3 | flip4 | min(flip 0,1,2,3,4) |
| HHHHH | 1/32 | 0 | E1 | E2 | E3 | E4 | 0 |
| HHHHT | 5/32 | 1 | E1 | E2 | E3 | E4 | E1 |
| HHHTT | 10/32 | 2 | 1+E1 | E2 | E3 | E4 | E4 |
| HHTTT | 10/32 | 3 | 2+E1 | 1+E2 | E3 | E4 | E4 |
| HTTTT | 5/32 | 4 | 3+E1 | 2+E2 | 1+E3 | E4 | E4 |
| TTTTT | 1/32 | 5 | 4+E1 | 3+E2 | 2+E3 | 1+E4 | 1+E4 |E5 = (1/32 * 0) + (5/32 * 1/2) + (10/32 * 79/128) + (10/32 * 79/128) + (5/32 * 79/128) + (1/32 * 207/128) = 1251/2048
Here’s the result that I found so surprising: that when flipping five coins, the optimal strategy is to reflip as many coins as possible (including heads) whenever our initial flip contains more than one coin showing tails. In fact, with the help of a computer, I’ve extended these tables up to 20 coins and found that the same pattern and optimal strategy applies to all hand sizes 5-20 (and beyond, I would expect). The expected value slowly creeps around 0.6. I calculated E20 to be ~0.5996 (or 3764286286996839944308718867837813911129194142252135609361 / 6277101735386680763835789423207666416102355444464034512896 to be more precise). And the intuitive strategy of only reflipping your tails is only optimal when you have zero or one tails to flip, or owhen you are down to four or fewer available coins.
I like the idea of a game with an optimal strategy of rolling and playing more, in contrast to the lessons from the classic 80s movie War Games. But repeatedly flipping all but one coin until you either hit the jackpot or run out of coins may appeal more to our inner gamblers than our inner strategy gamers. I think the lesson here for strategy game designers here is to avoid positioning players too far above this inflection point.
I hope that this article has shed some light on the value of getting extra flips, rolls, or other kinds of attempts in games. In the next installment, we’ll address the confound of ACEs negating FAILs within these kinds of expected value computations. Feel free to work ahead or in parallel with me, as I continue to analyze this games. I’ll be interested to hear whether your results align with mine, and whether you notice any other patterns or convenient perspectives for working through this kind of analysis.
:: Cross-posted from http://sugarpill.games/articles/2021/06/01/roll-more-to-win-...
Pondering the problems of probability in games.
Archive for Gary Dahl
02 Jun 2021
- [+] Dice rolls
The astute reader of this series has probably picked up on my enjoyment of push-your-luck style games like Farkle, Dead Man's Draw, Incan Gold, and many more. These games are full of exciting gambles: Do you want to keep what you have earned, or will you risk it all for a chance to win more? I've already written several posts about calculating the chances of different dice rolls and card draws. So in this installment of Probability in Games, I will focus on the expected value and utility of different rolls in Farkle.
Farkle is a classic push-your-luck game with many variations, so I'll assume the following basic rules. Each turn starts with a player rolling six dice. They must set aside one or more scoring dice, and then decide whether they want to keep the points on those dice, or try re-rolling their other dice in hopes of scoring even more points. The catch is that if none of the re-rolled dice can be scored, the player loses all of the points they have held so far, and ends their turn. If you score all six of your dice, the stakes go up as you can re-roll all six of your dice to accumulate even more points that will all be banked or lost together. Scoring varies widely, but generally you can earn 50 points each for each five that you hold, 100 points for each for each one, and then more points for various poker-style combinations. For the sake of simplicity, we'll ignore these extra combination scorings here, and leave adding them as an exercise for the most engaged readers. The game ends as soon as someone banks a total of 10,000 points or more.
In Farkle, each die has two sides that score and four sides that do not, for a total of six sides. So you have a 4/6 chance of busting, and a 2/6 chance of scoring when you roll one die. When you roll two dice, your chances of busting are 4/6 * 4/6 = 16/36, and the other 20/36 of the time you will score. In fact we can create a little table with these busting probabilities. As you'd expect, rolling more dice makes it more likely that at least one will score, and therefore less likely that you will bust.
Number of Dice Chance to Bust
But in Farkle, you can't just focus on the odds of busting. You need to weigh the odds of busting against how bad it will be for you to bust. And we'll get to looking at the scoring potential of each roll in the following paragraph. Let's use the number of points won or lost as a measure of how good or bad a roll might be (more on this choice below). We can now compare two situations, one where you are holding four 5s worth 200 points, and another where you are holding four 1s worth 400 points. In each case you have a 44% chance of busting according to the table above. But you are clearly risking more when you roll the 1s, because you have more points to lose. We can get a single number to quantify this risk by multiplying our odds of busting with the cost of busting: 0.44 * -200 = -88 with four 5s, versus 0.44 * -400 = -176 with four 1s.
Of course the only reason that we consider re-rolling and potentially busting is because it provides a chance for us to score points. When rolling one die, we have a 1/6 chance of scoring 50 points, a 1/6 chance of scoring 100 points, and 4/6 chance of scoring nothing. By multiplying these odds and point values together like we did above we get: 0.16 * 50 = 8, 0.16 * 100 = 16, and 0.66 * 0 = 0. The sum of these probability-weighted values is called the expected value. The expected value tells us that on average, we can expect to score 8 + 16 + 0 = 24 points per die. This actually comes out to 25 points, if you avoid rounding the percentages like I did above.
