$18.00
GeekGold Bonus for All Supporters: 134.69

7,855 Supporters

$15 min for supporter badge & GeekGold bonus
49.5% of Goal | left

Support:

Shaun Higgins
United Kingdom
London
flag msg tools
mbmbmbmbmb
I am designing a game with a simple push-your-luck dice mechanic, where each turn players roll five six-sided dice. Each die has:

3 Fail sides
2 Pass sides
1 Ace side

Players are attempting to end-up with a set of dice displaying as few fails and as many passes or aces, as possible. As with many traditional dice games, they can re-roll the dice, but whenever they do must set aside and lock at least one die. Any ace that is rolled stands for a pass and also cancels out any fail held amongst the final dice. The winning player is the one with the least amount of fails.

How can I use probability to calculate the best strategy for this game?

Many thanks!
1 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Don Lynch
United States
Boston
Massachusetts
flag msg tools
mbmbmbmbmb
maybe this guy can help:

https://www.youtube.com/watch?v=ZkVSRwFWjy0
 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Maarten D. de Jong
Netherlands
Zaandam
flag msg tools
Your question has two answers, really. One is that you don't need probability to determine the best strategy: just put aside anything pass or ace, and reroll anything fail; if there is nothing but fail to put aside just select 1 die and reroll what is left. Repeat until you can roll no more. This guarantees that the number of fails will be as small as possible in the end. Two is that you want to know the odds of 0, 1, 2, ... fails, but frankly I don't think they matter (hence me not bothering to calculate them). Someone who lucks out and rolls less fails simply wins. It doesn't matter if the odds are small.

The problem is that the result you check will always constitute all dice, and that there is no penalty on rolling more often.
3 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Shaun Higgins
United Kingdom
London
flag msg tools
mbmbmbmbmb
Maarten, why is it a problem that the result you check will always constitute all dice?
1 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Maarten D. de Jong
Netherlands
Zaandam
flag msg tools
Because it doesn't make sense to select a subset of dice when there are still F's waiting to be re-rolled. At worst, they stay F. At best they become P or A. Similarly, selecting a subset of dice with P's and A's waiting to be locked is just... silly.

Therefore: since most P + A's count for the win, you roll until you can roll no more, and all 5 dice have been locked.
 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Guest Starring...
Austria
flag msg tools
mbmbmbmbmb
cymric wrote:
Your question has two answers, really. One is that you don't need probability to determine the best strategy: just put aside anything pass or ace, and reroll anything fail; if there is nothing but fail to put aside just select 1 die and reroll what is left. Repeat until you can roll no more. This guarantees that the number of fails will be as small as possible in the end. Two is that you want to know the odds of 0, 1, 2, ... fails, but frankly I don't think they matter (hence me not bothering to calculate them). Someone who lucks out and rolls less fails simply wins. It doesn't matter if the odds are small.

The problem is that the result you check will always constitute all dice, and that there is no penalty on rolling more often.

I believe that this is the optimal strategy, and the one most players will instinctively employ, but you do need to use probability to prove that this is the case. Showing that, on a roll of ace-pass-pass-fail-fail, it isn't better to keep only the ace isn't entirely trivial.
1 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Maarten D. de Jong
Netherlands
Zaandam
flag msg tools
Since A counts as P in absence of F (or rather, it counts as non-F in the absence of F), and the probability of rolling F is higher than rolling P, it follows that keeping A is the optimal move.
1 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Guest Starring...
Austria
flag msg tools
mbmbmbmbmb
cymric wrote:
Since A counts as P in absence of F (or rather, it counts as non-F in the absence of F), and the probability of rolling F is higher than rolling P, it follows that keeping A is the optimal move.

I'm sorry, I have no idea what you mean by this. Could you please elaborate?
1 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Maarten D. de Jong
Netherlands
Zaandam
flag msg tools
Oopsie, looks like I misunderstood the rules to the game. Here's the start of a new analysis. The object of the game is to end up with the smallest Fail count.

I look at the die in terms of 'F-value': how much F does a result contribute to the total?
— In case of the result being F: 1 F (obviously).
— In case of the result being P: 0 F (because P does not count as F, but also doesn't discount the number of F's).
— In case of the result being A: -1 F (because A does not count as F, and subtracts 1 F from the total amount of F's).

The expectation value of the die is therefore (3 * 1 F + 2 * 0 F + 1 * -1 F) / 6 = 2 F / 6 = 1/3 F.

