uncooper 1 Posted August 2, 2008 Share Posted August 2, 2008 I played a dice game at a bar last night that I am trying to do some math on. The part I'm getting stuck on is this:How do I calculate the expected value of the best of n dice?For example, if by definitionrolling 6-4-2 has a value of 2rolling 5-3 has a value of 3rolling 5-4-2-1 has a value of 1, etc.How can I calculate what the average best of n dice will be?edit: poker poker poker Link to post Share on other sites

Sheiky 0 Posted August 3, 2008 Share Posted August 3, 2008 I'm a bit confused about your definitions here, could you try explaining it a bit more?Currently, it seems like you're terming n as the total number of dice rolls, then giving the multiple dice rolls the value of the last roll, but I think i'm missing something there as that would be incredibly pointless if so. Link to post Share on other sites

uncooper 1 Posted August 3, 2008 Author Share Posted August 3, 2008 Oh. Yea. I didn't define best, which in this game is lowest. So yea, n is the number of rolls, and I want to know what the average lowest of n rolls will be.edit: i feel like it might be (n - 1) / 2... Link to post Share on other sites

AKProdigy 0 Posted August 3, 2008 Share Posted August 3, 2008 I can't really prove this or try it out, but I think this is right just the way I think it in my head (I know thats not very reassuring lol). Basically thinking behind it is that the dice roll probabilities make up a matrix that is 6^{n} large because every possible roll adds a new dimension. So one roll is a single row matrix of 6 cells, two rolls is a 6 x 6 matrix, three rolls is a 6 x 6 x 6 matrix etc. Average score for n rolls:(6^{n} + 5^{n} + 4^{n} + 3^{n} + 2^{n} + 1^{n}) / 6^{n}Someone actually try it out and let me know if it works out. Nite Link to post Share on other sites

Actuary 3 Posted August 7, 2008 Share Posted August 7, 2008 This is a cool problem.AKProdigy, while I can't get my head around your soluation, but it seems to work, matches up with n=1,2,3,4 at least. - And it's correct as n goes to infinityGotta assume its correct, but I want to understand your formula, as I convert my brute force method to a formula as wellSo far, I'm just using counting methods and trying to find patterns.My probability shortcuts are buried too deepI'll avoid this thread for a day, just so I can work it out on my own.well done Link to post Share on other sites

Mr. Sparco 2 Posted August 7, 2008 Share Posted August 7, 2008 Here's the proof of AKProdigy's conjecture:Say we throw n dice. Let's count the number of ways in which we can get a possible best (lowest) score:Best score of 6: For the best score to be 6, all of the dice have to show a 6, so this can only be done in one way.Best score of 5: For the best score to be 5 or better, all of the dice have to show a 5 or a 6, so there are two possibilities for each die. This makes a total of 2^{n}, but these include the single case where the best score is 6, so to have a best score of exactly 5, there are 2^{n} - 1 possibilities.Best score of 4: For a best score of 4 or better, there are 3^{n} possibilities. (Each die has a 4, 5 or 6.) Subtracting the numbers above, we get 3^{n} - (2^{n} - 1) - 1 = 3^{n} - 2^{n} possibilities with a best score of exactly 4.Best score of 3: In the same way, we get 4^{n} - 3^{n} ways to have such a best score.Best score of 2: 5^{n} - 4^{n} ways.Best score of 1: 6^{n} - 5^{n} ways.Now we calculate the expected value. That is, we multiply each score by the number of ways in which it can be obtained, and divide by the total number of ways. Result:6 x 1 + 5 x (2^{n} - 1) + 4 x (3^{n} - 2^{n}) + 3 x (4^{n} - 3^{n}) + 2 x (5^{n} - 4^{n}) + 1 x (6^{n} - 5^{n})divided by1 + (2^{n} - 1) + (3^{n} - 2^{n}) + (4^{n} - 3^{n}) + (5^{n} - 4^{n}) + (6^{n} - 5^{n}).Simplifying each line gives AKProdigy's answer:EV = (1^{n} + 2^{n} + 3^{n} + 4^{n} + 5^{n} + 6^{n}) / 6^{n} Link to post Share on other sites