By combining these expected winnings and losses, we can calculate the expected value for an entire roll in Farkle. Let's return to the example of holding four 5s worth 200 points. The points that we can expect to win on our next roll of two dice is 2 * 25 = 50 points. And we previously calculated the risk of this roll to be -88. Adding these numbers together we find that the expected value of a roll from this position is 50 + -88 = -38 points. Some times this roll will score us points and sometimes it will lose, but we can now see that on average this decision will lose us more points than it will win. This calculation might lead you to wonder about the threshold between rolls with positive and negative expected values. Here's a summary of the calculation that we are using to calculate the expected value of a single roll in Farkle:
expectedValue = numberOfDice * 25 + (4/6)^numberOfDice * (-heldPoints)
The threshold between positive and negative expected values is obviously at zero. So lets put zero in for the expected value in this equation, and then solve for heldPoints based on the numberOfDice. This will allow us to create a table containing the maximum number of held points that you should ever risk, if you want to ensure you that you only make re-rolls with positive expected values.
heldPoints < numberOfDice * 25 / (4/6)^numberOfDice
6 < 1708
5 < 949
4 < 506
3 < 253
2 < 112
1 < 37
So one strategy for playing Farkle is to check this table, and then only roll when the expected value of a roll is positive. While this seems like a rational (and boring) strategy to pursue, it fails to account for one very important (and interesting) factor that is common among many press-your-luck games. As you are racing your opponents toward a target number of points, all points are not valued equal. The changing value of these points as the game plays out are often described in terms of utility. Fifty points is much more useful and therefore has more utility when you are fifty points from winning, than when you have zero points at the beginning of a game.
When you are losing, points will have more utility than when you are in the lead, because you need them to catch up before you can win. If you imagine replacing the 25 points per die in our equations above with a number that increases to reflect this increasing utility, you'll see that this results in increased maximum held point thresholds in our table. In an extreme case, you can imagine one of your opponents being very likely to win on their next turn. This should lead to a huge increase in your point valuation and maximum held point thresholds. And the result of these increases will be that you choose to roll every chance you get until you either bust or win yourself.
It's tempting to think about these utility functions as being dependent entirely on game state including players' scores. However, I believe that these functions can vary not only from player to player, but also from turn to turn as players attempt to mislead and confound the expectation of their opponents. However that sounds like a good potential topic for a future post. I hope you've enjoyed this investigation of probability in games! Please feel free to comment with questions and requests for future posts. Also please thumb and share this blog, if you find it interesting or useful. Thanks for reading!
- [+] Dice rolls
15 May 2014
In this series, we’ve already looked at the probabilities of rolling straights versus multiple dice of a kind, and the probabilities of rolling different sums with multiple dice. In this installment of Probability in Games, we’ll combine aspects from each of these techniques to analyze a mechanic from classic games like Can’t Stop and Kingsburg (both available on iOS). In each of these games, you roll some dice, choose a subset of those dice to sum, and that sum then influences your turn in some way. In the example of Can’t Stop, players roll four dice. They then divide those dice into two groups of two dice, and calculate the sum of each pair. These two sums determine which, if any, scoring track(s) they can progress along on their turn. What do you think this probability distribution should look like? In other words, how often will your roll allow you to choose two dice with a specific sum, for each possible sum?
Modern gamers tend to be familiar with the probability distribution of sums rolled on two six-sided dice. An abbreviated histogram of this distribution curve is depicted in the small pips below each numbered token in the Settlers of Catan.
Here's the full probability distribution for rolling 2d6:
And at first glance, this graph also looks similar to the board in the game Can’t Stop. Are the probability distributions in these two games related? And if so, how?
In this article, we’ll analyze the player’s chances of being able to add two of four randomly rolled dice to a particular sum. In fact we’ll do this calculations for every possible sum to explore the distribution of these probabilities. To do this, I’d like to start with a somewhat naïve approach that will lead us to a rough approximation of the answer. From there, we can discuss the short-fallings of this naïve approach and use those insights to develop a more complete solution.
Let’s think about the number of ways that a player might choose two of the four dice that they have rolled. You might recall our discussion from a previous blog post about calculating combinations, and recognize that this is 4-choose-2 = 4!/(2!*2!) = 6 ways to choose two of the four dice. If not, you can just as quickly list out the six ways of choosing two of four dice. From this, we might naïvely expect that choosing any two of four randomly rolled dice should be equivalent to choosing any one of six randomly rolled pairs of dice. In both cases we have six choices, and each choice consists of two randomly rolled dice.
So let’s find the likelihood that one of six randomly rolled pairs of dice hit each possible sum… starting with the most common sum of seven. We’ll start by working out the odds that a single pair of dice will add to seven. No matter what you roll on the first die in any pair, there is always one value for that second die that can be added to the first to reach seven (in fact, it is always the value on the bottom of the first die). So there is a 1/6 chance that any pair of randomly rolled dice will add up to seven.
But what if we roll six pairs of dice? This calculation might remind you of the previous blog post where we wanted to know how often we’d roll multiple dice with a particular value. We have a 1/6 chance that the first pair adds to seven, and the other 5/6 of the time we need to check another pair. The same goes for each subsequent pair. Expanding this out, we get:
1/6 + 5/6*(1/6 + 5/6*(1/6 + 5/6*(1/6 + 5/6*(1/6 + 5/6*(1/6))))) = 66.5%
So we have a 66% chance of rolling a pair of dice that add up to seven, when rolling six pairs of dice. We can substitute into this equation, the probabilities of rolling other sums (as listed on the table above), to find our chances of rolling them across six different attempts to build the following table.
With this distribution in mind, let’s now return to our initial assumption about how choosing two of four randomly rolled dice should be equivalent to choosing one of six randomly rolled pairs of dice. Can you think of any ways or reasons why these two mechanisms should differ?