Then the strategy: anything with an actual F-value above the die's expectation value must be re-rolled, anything less is kept. In other words: re-roll F, keep A and P. A is best of course, F is worst... and P is kept because although it doesn't have the benefit of A, you're likely to end up with more F than you started out with upon re-rolling.
2 
 Thumb up
0.05
 tip
 Hide
  • [+] Dice rolls
Nathaniel Grisham

Indiana
msg tools
mb
I honestly think that calculating the probabilities is a little over the top for a game as simple/light as this one. (Unless it's going to be used for gambling, in which case, I won't help with that anyway.)

cymric wrote:
Your question has two answers, really. One is that you don't need probability to determine the best strategy: just put aside anything pass or ace, and reroll anything fail; if there is nothing but fail to put aside just select 1 die and reroll what is left. Repeat until you can roll no more. This guarantees that the number of fails will be as small as possible in the end. Two is that you want to know the odds of 0, 1, 2, ... fails, but frankly I don't think they matter (hence me not bothering to calculate them). Someone who lucks out and rolls less fails simply wins. It doesn't matter if the odds are small.

The problem is that the result you check will always constitute all dice, and that there is no penalty on rolling more often.


I think that this is the obvious strategy, but the OP is also asking about the strategy for determining when to stop rolling.

First, if you know the final result of any other players' rolls, and your roll is worse than any of them, then you keep rolling until either you have a better result, or it you realize that it's impossible for you to win.

I'm making the following assumptions based on what you said about the rules.
1. The player who rolls the fewest fails wins.
2. In case of a tie, the player who rolls the most successes wins.
3. An ace cancels a fail, but does not turn the fail into a success.

I'll just name a few points that I believe would be most helpful:

* If you have a roll with failures as the only options to lock, then you probably need to. It'll be good to try to get something better on the reroll.
* If you have a roll with zero failures (after all aces are accounted for), it might be a good idea to keep it. Unless you already know that you need more passes to win, You are better off minimizing failures before maximizing passes.
* If you know that you are still trying to beat other players, whose final results you don't know yet, then the strategy is still fairly simple.

Basically, if you are the first player, then you have to stop whenever you feel like you have the fewest failures you are going to get. Anyone after the first player has to keep trying until they have beat everyone ahead of them. The advantage in this game goes to the last player, since he has the easiest time choosing when to stop.

So, after all is said, maybe it would be helpful for the first player to know how likely it is to end up with X number of failures and Y number of successes. But after that, the best strategy is "re-roll until I beat the other results."

If players are supposed to play simultaneously, or secretly, then the design doesn't really work, since there isn't really an easy way to keep the other players honest.
1 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Brent Gerig
United States
Upland
Indiana
flag msg tools
mbmbmbmbmb
I don't see how this is a press-your-luck game. There's no penalty or potential to lose anything be re-rolling. Keep/lock any Aces or Passes on each roll, unless you roll all Fails and have to keep a single Fail. I haven't calculated odds, but given the 1/6 chance at an Ace vs. 1/3 chance of Pass or 1/2 chance of Fail, it doesn't seem at all logical to re-roll Passes hoping to get Aces.
3 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Brent Gerig
United States
Upland
Indiana
flag msg tools
mbmbmbmbmb
Grishhammer wrote:
What's missing here is that when you re-roll, you lock exactly one die.


OP says:
Shaun Higgins wrote:
they can re-roll the dice, but whenever they do must set aside and lock at least one die.
 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Nathaniel Grisham

Indiana
msg tools
mb
flaquito wrote:
Grishhammer wrote:
What's missing here is that when you re-roll, you lock exactly one die.


OP says:
Shaun Higgins wrote:
they can re-roll the dice, but whenever they do must set aside and lock at least one die.

I see, thanks for the correction.
 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Guest Starring...
Austria
flag msg tools
mbmbmbmbmb
cymric wrote:
Oopsie, looks like I misunderstood the rules to the game. Here's the start of a new analysis. The object of the game is to end up with the smallest Fail count.

I look at the die in terms of 'F-value': how much F does a result contribute to the total?
— In case of the result being F: 1 F (obviously).
— In case of the result being P: 0 F (because P does not count as F, but also doesn't discount the number of F's).
— In case of the result being A: -1 F (because A does not count as F, and subtracts 1 F from the total amount of F's).

The expectation value of the die is therefore (3 * 1 F + 2 * 0 F + 1 * -1 F) / 6 = 2 F / 6 = 1/3 F.