Mr. Sparco 2 Posted August 7, 2008 Share Posted August 7, 2008 And for those who don't like long calculations and want something slightly more intuitive: Indeed, you can think of all possibilities as presented in a 6 x 6 x 6 x ... x 6 matrix, as AKProdigy mentioned. Then all possibilities with a best score of 2 or better are in a 5 x 5 x 5 x ... x 5 submatrix; all possibilities with a best score of 3 or better are in a 4 x 4 x 4 x ... x 4 submatrix of that, and so on.Now put one penny on each position in the 6 x ... x 6 matrix. Put an extra penny in the 5 x ... x 5 submatrix. Put another extra penny in the 4 x ... x 4 submatrix, and so on. Each position in the matrix now contains a number of pennies that equals its score.You have put in a total of 6^{n} + 5^{n} + 4^{n} + 3^{n} + 2^{n} + 1^{n} pennies in a matrix of size 6^{n}, so the average number of pennies per matrix entry is the answer that we found above. Link to post Share on other sites

rdtedm 0 Posted August 7, 2008 Share Posted August 7, 2008 Last year i took a probability course and had to find probabilities of X happening in a roll of five dice. Poker was easy because every card is distinct, but for dice the fact that there are "identical fives", etc, really threw me off. Moral: **** dice. Link to post Share on other sites

uncooper 1 Posted August 8, 2008 Author Share Posted August 8, 2008 Now we calculate the expected value. That is, we multiply each score by the number of ways in which it can be obtained, and divide by the total number of ways. Result:6 x 1 + 5 x (2^{n} - 1) + 4 x (3^{n} - 2^{n}) + 3 x (4^{n} - 3^{n}) + 2 x (5^{n} - 4^{n}) + 1 x (6^{n} - 5^{n})divided by1 + (2^{n} - 1) + (3^{n} - 2^{n}) + (4^{n} - 3^{n}) + (5^{n} - 4^{n}) + (6^{n} - 5^{n}).Simplifying each line gives AKProdigy's answer:EV = (1^{n} + 2^{n} + 3^{n} + 4^{n} + 5^{n} + 6^{n}) / 6^{n}OK, thanks for doing the hard part. In this particular game, each number has its face value EXCEPT threes, which count as zero. Am I correct that I can ammend the formula to:EV= 6 x 1 + 5 x (2^{n} - 1) + 4 x (3^{n} - 2^{n}) + 2 x (5^{n} - 4^{n}) + 1 x (6^{n} - 5^{n}) / 6^{n}OREV= (1^{n} + 2^{n} + 4^{n} + 5^{n} + 6^{n}) / 6^{n} Link to post Share on other sites

Mr. Sparco 2 Posted August 8, 2008 Share Posted August 8, 2008 OK, thanks for doing the hard part. In this particular game, each number has its face value EXCEPT threes, which count as zero. Am I correct that I can ammend the formula to:EV= 6 x 1 + 5 x (2^{n} - 1) + 4 x (3^{n} - 2^{n}) + 2 x (5^{n} - 4^{n}) + 1 x (6^{n} - 5^{n}) / 6^{n}OREV= (1^{n} + 2^{n} + 4^{n} + 5^{n} + 6^{n}) / 6^{n}I am assuming here that by a three counting as zero, you mean that as soon as you throw at least one three, you get 0 points. This does not mean that you can simply remove the "3-term" form the original formula; the reason being that in the derivation, it was important that a 3-low was better than a 2-low and worse then a 4-low, and this is no longer the case. To find the correct answer, it is easiest to just relabel the "3" as "0" and do the original calculation once again, in the same decreasing order. That is, in the same way as above we findBest value 6: 1 possibilityBest value 5: 2^{n} - 1 possibilitiesBest value 4: 3^{n} - 2^{n} possibilitiesBest value 2: 4^{n} - 3^{n} possibilities (note that we skipped 3, since the next highest value after 4 is now 2)Best value 1: 5^{n} - 4^{n} possibilitiesBest value 0: 6^{n} - 5^{n} possibilities (counting all throws in which we throw at least one "0", i.e. a 3)EV = [ 6 x 1 + 5 x (2^{n} - 1) + 4 x (3^{n} - 2^{n}) + 2 x (4^{n} - 3^{n}) + 1 x (5^{n} - 4^{n}) + 0 x (6^{n} - 5^{n}) ] / 6^{n}or after simplifying by adding terms of the same form a^{n}:EV = (1^{n} + 2^{n} + (2 x 3^{n}) + 4^{n} + 5^{n} ) / 6^{n} Link to post Share on other sites

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now