In my mind, there are two reasons that these distributions should diverge. The first has to do with rolling multiple dice of a kind. For instance when you roll four of a kind, you have no choice about the sum that you choose. And although this could happen when rolling six pairs of dice, it will happen much less often with unlikely sums like two and twelve. Also notice that the sum of two equal dice is always even, so it seems reasonable to expect some more variation between odd and even sums. The second issue in my mind has to do with rolling dice with different values. For instance imagine rolling dice with four different values. Whenever this happens, you are guaranteed to have rolled two dice that add up to seven. There is just no way to pick four different numbers between one and six without getting at least one pair that adds up to seven. Give it a try, if you don’t believe me. In contrast to this, it is quite possible to roll six pairs of dice with different sums without any of them being equal to seven.
Alright, so if these techniques result in different probability distributions, how do we formulate a better solution? To get at the issue of our options being limited by rolling multiple equal dice, let’s start by breaking down the different ways in which we might roll duplicate values. For each type of roll, we’ll make a note of our chances of rolling it.
Four Singles (ABCD): 6/6 * 5/6 * 4/6 * 3/6 = 27.7%
In order to roll four singles, the first die can have any value (6/6), the second can be anything besides the first (5/6), the third dice can be anything other than the previous two (4/6), etc.
One Pair (AABC): 6/6 * 1/6 * 5/6 * 4/6 * 4C2 = 55.5%
The first die can be anything (6/6), another must match that (1/6), and we need two more that are not equal to the first nor to each other (5/6 and 4/6). However the pair can really fall between any two of the four dice (4C2).
Two Pairs (AABB): 6/6 * 1/6 * 5/6 * 1/6 * 4C2/2 = 6.94%
The first die can be anything (6/6), and the second must match (1/6). The third die must be different from the first two (5/6), and the fourth must match the third (1/6). However each pair could really fall between any two of the four dice (4C2). Since choosing one pair automatically fixes the other, this is actually counting every possible order twice so we divide by two (/2).
Three of a Kind (AAAB): 6/6 * 1/6 * 1/6 * 5/6 * 4C1 = 9.25%
The first three dice must all be the same (6/6, 1/6, and 1/6), and the fourth die must be different (5/6). However the single unique value rolled could appear on any of the four dice (4C1).
Four of a Kind (AAAA): 6/6 * 1/6 * 1/6 * 1/6 = 0.462%
The first die can be anything (6/6), and every other dice must be the same (1/6, 1/6, and 1/6).
With these different types of rolls and their likelihoods in mind, we’ll now turn to the distribution of sums available within each type of roll. These distributions can be broken down into four basic distributions: the sum of two equal values (Equal), the sum of two unique values (Two Unique), the sum of two from three unique values (Three Unique), and the sum of two from four unique values (Four Unique). Here’s how each of our roll types can be decomposed into these distributions.
Four Singles (ABCD): Four Unique (AB, AC, AD, BC, BD, CD)
One Pair (AABC): Equal (AA) + Three Unique (AB, AC, BC) – overlap (AA = BC)
Two Pairs (AABB): 2 * Equal (AA, BB) + Two Unique (AB)
Three of a Kind (AAAB): Equal (AA) + Two Unique (AB)
Four of a Kind (AAAA): Equal (AA)
Hopefully these letters designating unique dice values help you see how these rolls can be decomposed into these different distributions. Once we work out these four distributions, we’ll be able to calculate the full probability distribution for every possible sum in Can’t Stop with them. As an example of how this is done, let’s consider computing the probability of being able to choose two dice that sum to seven. In terms of expressing this, I’ll write Four Unique(7) below to mean the chances of four unique values including two that add up to 7, Equal(7) to mean the chances of two equal values summing to seven, etc. Also, don’t worry about that overlap term above for now. It’s zero in this case, and we’ll discuss it in more detail near the end of this discussion.
27.7% * Four Unique(7) +
55.5% * (Equal(7) + Three Unique(7)) +
6.94% * (2*Equal(7) + Two Unique(7)) +
9.25% * Equal(7)
Fortunately these distributions are relatively easy to write out. Equal is the easiest, since any die value added to itself is an even value. So each even value have a likelihood of 1/6, and every other value has a chance of 0/6. For the sums of unique values, there fortunately aren’t too many of them. There are 6C2=15, 6C3=20, and 6C4=15 ways to choose two, three, and four unique values from the six possible dice values. So I wrote each of these combinations out, and counted the number of times each sum could be made from a combination. I’d certainly love to hear, if anyone knows of a formula to generate these distributions. In the mean time, here are the values that I calculated:
By plugging these values into our formula above, we can find the chances of rolling four dice so that a pair of them add to seven.
27.7% * 100% +
55.5% * (0% + 60%) +
6.94% * (2*0% + 20%) +
9.25% * 0%
= 62.3% (or without the rounding errors is 64.35%)
We can perform a similar calculation for the ten other possible sums, but first lets discuss that overlap that we’ve been ignoring. When rolling one pair, it’s possible that the two unique values have the same sum as the the pair of matching dice values. The formula above is counting each of these occurrences twice: once in the Equal distribution, and once in the Three Unique distribution. However this can only happens with even sums. Specifically, this happens each time the two non-matching dice add up to the even sum in question. And you might recall that we just counted the number of ways that two unique values can add up to each possible sum while generating the distribution for Two Unique. So for even sums, the overlap that we must subtract is Two Unique*15/60. Multiplying by fifteen gets us back to the number of ways that two unique values can add up to the sum, and dividing by sixty ensures that we only count each way once out of the sixty possible ways to roll exactly one pair among four dice: 6C3 * 3 = 60. With all of this work behind us, we can now finally compute the probability distributions for rolling every possible sums in Can’t Stop:
This post has now gotten much longer than I originally anticipated, so I will stop here. If you’ve enjoyed this exploration, I encourage you to take it further. What really matters when playing Can’t Stop is that your roll produces one of the three values that you are currently scoring. Try extending these techniques to come up with a distribution that covers every possible combination of three sums, instead of for each sum individually. As always, please post any questions or request that you have for this or future installments of the blog in the comments below. Thanks for reading!