Then the strategy: anything with an actual F-value above the die's expectation value must be re-rolled, anything less is kept. In other words: re-roll F, keep A and P. A is best of course, F is worst... and P is kept because although it doesn't have the benefit of A, you're likely to end up with more F than you started out with upon re-rolling.

flaquito wrote:
I don't see how this is a press-your-luck game. There's no penalty or potential to lose anything be re-rolling. Keep/lock any Aces or Passes on each roll, unless you roll all Fails and have to keep a single Fail. I haven't calculated odds, but given the 1/6 chance at an Ace vs. 1/3 chance of Pass or 1/2 chance of Fail, it doesn't seem at all logical to re-roll Passes hoping to get Aces.

This isn't quite how it works.

Take, for example, an initial roll of pass-pass-pass-pass-fail.

If you keep all four passes and reroll only the failure, you have a 50% chance of finishing the round without any failures, and a 50% chance of finishing with one failure.

But if you keep only three passes, and reroll a pass and a failure, you now have (if my hastily scrawled back-of-the-napkin math is correct) better odds of finishing without a failure (62.5%), although you incur the risk of finishing with two failures (12.5%).

It may not seem logical to reroll passes in the hope of getting aces, but that's not all you're doing. You're also rerolling passes in order to get additional rerolls on your failures.
2 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
PenumbraPenguin
Australia
Sydney
NSW
flag msg tools
mbmbmbmbmb
cymric wrote:

The expectation value of the die is therefore (3 * 1 F + 2 * 0 F + 1 * -1 F) / 6 = 2 F / 6 = 1/3 F.

Then the strategy: anything with an actual F-value above the die's expectation value must be re-rolled, anything less is kept.


The problem with this analysis is that while the expected value of a roll is 1/3, that doesn't take into account rerolls.

For example, let's work out the expected value of a die that you're allowed to reroll up to once. It's 1/6*(-1 + 3*0 + 2*1/3) = -1/18, because you keep aces and passes, and reroll the fails, and the expected value of a reroll is 1/3.

Now, this means that the expected value of a die that you can reroll once more is negative - that is, better than a pass. So if you roll a pass, and are going to have the option to reroll this die at least twice, you should reroll it (to maximise the given expected value).

It's not simple to work out the number of rerolls you'll end up taking, but I think this should convince you that on the very first roll, you should only keep aces, and reroll passes and fails both.
2 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Guest Starring...
Austria
flag msg tools
mbmbmbmbmb
PenumbraPenguin wrote:
cymric wrote:

The expectation value of the die is therefore (3 * 1 F + 2 * 0 F + 1 * -1 F) / 6 = 2 F / 6 = 1/3 F.

Then the strategy: anything with an actual F-value above the die's expectation value must be re-rolled, anything less is kept.


The problem with this analysis is that while the expected value of a roll is 1/3, that doesn't take into account rerolls.

For example, let's work out the expected value of a die that you're allowed to reroll up to once. It's 1/6*(-1 + 3*0 + 2*1/3) = -1/18, because you keep aces and passes, and reroll the fails, and the expected value of a reroll is 1/3.

Now, this means that the expected value of a die that you can reroll once more is negative - that is, better than a pass. So if you roll a pass, and are going to have the option to reroll this die at least twice, you should reroll it (to maximise the given expected value).

It's not simple to work out the number of rerolls you'll end up taking, but I think this should convince you that on the very first roll, you should only keep aces, and reroll passes and fails both.

This too isn't entirely accurate. The value of an ace is only conditionally better than a pass, and that changes the calculations significantly.
1 
 Thumb up
1.00
 tip
 Hide
  • [+] Dice rolls
Shaun Higgins
United Kingdom
London
flag msg tools
mbmbmbmbmb
Thanks, E Decker, for your response! The tactical approach you have highlighted, of rerolling Passes but not Aces to increase the available number of rerolls a player has to remove Fails, is what I was wanting to demonstrate via probability. You have described something that I was unable to articulate, having less knowledge of probability.

And, yes, this is also the reason why I presented this game as a push-your-luck variant, as in rerolling an already achieved Pass to attempt to void one Fail also runs the risk of a player gaining two Fails.

How can I use probability to test for this process in other situations?
 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Maarten D. de Jong
Netherlands
Zaandam
flag msg tools
E Decker wrote:
This too isn't entirely accurate. The value of an ace is only conditionally better than a pass, and that changes the calculations significantly.

