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Subject: Probability charts for Perudo/Liars Dice? rss

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Ender Wiggins
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An earlier thread pointed out that the value of "Wild/Ace" bids in Perudo is different from "Bluff/Liars Dice": http://www.bgg.cc/thread/61722

It seems to me that the "Bluff" order best fits what is suggested by probability and statistics. The Perudo order is as follows:
Quote:
1 of anything
2 of anything
1 WILD
3 of anything
4 of anything
2 WILD
5 of anything

But probability wise, isn't there a greater probability of rolling "1 WILD" than there is of rolling "2 of anything"? Similarly, isn't there a greater probability of rolling "2 WILD" than there is of rolling "4 of anything"?

If I am correct, then it would logically follow that a bid of "1 WILD" should precede a bid of "2 of anything", since the "1 WILD" is statistically an "easier" bid. Similar, a bid of "2 WILD" should precede a bid of "4 of anything", since the "2 WILD" bid is easier. Thus the "Bluff" (or "Liar's Dice") order seems to me to be correct, and best fits the order of probability:
Quote:
1 of anything
1 WILD
2 of anything
3 of anything
2 WILD
4 of anything
5 of anything


Is there a probability chart available anywhere suggesting the stastical likelihood of various rolls?

For example, if there are 2 dice, I calculated the statistical odds of achieving the following bids:
0.505 - 1 of anything
0.305 - 1 WILD
0.111 - 2 of anything

It would be helpful and interesting to see some charts which list the exact odds of getting bids for 3 dice, 4 dice etc in a similar manner. An exercise for a mathematician among us perhaps? Or does a chart like this already exist?
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Matthew M
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What any such statistical chart shows would miss the fact that you have some information to go on. If my opponent and I each have one die left and I have a 4 showing I would much sooner bid two 4s than I would one wild. Yes, if I had no information making the one wild bet gives better odds, however due to knowing what my die is showing I am twice as likely to be correct with the two-4s bet.

-MMM
 
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Ender Wiggins
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Agreed, but not entirely. You are right that a chart wouldn't be a complete solution, because you would have to mentally plug in several factors:
1. the information from your own dice, this concrete data will either increase or decrease the "normal" odds.
2. the psychological data of how others are bidding, which might be bluffing, but could give some indication of what else is being rolled.

But a chart will also give you some idea of the odds of the results of the rest of the dice aside from your own. For example, if there are 15 dice being thrown aside from your own five, you will know the odds for what those other 15 dice are showing. Knowing the probabilities is one helpful ingredient for playing the game well, although clearly there are many other necessary elements.

At the very least, a statistical table should prove that the value that the Perudo rules allocate to the wild (aces) is incorrect, and that the Bluff/Liars Dice rules are correct in determining the order/value of different bids.

It could look like this:

2 DICE
0.555 - 1 of anything
0.306 - 1 WILD
0.111 - 2 of anything
XXXXX - 3 of anything
0.028 - 2 WILD

3 DICE
0.704 - 1 of anything
0.421 - 1 WILD
0.??? - 2 of anything
0.037 - 3 of anything
0.??? - 2 WILD
XXXXX - 4 of anything
XXXXX - 5 of anything
0.005 - 3 WILD

4 DICE
0.802 - 1 of anything
0.518 - 1 WILD
0.??? - 2 of anything
0.??? - 3 of anything
0.??? - 2 WILD
0.012 - 4 of anything
XXXXX - 5 of anything
XXXXX - 3 WILD

5 DICE
etc

The probability figure should not be for an exact bid, but for a bid that would be correct, i.e. the value of 0.802 for "1 of anything" with 4 dice is the probability of getting at least 1 of anything.

Unfortunately I'm not good enough at probability and statistics to figure out more results other than those listed above. For example, what is the exact probability of getting 2 (or more) WILD when 4 dice are rolled? Am I correct that it is a greater probability than getting 4 of anything (0.012)? And thus should be a less risky bid, and that it should come before, not after a bid of "4 of anything", as is incorrectly the case in the rules of the Perudo edition?
 