- [+] Dice rolls
26 Feb 2014
This installment of Probability in Games will explore conditional probabilities using Bayes Rule (aka Bayes Law and Bayes Theorem). Bayes Rule can be used to analyze probabilities in many popular bluffing and deduction style games. However, many of the interesting probabilities in these games are also accessible through more straight forward calculations. In this article, I will propose a variant of the popular game Zombie Dice to help illustrate the concept and application of Bayes Rule.
Let’s begin by reviewing the notion and notation of conditional probabilities, which Bayes Rule will express a relationship between. Conditional probabilities are probabilities that rely on a specific condition in order to be true. In the game Zombie Die, there are three different colors of six-sided dice: green, yellow, and red, that show brains on different number of sides: three, two, and one, respectively. We can use conditional probabilities to express the chances of rolling brains in a way that relies on the color of the die being rolled. The notation P(X|Y) is the probability of X, when Y is guaranteed to be true. So in Zombie Dice:
P(Rolling A Brain | Rolling A Green Die) = 3/6
P(Rolling A Brain | Rolling A Yellow Die) = 2/6
P(Rolling A Brain | Rolling A Red Die) = 1/6
Notice that the order here is critical. There is an important difference between P(Rolling A Brain | Rolling A Green Die) and P(Rolling A Green Die | Rolling A Brain). A common mistake is for people to expect these conditional probabilities to always be equal. These conditional probabilities are illustrated in the diagram below, using a blue background to indicate all outcomes that satisfy the condition, and a white background to indicate which of those outcomes represent a success. On the left, P(Rolling a Brain | Rolling One Green Die) illustrates the chances of rolling a brain when you know that you are rolling a green die. On the right, P(Rolling One Green Die | Rolling A Brain) illustrates the chances that you are rolling a green die when a brain is rolled.
Notice that the successful events on each side of this diagram are the same: they include the green brain. The difference between these conditional probabilities comes from the different groups of possible outcomes that satisfy each probability’s condition. Here’s a classic Venn diagram showing this same relationship. The intersection of these two groups of events (X∩Y) represents the shared successes of both P(X|Y) and P(Y|X). Any difference in these conditional probabilities comes exclusively from the relative sizes of X and Y. Can you see in this diagram which will be larger between P(X|Y) and P(Y|X)?
Another way to express the conditional probabilities illustrated in this diagram is as follows:
1. P(X|Y) = P(X∩Y) / P(Y)
2. P(Y|X) = P(X∩Y) / P(X)
From these two formulas, we can derive Bays rule, which relates P(X|Y) to P(Y|X). We do this by solving for P(X∩Y) in the second equation, and then substitute that expression in for (X∩Y) in the first equation.
3. P(X∩Y) = P(Y|X) * P(X) <= solve for P(X∩Y) in equation 2
4. P(X|Y) = P(Y|X) * P(X) / P(Y) <= substitute result from 3 into equation 1 (Bayes Rule)
With Bayes in mind, let's return the our Zombie Dice example. We've already seen how this game's dice are color coded to bring attention to the conditional probabilities of their results, and it was pretty easy to determine P(Rolling a Brain | Rolling a Green Die). So now let's try using Bayes Rule to calculate P(Rolling a Green Die | Rolling a Brain). To save space, I'll abbreviate “Rolling a Green Die” as G, and “Rolling a Brain” as B. Bayes tells us that:
P(G|B) = P(B|G) * P(G) / P(B)
P(B|G) = 3/6 <= as discussed above
P(G) = 6/13 <= because six of the thirteen dice are green
P(B) = (3*6 green dice + 2*4 yellow dice + 1*3 red dice) / (13dice*6sides) = 29/78
Therefore, P(G|B) = (3/6) * (6/13) / (29/78) = 1404 / 2262 = 62%
Now, can you imagine an expansion for Zombie Dice that would encourage players to consider conditional probabilities in this direction: asking how likely a particular color would be giving a specific symbol? I have a proposal that is completely untested, but absolutely inspired by thinking about how games might challenge players to consider some of the implications of Bayes Rule. I'd love to hear if anyone can think of an existing game that already does this, or of a fun new game concept that you or someone else reading this might try.
My idea for this expansion is to give players the option of spending some of their their hard earned brains to buy one of three different colors of armors. Each armor protects you from one future shotgun blast during the current round, but only protects you from a blast on a die with a matching color. Remember that green, yellow, and red dice have one, two, and three sides showing shotgun blasts respectively. So when you are considering the color of the armor you should purchase you are estimating P(Rolling a Specific Color of Die | Rolling a Shotgun). This seems like an interesting decision, since it will change depending on the number and colors of dice that have already been drawn and rolled. However it could also completely unbalance and break the game, since it is untested.
If you'd like more practice working with Bayes Theorem, I encourage you to try calculating P(Rolling a Red Die | Rolling a Shotgun). From there, you might also try to calculate some strategic thresholds for when each color of armor is most useful, based on the composition of dice left to be drawn and rolled. As always, post any questions or requests you have for this or future installments of the blog in the comments below. Thanks for reading!
::Cross-posted from http://sugarpillstudios.com/wp/?page_id=1042
- [+] Dice rolls
05 Dec 2013
This installment of Probability in games focuses on the concept of variance as it relates to rolling lots of dice. Rather than looking at the probability of rolling specific combinations of dice (as we did in Probability in Games 02), this article is focused on the probability of rolling dice that add up to different sums. The inspiration for this topic comes from two different sources. The first was a statement by Geoff Engelstein that more dice can mean less luck in a game (during the Dice Tower podcast on Nov 12th, 2013). And the second was an assertion by James Ernest about the long lasting advantage of rolling well early in a game (during a GenCon2012 lecture). Lets try to shine some light on both of these observations by digging into the concept of variance.