Indeed. I owe an apology to the OP for misjudging a problem which isn't quite as trivial as it turns out to be at first sight. And after many an hour spent staring at decision trees I actually believe that no answer given above is entirely correct. It's very subtle, all in all.

The problem with my approach for example is that it grades negative values of F, which are not rewarded by the game. In other words, 5 A's are not better than 5 P's whereas their F-value is distinctly smaller. In essence my approach allows for rolling 'too well' which is simply not necessary. E Decker's approach on the other hand does not take into account graded wins and losses if no 0 F result is present. As it turns out, P P P + re-roll P F still gives a better result than P P P P + re-roll F, but only in 1/16th of all cases which is smaller by half than the difference between 50 and 62.5%. That is, if both results are arrived at simultaneously, i.e. without waiting for one roll to complete before the other, because then too the probabilities change. (The player with P P P + re-roll P F ending up at P P P F F will lose immediately to the player still needing to begin locking P and re-rolling F.) At the same time, if the result now is P P + P F F with the decision being locking F and re-rolling P F, or locking P and re-rolling F F, then the second option is distinctly better. So one shouldn't blindly assume that a re-roll of P F is the default strategy to go for.

In any case, at this point the calculations begin to grow quite tedious, with the outcome dependent on crucial rules which are not (yet) given. However, I can say that previously locked in A's plus the fact that they are not graded when resulting in an excess allows one to take more risks than a cursory analysis suggests.
 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Shaun Higgins
United Kingdom
London
flag msg tools
mbmbmbmbmb

Maarten, thank you for yet another thorough response. Given your analysis of this early-stage design, what would you like to see implemented around these mechanics, in terms of crucial rules, that could give rise to in-game decisions that would grab your interest?
 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Laura Jula

Göttingen
Niedersachsen
msg tools
mbmbmb
As some people have mention, this problem doesn't have an straight forward answer. If you want to continue with the calculations I will like to provide a framework where the calculations can be made. I will exemplify this with an example where you have the rules you mention but WITHOUT the rule of keeping at least one for each re roll.

First of all, you always want to define in a clear way, using random variables what quantities are important.

Let X(t) be the number of fails at time t, Y(t) the number of aces, Z(t) the number of passes, S(T) the score in time t (S(t)=X(t)-Y(t))

NOTE that I defined these r.v depending on time, the time is going to be the number of the round. For example X(1) is a random variable that tell you how many fails there are in the first round. This kind of r.v indexed on t are called STOCHASTIC PROCESSES.

SOME PROPERTIES:
E(X(1))=5/2
E(Y(1))=5/6
E(Z(1))=5/3

EXAMPLE.
Now I would like to give an example on how to handle this kind of problems. Assume t=1,2, this means that we only have two rounds. To simplify I will ignore the rule of always leave one of the dice if you want to re roll.

I will compare two strategies:
1. Keep the non-fails (passes and aces)
2. Keep only the aces

To compare this strategies you will need some knowledge of CONDITIONAL EXPECTATION. One of the important properties is that E(E(X|Y))=E(X) for any r.v X, Y.

We will see in which strategy E(S(2)) is bigger

STRATEGY 1:
First we calculate the expected value of the r.v given the knowledge of what happened in the first round.
E(X(2)|X(1)=x)= (1/2)x for x=0,...,5
This is because the number of fails will be the fails on the rerolled dice (in this case we reroll x dice).

This mean that the r.v. E(X(2)|X(1))=(1/2)X(1)
and we can conclude E(X(2))=E((1/2)X(1))=(1/2)E(X(1))=(1/2)(5/2)= 5/4

And

E(Y(2)|Y(1)=y and X(1)=x)=y+(1/6)x
Because you already had y aces and reroll x dices.

Then: E(Y(2))=(5/6)+(1/6)(5/6)=5/4
Here I omitted the details but is the same as above.

This means that E(S(2))=E(X(2)-Y(2))=0
In average, with this strategy you will arrive to 0 points.

STRATEGY 2:

E(X(2)|Y(1)=y)=(1/2)(5-y)
Because we reroll 5-y dices
Then E(X(2))=25/12

E(Y(2)|Y(1)=y)=y + (1/6)(5-y)
Then E(Y(2))= 55/36

Then E(S(2))=25/12 - 55/36 > 0

CONCLUSION: The first strategy is better because on average it will lead to less points.

COMMENTARY:

1. If you want to implement this reasoning to your situation with the extra rule things will get more complicated because now you need to distinguish the case were all rolls are fails.