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Matthew M
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But making such a chart while ignoring the fact that players will have information makes such a chart useless. The reason the stars come after the even numbers is because it is always more likely that you will have more of some other number than the number of stars you have, making it easier to make a number bid than a stars bid because of that already-known info. A statistical likelihood chart misses that fact.

-MMM
 
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Ender Wiggins
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Quote:
But making such a chart while ignoring the fact that players will have information makes such a chart useless.

But a statistical likelihood chart will give you an indication of the dice results that the other players have rolled, and in that respect it is still useful!
Quote:
The reason the stars come after the even numbers is because it is always more likely that you will have more of some other number than the number of stars you have, making it easier to make a number bid than a stars bid because of that already-known info. A statistical likelihood chart misses that fact.

Please help me understand you here - I read this over several times but couldn't quite figure out what you meant, for which I apologise! I'm not sure if you're defending the Perudo order of bids or the Liars Dice order (the difference is listed here: http://www.bgg.cc/thread/61722)? Surely the order can be defended statistically? And that the statistical odds for getting a certain roll (whether x of a certain number, or x wild) can be accurately and precisely calculated, as I've tried to do above?
 
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Matthew M
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EndersGame wrote:

Surely the order can be defended statistically? And that the statistical odds for getting a certain roll (whether x of a certain number, or x wild) can be accurately and precisely calculated, as I've tried to do above?


The order given no information can be defended statistically, sure. However you never have no information. Given that, imagine the start of a game - everyone has 5 dice in their cup. What is more likely? That they have more stars or that they have more of any other number? The latter, of course. So a high bid will be first be based upon that, plus any additional stars that were rolled.

So though you are correct in saying that without information the order of "1, 1, 2, 3, 2, 4, 5, etc..." is closer to being correct, the fact of the matter is that you will almost always have more information regarding the number totals than you will have regarding the star totals.

Truth be told, I'm not very invested in saying one version of the bid progression is better than the other. I am saying that simply listing the statistical likelihoods of the various outcomes does not prove one more right than the other because it completely ignores the typical game-state.

-MMM
 
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Ender Wiggins
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Octavian wrote:
So though you are correct in saying that without information the order of "1, 1, 2, 3, 2, 4, 5, etc..." is closer to being correct, the fact of the matter is that you will almost always have more information regarding the number totals than you will have regarding the star totals.

Thank you for clarifying that Matthew. You raise a valid point that there is more to determining the difficulty of a bid than the mere statistical probability, in view of the information you do have about some of the dice. As you correctly point out, this will usually be more information about a number bid than about a wild bid. Quite frankly, I'm not sufficiently experienced enough to determine whether this information warrants changing the bid chart order so that it doesn't follow what is suggested by probability, or that this information justifies the Perudo bid order. Personally, I'm not convinced it does, but I wouldn't want to press the issue, you could well be right, although the burden of proof lies on those who would want to reverse the order suggested by probability.

But point taken, perhaps I should temper my argument somewhat, and instead state that the Bluff/Liars Dice bid chart order best reflects the order of bid difficulty that is suggested by probability alone. And that the information gleaned from the dice you roll is a complicating factor which means that the bid difficulty can't be determined merely by statistics, as you rightly point out.

Regardless, I'd still be interested in seeing detailed statistical charts of probability, and I think having an improved sense of the probability would equip a player to bid more accurately. If anyone has any, or wants to take on the challenge, I'd love to see them!

Thanks again for taking the time to give a useful contribution to the discussion Matthew, I certainly benefited from reading your posts on the subject!
 
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Dan Smith
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Here are the probability charts. It's complicated because with wild 1s you can throw (e.g.) 3 "sixes" in 4 different ways (111,116,166,666) but you can only throw "2 aces" in one way (11)

These charts give the probability of N UNSEEN dice meeting OR EXCEEDING the given quantity of a particular denomination, assuming 1s are wild. I have cut each chart off when no higher bid has a probability of over 10%. They sum to more than 100% because any throw can satisfy more than 1 bid.