Having players add up the numbers rolled on several dice is a common mechanism in many games. Part of what makes this mechanism interesting is that different sums often have different probabilities of being rolled. For instance, you are about twice as likely to roll a sum of 7 as you are to roll a sum of 4 on two six sided dice. Yet, a sum of 7 is over four times more likely than a sum of 4 when rolling three six sided dice. This is in contrast to rolling a single die, where every side (and every possible sum) is equally likely.
When working with two or three dice, it’s not too hard to write an exhaustive table (or graph) for the probabilities of every sum. However, this becomes tedious for larger numbers of dice. Let’s look at some graphs for the distribution of sums when rolling multiple dice. The “n=” title for each graph tells you how many dice are being rolled. Below that, the height of each bar indicates the likelihood of rolling a specific sum with that many dice.
Notice that there are a few important changes to these graphs as the numbers of dice increase. First, the central and most frequent sum of each graph moves to the right (getting larger) as more dice are added. We’ll look at calculating the mean, which is the measurement of this central and most common sum. Second, the curve gets closer and closer to the common bell shape of a normal distribution. In fact, the Central Limit Theorem provides some insight into why the sum of a bunch of random dice must always approximate this normal distribution. This is part of what makes that bell curve so common. The width and steepness of this bell can be quantified with a measurement called variance that we’ll explore in more detail below. And third, the entire curve gets wider by five extra sum possibilities per die. The measurement of this space of possible sums is called the range, and the only reason that I mention it is to help distinguish it from variance. Let’s jump right into calculating the mean and variance when rolling several six sided dice.
The mean of each graph is the average of all possible sums. This average sum is also the most common sum (the mode), and the middle most sum (the median) in a normal distribution. In terms of looking at bell curves, the mean is how far left or right on the x-axis you’ll find the highest point of the curve. To calculate this mean for a single die, we can take the weighted average of every possible sum. However, the symmetry in a bell curve provides us with a nice shortcut of averaging only the smallest and largest possible sums.
Mean(1D6): (1 * 1/6) + (2 * 1/6) + (3 * 1/6) + (4 * 1/6) + (5 * 1/6) + (6 * 1/6) = 21/6 = 3.5
Mean(1D6): (1 + 6) / 2 = 7/2 = 3.5
Mean(2D6): (2 + 12) / 2 = 7
Mean(3D6): (3 + 18) / 2 = 10.5
Mean(nD6): (n + 6*n) / 2 = n * 7/2
Now that the mean is out of the way, we can discuss variance. Variance is a measure of how spread out the values in a distribution are. In our example, a low variance means the sums that we roll will usually be very close to one another. By contrast, the variance is large when the sums that we roll are frequently distant values. The way that we calculate variance is by taking the difference between every possible sum and the mean. Then we square all of these differences and take their weighted average. This gives us an interesting measurement of how similar or different we should expect the sums of our rolls to be.
Variance(1D6): (1 - 3.5)^2 * 1/6
+ (2 - 3.5)^2 * 1/6
+ (3 - 3.5)^2 * 1/6
+ (4 - 3.5)^2 * 1/6
+ (5 - 3.5)^2 * 1/6
+ (6 - 3.5)^2 * 1/6
= 70/24 = 2.91
This was a bit more involved than calculating the mean. But fortunately, variances (like means) can simply be added up to account for extra dice (this is because each random die roll is an independent event).
Variance(2D6): 70/24 + 70/24 = 140/24 = 5.83
Variance(3D6): 70/24 + 70/24 + 70/24 = 210/24 = 8.75
Variance(nD6): n * 35/12
We now have a nice way of calculating the mean and variance for the sums of any number of six sided dice. The mean is easy to see in each graph, but the variance is a bit trickier to wrap our heads around. A more natural way to think about variance is to think about the percentage of rolls that share a small range of sums. Something like, most (68%) of rolls should sum to a value between 13 and 22. But there is one step we must take between finding the variance and relating it to a percentage like this, and that is to calculate something called the standard deviation.
Think of the standard deviation as another way of measuring variance; much like the way that distance can be measured in inches, meters, and light years. The standard deviation is easy to calculate once you know the variance, it’s just the square root of the variance. Another benefit of the standard deviation is that it is in units that we can visualize in relation to our graphs. Approximately 68% of our rolls will have sums that land within one standard deviation of the mean. And about 95% of our rolls will fall within two standard deviations of the mean. These magic percentages are common to all normal distributions. Here’s a graph that shows how these standard deviations relate to the chances of different sums. The greek letter mu is used here to label the mean, and the greek letter sigma represents the standard deviation.
We can now find the ranges of sums that will be most commonly rolled with any number of dice. Let’s go through the example of finding the range of sums that will account for 68% of all six die rolls. We start by calculating the mean, the variance, and the standard deviation for the sums of six dice.
Mean(6D6): 6 * 3.5 = 21
Variance(6D6): 6 * 35/12 = 17.5
StandardDeviation(6D6): SquareRoot(17.5) = 4.18
Because 68% of a normal distribution is always within one standard deviation of the mean, we now know that 68% of the time that we roll six dice, those dice will have a sum between 21 – 4.18 = 16.82, and 21 + 4.18 = 25.18. Obviously we can only roll sums that are whole numbers, so it’s 17 to 25. But remember that this is only an estimate, and that the distribution of sums for six dice are merely and approximation of the normal distribution.
It’s natural to look at this relationship between standard deviations and percentages, and wonder about the percentages that lie between each multiple of the standard deviation. For instance, you might want to calculate the percentage of rolls that sum to a value within 3 of the mean. This is a fairly complex calculation to perform by hand, but it is common enough to warrant look-up tables and calculator functions (like logarithms and various trigonometric functions). While preparing this article, I came across the following link to an online table / calculator of what are often called z-scores: http://davidmlane.com/hyperstat/z_table.html. Let’s try entering the mean and standard deviation that we just calculated for the sum of six dice into this webpage. Now, we can find out the percentage of rolls that will fall above, below, between, or outside of any particular sum(s). For instance, we can find the chance of rolling six dice to sum a value within 3 of the mean by entering “Between: 18 and 24”. The area (probability) field then populates with the value 0.5267, which tells us that 52.67% (or just over half) of our rolls should fall in this range.