For example in strategy one
E(X(2)|X(1)=x)= (1/2) x if x<5
= 1 + (1/2)(x-1) if x=5

2. There are more strategies possible, for example to re roll the passes if Z(t)>2 or something like that. This will also add to the complexity and possibly you will need to consider the JOINT P.D.F of X(t) and Y(t). This is a MULTINOMIAL DISTRIBUTION.

3. Also in this example we consider only two periods of time. In our example this is sufficient, it is always better to reroll only the negative, independently of the number of dice. But with your additional rule things will become more complicated, specially in strategy 2.


This was a little bit technical but hopefully clear enough. In conclusion it is a hard problem but doing the math is always possible. If you have any questions about the reasoning I will be happy to answer.





2 
 Thumb up
1.00
 tip
 Hide
  • [+] Dice rolls
Brent Gerig
United States
Upland
Indiana
flag msg tools
mbmbmbmbmb
Very interesting to see the various statistical analyses showing that the solution which appears to be obvious and simple in fact is much more subtle and complex. Nice work, all.
 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Lluluien
United States
Missouri
flag msg tools
mbmbmb
RainbowJ's post was really good, but I was disappointed that it mentioned stochastics without mentioning Monte Carlo Simulation (https://en.wikipedia.org/wiki/Monte_Carlo_method). I would argue that the thread is already gotten sufficiently complicated that the fastest way to figure this out is spend 30 minutes writing a computer program that simulates the game, put in whatever selection strategies you wish to test, then run 10 million games and see which strategy performs better.

This may be a hard mathematical problem, but it's a trivial programming problem - this would make a good project for a first intro college course.
1 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Franz Kafka
United States
St. Charles
Missouri
flag msg tools
mbmbmbmbmb
I appreciate the analysis that has been presented so far. However, I think there are a couple related questions:
1. How does a player maximize expected points? (This is addressed above.)
2. How does a player maximize the chance of victory?

This second one would depend on the rules of the game and the other players' strategies and results so far. For example, other players rolling poorly might encourage conservative play, while other players rolling well might encourage more aggressive play.
1 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Shaun Higgins
United Kingdom
London
flag msg tools
mbmbmbmbmb
This is a fair point, JosefK, but at this moment I am just looking at how to identify a pure strategy or strategies, before considering how this will then be influenced by further factors.
 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
Guest Starring...
Austria
flag msg tools
mbmbmbmbmb
cymric wrote:
E Decker wrote:
This too isn't entirely accurate. The value of an ace is only conditionally better than a pass, and that changes the calculations significantly.

Indeed. I owe an apology to the OP for misjudging a problem which isn't quite as trivial as it turns out to be at first sight. And after many an hour spent staring at decision trees I actually believe that no answer given above is entirely correct. It's very subtle, all in all.

The problem with my approach for example is that it grades negative values of F, which are not rewarded by the game. In other words, 5 A's are not better than 5 P's whereas their F-value is distinctly smaller. In essence my approach allows for rolling 'too well' which is simply not necessary. E Decker's approach on the other hand does not take into account graded wins and losses if no 0 F result is present. As it turns out, P P P + re-roll P F still gives a better result than P P P P + re-roll F, but only in 1/16th of all cases which is smaller by half than the difference between 50 and 62.5%. That is, if both results are arrived at simultaneously, i.e. without waiting for one roll to complete before the other, because then too the probabilities change. (The player with P P P + re-roll P F ending up at P P P F F will lose immediately to the player still needing to begin locking P and re-rolling F.) At the same time, if the result now is P P + P F F with the decision being locking F and re-rolling P F, or locking P and re-rolling F F, then the second option is distinctly better. So one shouldn't blindly assume that a re-roll of P F is the default strategy to go for.

In any case, at this point the calculations begin to grow quite tedious, with the outcome dependent on crucial rules which are not (yet) given. However, I can say that previously locked in A's plus the fact that they are not graded when resulting in an excess allows one to take more risks than a cursory analysis suggests.

You are overlooking that with pass-pass-pass locked and pass-fail rerolled to fail-fail, you can still reroll one of the fails. This means that this roll doesn't immediately lose, and changes the probabilities you arrive at.
 
 Thumb up
 tip
 Hide
  • [+] Dice rolls
1 , 2 , 3 , 4  Next »   | 
Front Page | Welcome | Contact | Privacy Policy | Terms of Service | Advertise | Support BGG | Feeds RSS
Geekdo, BoardGameGeek, the Geekdo logo, and the BoardGameGeek logo are trademarks of BoardGameGeek, LLC.