The probabilities are in Perudo bidding order. Perudo order is right for the higher probability bids and becomes wrong at the lower probabilities, whereas Bluff order is wrong at the higher probabilities and goes right at the lower ones. Since in practice the higher probabilities are used more often I think the perudo order has the edge, but it is a moot point.

For the odds with no wilds, just take the "ace" probabilities,

1 Die
1 of anything 33%
1 ace 17%

2 Dice
1 of anything 56%
2 of anything 11%
1 ace 31%

3 Dice
1 of anything 70%
2 of anything 26%
1 ace 42%

4 Dice
1 of anything 80%
2 of anything 41%
1 ace 52%
3 of anything 11%
4 of anything 1%
2 aces 13%


5 Dice
1 of anything 87%
2 of anything 54%
1 ace 60%
3 of anything 21%
4 of anything 5%
2 aces 20%


6 Dice
1 of anything 91%
2 of anything 65%
1 ace 67%
3 of anything 32%
4 of anything 10%
2 aces 26%


7 dice
1 of anything 94%
2 of anything 74%
1 ace 72%
3 of anything 43%
4 of anything 17%
2 aces 33%
5 of anything 5%
6 of anything 1%
3 aces 10%


8 Dice
1 of anything 96%
2 of anything 80%
1 ace 77%
3 of anything 53%
4 of anything 26%
2 aces 40%
5 of anything 9%
6 of anything 2%
3 aces 13%

9 Dice
1 of anything 97%
2 of anything 86%
1 ace 81%
3 of anything 62%
4 of anything 35%
2 aces 46%
5 of anything 14%
6 of anything 4%
3 aces 18%


10 Dice
1 of anything 98%
2 of anything 90%
1 ace 84%
3 of anything 70%
4 of anything 44%
2 aces 52%
5 of anything 21%
6 of anything 8%
3 aces 22%


11 Dice
1 of anything 99%
2 of anything 92%
1 ace 87%
3 of anything 77%
4 of anything 53%
2 aces 57%
5 of anything 29%
6 of anything 12%
3 aces 27%


12 Dice
1 of anything 99%
2 of anything 95%
1 ace 89%
3 of anything 82%
4 of anything 61%
2 aces 62%
5 of anything 37%
6 of anything 18%
3 aces 32%
7 of anything 7%
8 of anything 2%
4 aces 13%


13 dice
1 of anything 99%
2 of anything 96%
1 ace 91%
3 of anything 86%
4 of anything 68%
2 aces 66%
5 of anything 45%
6 of anything 24%
3 aces 37%
7 of anything 10%
8 of anything 3%
4 aces 16%


14 Dice
1 of anything 100%
2 of anything 97%
1 ace 92%
3 of anything 89%
4 of anything 74%
2 aces 70%
5 of anything 52%
6 of anything 31%
3 aces 42%
7 of anything 15%
8 of anything 6%
4 aces 19%


15 Dice
1 of anything 100%
2 of anything 98%
1 ace 94%
3 of anything 92%
4 of anything 79%
2 aces 74%
5 of anything 60%
6 of anything 38%
3 aces 47%
7 of anything 20%
8 of anything 9%
4 aces 23%
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Dan Smith
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Octavian wrote:
What any such statistical chart shows would miss the fact that you have some information to go on. If my opponent and I each have one die left and I have a 4 showing I would much sooner bid two 4s than I would one wild. Yes, if I had no information making the one wild bet gives better odds, however due to knowing what my die is showing I am twice as likely to be correct with the two-4s bet.

-MMM


Hmm, if you bid 1 wild and he has no wilds he will definitely call you, so you only have a 1 in 6 chance of winning that.

If you bid 2 fours then you will only win if he has a 4. You will lose if he has a wild (becaue he will bid one wild and you can't overcall), You will also lose if he has a 2,3,5 or 6 because he will call your bid. So your chances are again 1/6

Say you bid one 4...
with a wild or a 4 he will bid two fours and win.
with a 2 or a 3 he will lose and with a 5 or a 6 he will overcall you. Your chances increase to 2/3

However...
Say you bid one 2.
With wild he can overcall one wild, and will win
With another two he will probably call two 2s, and you will win.
With a 3 he will probably call one 3 which you can overcall with one 4, and you will win.
With a 4 he will probably call one 4 which you can overcall with two 4s and win.
With a 5 or a 6 he will win again,
So this bid gives you a chance of 1 in 2 of winning, and may well be best As long as he doesn't think you are lying!
 