If you think of this conversion from mean, standard deviation, and range to a percentage as a table look-up, you might correctly guess that you can perform the look-up in reverse. Instead of looking up a percentage based on a range of sums, you can just as easily look up a range based on a desired percentage. For instance, maybe you’d like to find a range of sums that account for 30% of the rolls of six dice. To do so, change the radio button at the top of the webpage linked above, from “Area from a Value” to “Value from an Area”. Then enter the mean and standard deviation for six dice, followed by the area (which we’ve been calling the probability) of 0.3 for 30%. The radio buttons at the bottom should now allow you to calculate the following six die 30% chance rolls.
Sum < 18.8
Sum > 23.1
19.3 < Sum < 22.6
Based on the length of this post, I believe this will be a good place to end. Hopefully you are now more comfortable calculating probabilities for rolling any range of sums on any number of dice. In order to reach this point, we've had to wrap our brains around the concept of variance, and acquire some experience working with normal distributions. To keep your brain going until next time, here's the problem posed by James Ernest (see the link above):
In a dice-driven horse race where each player will roll a 6-sided die 50 times, suppose the results after turn 1 are 1 versus 6. This early in the game, with 49 rolls to go, you would hope that the game is not already tilted heavily in one player's favor.
This takes a bit of clever-ness, but see if you can use this concept of variance to figure out each players' chance of winning in this game. Please post your solutions and how you got them below, along with any questions or requests you have for this or future installments of the blog. Thanks for reading!
::Cross-posted from http://sugarpillstudios.com/wp/?page_id=1004
- [+] Dice rolls
Combinations, and their younger siblings Permutations, come up frequently when analyzing game designs and strategies. These concepts are also frequently taught and explained poorly, as in “here, use this formula to solve that problem”. In this blog we’ll investigate the mathematics behind counting combinations, and see how it can be used to determine the odds of drawing specific cards in the deck building game Ascension. If you’re not familiar with this game, there is a great implementation for iOS devices.
In Ascension, each player starts with their own deck of ten cards. Before each turn they randomly draw five cards from their own deck. Eight of the cards in each starting deck are called Apprentices (A), and the other two are Militia (M). Different combinations of Apprentices and Militia may make more or less useful starting hands depending on a separate center row. So we’re going to figure out how likely different starting hand advantages might be in this game.
We’ll begin by considering the different ways that a deck of ten cards can be arranged after shuffling (another name for each possible arrangement is a permutation). There are ten different cards that could end up on the top of the deck after shuffling. With each of those possible deck tops, there are nine other cards that might follow. So there are a total of 10 * 9 = 90 different possibilities for the top two cards in a shuffled ten card deck. Notice how this differs from the number of ways you might roll two ten-sided dice (10*10=100): since each card can only occupy one position in the deck, but every value can be rolled on every die. Each of our 90 possibilities might be followed by any the 8 remaining cards, and so on. The name for the mathematical operation of multiplying every number from a starting value down to one is factorial, and it is typically expressed with an exclamation mark. We can use this operation to calculate the number of ways that a deck of ten cards can be shuffled.
10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 3,628,800
That’s a lot of permutations, but many of them are redundant for our purposes. For instance, imagine swapping the first two cards from any of these possible orderings. This will result in a new permutation (aka ordering), but it will not change which cards we draw in our first hand.
So how can we reduce these 3.6 Million+ permutations down to the number of orderings that result in unique starting hands? We divide out the redundancies. We need to figure out how many different permutations there are for each unique starting hand, and then divide our 3.6 Million by that number. We’ll start by calculating the number of ways that any five cards can be arranged within the top five positions of the deck. Notice that this is similar to the problem of calculating the number of ways the entire deck can be shuffled, but with only five cards instead of ten.
5! = 5 * 4 * 3 * 2 * 1 = 120
It’s tempting to divide by 120 and think we’re done. But for each of these 120 ways that five cards can be arranged at the top of a starting deck, the bottom five cards can also be arranged in just as many ways. So there are a total of 120 * 120 = 14,400 different ways that a deck of ten cards can be rearranged without changing which five cards are on the top and bottom of the deck. Dividing our initial 3.6 Million+ permutations for a shuffled deck by this 14,400 will tell us the number of unique starting hands we might draw from a deck of ten cards.
10! / (5! * 5!) = 3,628,800 / 14,400 = 252
It’s possible that you recognize these calculations from the general formula for combinations.
n C k = n! / (k! * (n-k)!)
The left hand side of this equations just states that this is the formula for calculating the number of ways that you can choose k objects from a set of n objects (the C stands for combinations and choose). In our case, we are finding the number of ways to choose k=5 cards from a set of n=10 cards. To calculate this, we started with the number of ways all n=10 cards could be arranged: n!. We then divided out the redundant ways that the first k=5 cards that we are drawing could be ordered: k!, and then finally divided out the ways that the remaining n-k=5 cards at the bottom of the deck might be arranged: (n-k)!.
Back to our Ascension example. We know that there are 252 unique ways to draw five cards from a shuffled deck of ten. But in Ascension, 8 of these cards are Apprentices and the other 2 are Militia. So if we only consider the types of cards that we draw (rather than which of those cards we draw), there are really just three types of opening hands we might draw. The likelihood of each opening is the percentage of hands, out of the 252 possible, that match that opening’s pattern.
Opening: #1 #2 #3
Hand One: AAAAA AAAAM AAAMM
Hand Two: AAAMM AAAAM AAAAA
Probability: ?/252 ?/252 ?/252
Consider the number of ways that you can draw five apprentices in your first hand. There are eight apprentices total, so finding the number of ways you can draw five of them can be formulated as a combinations problem: 8 C 5.
8 C 5 = 8! /(5! * (8-5)!) = 40,320 / 720 = 56
So the probability of drawing Opening #1 is 56/252. This is also the probability for Opening #3, because there are the same number of ways to select five cards for the top of the deck, as there are ways to select five cards for the bottom of the deck. All that’s left is to calculate the number of ways that we might draw the hands in Opening #2. This involves choosing four of the eight Apprentices, and one of the two Militia. Each way of selecting Apprentices can occur with each way of selecting the Militia, so we should multiply the the two combinations: (8 C 4) * (2 C 1)
8 C 4 = 8! / (4! * (8-4)!) = 40,320 / 576 = 70
2 C 1 = 2! / (1! * (2-1)!) = 2 / 1 = 2
(8 C 4) * (2 C 1) = 70 * 2 = 140
This helps us complete our probability table with a 140/252 chance of drawing Opening #2.
Opening: #1 #2 #3
Hand One: AAAAA AAAAM AAAMM
Hand Two: AAAMM AAAAM AAAAA
Probability: 22.2% 55.5% 22.2%
I originally intended to take this example a bit further, but don’t want this post to get too long (if it isn’t already). As food for thought, you might consider further calculating how often each of these openings should coincide with valuable starting center row cards that require four or five Apprentices to be purchased. As always, please let me know what you think of this series, and whether you have any questions or comments that might help improve it going forward.
::Cross-posted from http://sugarpillstudios.com/wp/?page_id=990
- [+] Dice rolls
The idea for this article was inspired by a tweet from WatchItPlayed’s Rodney Smith, while playing the game Roll For It: https://vine.co/v/hTMDn5BjJY7. In fact I just heard from Chris Leder (the designer of Roll For It), that he experienced the exact same luck on his first roll in an early prototype of the game. So what is the probability of rolling six different numbers on six dice? And are people justified in how surprised they are when they see this roll? Whether you are familiar with Roll For It or not, now is a great time to play a few rounds using the free online app at http://www.rollforitgame.com/online.html.
Before we get to the 15pt card from the video, let’s consider the 2pt card that is won by rolling two ones. What are the chances of winning this card, when you have only two dice to roll? Most gamers understand that the probability of rolling a one on a six sided die is 1/6. But we need to roll two ones to win this card. To help illustrate this, lets use the function P(x,y) to describe the probability of rolling at least x ones with y dice. So P(1,1) is the probability of rolling a one with a single die, or 1/6. And P(1,0) is the probability of rolling a one with zero dice. This is impossible, and we’ll call this P(x,y) = 0 whenever x > y. What we’d really like to calculate next is the probability of rolling two ones with two dice, or P(2,2). Note that I’m only using this P(x,y) function as a place holder for probabilities that still needs to be calculated. I don’t intend to derive or use any “magic” formulas for calculating this below.
It’s often helpful to consider rolling dice one at a time to help simplify problems like this. The first die that we roll has a 1/6 chance of being a one, which means that winning depends entirely on the other die. However, the 5/6 of the time that we don’t first roll a one, we cannot win because there’s only one die left to roll and we need two ones to win the card. Here’s how we can write this using the function P(x,y) as described above.
P(2,2) = 1/6 * P(1,1) + 5/6 * P(2,1)
P(2,2) = 1/6 * 1/6 + 5/6 * 0 <= simplify P(1,1) and P(2,1) as described above.
P(2,2) = 1/36 = 2.77%
Can you imagine the 36 outcomes that this probability represents? Think about matching every value on one die with every possible value on the other: 6 * 6 = 36.
Next we can look at a slightly more complex problem. What if we have three dice that we are trying to roll two ones with. That extra die should give us a better chance at winning the card. We'll again start by accounting for the different possibilities that rolling the first die could lead to. We'll roll a one on the first die 1/6 of the time, in which case we only need one more one from the two remaining dice. The other 5/6 of the time, we'll need to roll two ones with our last two dice (and we just calculated this probability to be 1/36). Here's what this looks like using our function notation:
P(2,3) = 1/6 * P(1,2) + 5/6 * P(2,2)
P(2,3) = 1/6 * P(1,2) + 5/6 * 1/36 <= simplify P(2,2) as calculated above
P(2,3) = 1/6 * P(1,2) + 5/216
We still have one more step to break down P(1,2): the probability of rolling at least a single one with two dice (Note: it's ok if we roll extra ones, since the Roll For It cards specify the minimum rather than exact dice necessary for scoring). So if the first and second dice that we roll are ones, we are successful no matter what is rolled on the third die. If the first die is a one and the second die isn't, then we still have the third die and final chance at winning the 2pt card. Here's what this looks like using the function P(x,y).
P(2,3) = 1/6 * ( 1/6 * P(0,1) + 5/6 * P(1,1) ) + 5/216
P(2,3) = 1/6 * ( 1/6 * 6/6 + 5/6 * 1/6 ) + 5/216 <= simplify P(0,1) to 6/6, and P(1,1) to 1/6
P(2,3) = 16/216 = 7.40%
So that one extra die nearly triples our chances of winning.
Next, lets move onto a couple of the harder cards: starting with the 15pt card that requires rolling all six sixes. The probability of this is P(6,6), since the P(x,y) should be the same whether you're trying to roll x ones versus x sixes. Using this function notation, we can break the problem down as we have been. We can also save ourselves some effort by ignoring the 5/6 chance of any die being a non-six, because we know that the probability of winning with any non-six die is 0.