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Andy Eardly
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Great work on the probability table for Perudo. Hope you don't mind if I ask what logic steps and formulas you are using to come up with the odds? I'm hoping come up with a universal formula for a spreadsheet to calculate the odds up to 30 dice. Calculating the odds for one of anythings is straight forward, but how are you altering the formulas for 2 or more dice?
 
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Lewis Wagner
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Perudo has a different reason for the skewed wild bids. It has different rules.

It's against the rules to open with a wild bid in Perudo, and you can only lose one die. I think the purpose is to given the weaker player a slightly better chance when down to a small number of dice.

I own Perudo, because I think it has the nicest and most compact set of bits. Compare it to Liars Dice which has a gigantic box, unwieldy cups and only bits for 4 players - grossly inferior to Perudo. I play with Bluff rules including a Bluff style rollup chart of bids, because I prefer Bluff rules.

I know multiple gamers who own Perudo. I don't know any who play by Perudo rules.
 
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Dan Smith
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easyglider wrote:
Great work on the probability table for Perudo. Hope you don't mind if I ask what logic steps and formulas you are using to come up with the odds? I'm hoping come up with a universal formula for a spreadsheet to calculate the odds up to 30 dice. Calculating the odds for one of anythings is straight forward, but how are you altering the formulas for 2 or more dice?


There's probably a better way but I did it as follows:
In excel I created a 2d table with "n in exactly n aces" in the top row, ranging from 0 to 20 and "m in exactly m x's" in the lefthand column, where x represents a specific non-ace value, also ranging from 0 to 20

I also defined n_dice on the sheet, as any particular calculation in this table is dependent on a given number of dice (otherwise I would need a 3d table, which is hard to do on an excel sheet).

In practice, the n in n_aces is "n aces from n_dice given that m of those dice have already shown x, which is not an ace"

Then, the probability for each cell of the table can be calculated as follows:

If ndice < M + N then prob = 0
Else
prob = (ndice - m)!/(n!*(ndice-n-m)!) * ndice!/(n!*(ndice-n)!) * (1/6)^(n+m) * (4/6)^(n_dice-n-m)

That function is the number of ways of getting the throws in question multiplied by the chance of any given throw and can be described in words as:
number_of_ways_of_getting_n_aces * number_of_ways_of_getting_m_x's * probability_of_a_dice_being_an_ace^num_aces * probability_of_a_dice_being_an_x^num_x's * probability_of_a_dice_being_something_else^num_dice_that_are_not_aces_or_x's


Then I defined a columnar table, indexed again for 0 to 20 where each entry was the probability of getting exactly that number of (say) sixes, made up of sixes and aces. Each proability for index K was the sum of all the probabilities in the preceding table for which n+m=K.

Since I wanted the probability of getting "at least K" I added another column to the K table, for which each entry J was the sum of getting exactly K where K >= J.

I generated the table I posted by manually changing n dice and writing the results down.

I can send you the spreadsheet if you like!

If I was continuing this work I would probably now write a program to do the same calculations as I've done on the excel sheet (but for all numbers of dice) and would use the excel results to test that my program's answers were right.
 
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Confusion Under Fire
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iamatiger wrote:


However...
Say you bid one 2.
With wild he can overcall one wild, and will win
With another two he will probably call two 2s, and you will win.
With a 3 he will probably call one 3 which you can overcall with one 4, and you will win.
With a 4 he will probably call one 4 which you can overcall with two 4s and win.
With a 5 or a 6 he will win again,
So this bid gives you a chance of 1 in 2 of winning, and may well be best As long as he doesn't think you are lying!


If I have only one die left I usually do not give away what that number is by calling that number. That leaves my opponent(s) not knowing what I have left.
 
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