P(6,6) = 1/6 * P(5,5)
+ 5/6 * P(6,5)
P(6,6) = 1/6 * 1/6 * P(4,4)
+ 5/6 * P(5,4)
P(6,6) = 1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 1/6
P(6,6) = 1 / 46,656 = 0.00214%
Now, let's look at the example from Rodney's tweet that helped motivate this article: the 15pt card that requires rolling one die with each value from one to six. Do you think that this will be harder, easier, or the same difficulty as rolling all sixes with six dice? We can no longer use our P(x,y) function because every die needs a different value to win. However we can still break the problem apart by considering each die-roll in order. Since we need one die with each value to win, the first die that we roll can be anything: 6/6. The second die can be anything other than the value rolled on the first die: 5/6. The 1/6 of the time that the second die actually matches the first die, we lose. So we can leave this zero probability out of our calculations. The third die must be anything other than what was rolled on the first two dice, and so on.
One Die: 6/6
Two Dice: 6/6 * 5/6
+ 1/6 * 0
Three Dice: 6/6 * 5/6 * 4/6
+ 2/6 * 0
Six Dice: 6/6 * 5/6 * 4/6 * 3/6 * 2/6 * 1/6
Six Dice: 720 / 46,656 = 1.54%
So there's the answer to Rodney's question, and to our own question about whether this might happen more or less often than we expect. If we expect rolling six different numbers to happen as often as rolling six sixes, we should be very surprised. Rolling six different numbers should happen 720 times as often as rolling six sixes. Remember this next time you play Roll For It, since both cards are worth the same 15pts.
There are certainly many more probabilistic considerations within Roll For It as well as other games that I look forward to investigating in future articles. But this seems like a good place to stop for now. Please leave a comment below, if you're enjoying this series, if you have any questions, or if you have any requests for future topics. I've really enjoyed writing these first two installments, and hope people are finding them an interesting read.
::Cross-posted from http://sugarpillstudios.com/wp/?page_id=965
- [+] Dice rolls
Probability can be a useful tool in game design, but only when designers know how to use it. As someone who has struggled with probability in the past, frequently ponders its applications in games, and misses teaching: I’ve decided to start writing this series of articles on probability in games. To kick the series off, I’ll start with my earliest memory of frustration with probability. Looking back, echos of this early frustration have haunted many of my attempts to internalize probability over the years.
The concept of events having a certain probability or chance of occurring is not in itself a difficult one. Probabilities are typically expressed as the ratio of outcomes in which an event occurs, divided by the total number of possible outcomes. What often makes probability difficult is ensuring that you’re counting outcomes correctly: without missing, or re-counting any of them. To illustrate this difficulty, let me share the classic Monte Hall problem as it was described to me by a student teacher in grade school.
Imagine that you are on a game show (like Let’s Make A Deal) where you are told that you can keep a fantastic prize, if you can choose which of three doors it is hidden behind. You have a 1/3 chance of picking the correct door and winning the prize, because there are three doors you could possibly choose, and only one of them conceals the prize. However after you choose a door, the host opens one of the other two doors to reveal that there is no prize behind it. You are then given the option of changing your choice from the door you originally picked to the other unopened door. Are you more likely to win the prize by sticking with the door you picked in the first place, by changing your choice to the other door, or do you have the same chance of winning the prize either way?
My answer to this question was that it didn’t matter whether you switch doors or not. Since only one of the two doors conceals the prize, it’s a coin flip 1/2 chance of winning with either of the remaining doors. However my student teacher was adamant that you’re best chance at winning the prize would be to change doors after seeing the empty door revealed. This same answer was published in an issue of Parade magazine in 1990. I was outraged that this was the only acceptable answer, as were over 10,000 readers of Parade who wrote in to complain. Ever since this memorable math class, I’ve seen this solution posted on countless internet forums. And it’s often accompanied by a confused if not illogical, explanation.
The rational explanation of this solution requires that you know ahead of time that the game show host will always show you one incorrect door after you make your initial choice. Unfortunately, I don’t think I’ve ever seen or heard the problem explained with this crucial caveat. Once you know that the host will always reveal one of the two empty doors, you can start to look at your initial choice as a way of potentially limiting which non-winning door the host reveals. When your initial choice is a non-winning door, there is only one other non-winning door for the host to reveal. When this happens, you can win the prize by changing your choice of door after the reveal. Since your chance of first picking a non-winning door is 2/3, you can win 2/3 of the time by always changing doors after the revel. This is much better than the 1/3 or even 1/2 chances of blindly picking one of the available doors. However this strategy really depends on the host’s upfront honesty and predictably indifferent behavior.
So, what if the host does not explain ahead of time that they will always reveal one door? A greedy host might only give you the opportunity to change doors when you initially choose the winning door. A more philanthropic host might only give you the opportunity to change doors when you initially choose a non-winning door. You should never change your choice with a greedy host, and always change with a philanthropic judge. I suspect that many people’s frustration with the Monte Hall problem’s solution, comes from their skepticism about the human nature of game show hosts.
Host Type: | Indifferent Opener | Greedy Opener | Philanthropic Opener
Best Strategy: | Always Change | Never Change | Always Change
Chance to Win: | 2/3 | 1/3 | 3/3
This problem illustrates one of many difficulties in counting outcomes. Something as subtle as the motivation of a game show host can inform our strategies for changing doors, and result in different numbers of winning outcomes. We could extend our solution above by speculating on the probabilities that a random game show host falls into each category. However we might miss other categories of host motivations, For example, a game show host might try to maximize the perception that they are not acting greedily, while in fact trying to minimize the number of prizes they award across several game shows.
Hopefully this (and future articles in the series) leaves you with more questions than answers. I encourage you to further explore those questions and comment on any that catch your fancy. Some of those questions could lead to interesting followup comments, and others might be worth visiting in a full article format.
:: Cross-posted from http://sugarpillstudios.com/wp/?page_id=943
- [+] Dice rolls