Probability and Ten-Sided DiceDate: 06/23/2008 at 02:43:15 From: Vera Subject: The odds of a number greater than 95 from 10 10-sided dice This is from an online game. The challenge is to get a number higher than 95 by rolling a set of ten 10-sided dice. Originally, I thought that the dice only contained the numbers 0 to 9, but I've seen the number ten when others "rolled". What I want to know is, what are the odds of getting a number 95 or greater, and how do you come to figuring out these odds? So far, the numbers bounce in the mid-ranges, and seldom get higher than the eighties. Thanks for your time. :) Date: 06/23/2008 at 17:38:57 From: Doctor Tom Subject: Re: The odds of a number greater than 95 from 10 10-sided dice Hello Vera, This isn't a particularly easy question. Here's how I'd think about it. First imagine marking the dice, so we know which is die 1, which is die 2, and so on, up to die number 10. How many total outcomes are there? Well, each one can come up in 10 different ways, so there are 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 10^10 outcomes for rolling all the dice. Now, how many of those have a total of more than 95? We'll have to work it out for each possible score: 96, 97, ..., 100. There's only one way to get a score of 100; namely all ten dice show 10. To get a 99, there are ten ways to do it: one die shows a nine and the others all show 10, but there are 10 ways you can pick the particular die that has the 9. Things get ugly now, since to get 98 you can either have one 8 and all 10's, or two nines. There are 10 ways to get one 8 and nine 10's, and 45 ways to choose two dice that have the 9 and eight 10's for a total of 55. But looking at cases will quickly get out of hand, so here's the best way to work the problem. Let's say you want to find the number of ways to get exactly a score of 96. Imagine that you have ten boxes (one for each die) and you distribute 4 balls somehow into the boxes: maybe into 4 different boxes, maybe all 4 in one, or maybe two in one and two into two others, et cetera. Once the balls are distributed, subtract that number of balls from 10 and that's the number that comes up on the die corresponding to the box. In other words, a box with no balls in it comes up 10, one with three balls in it comes up 7 and so on. Thus, there are as many ways to get a 96 score as there are ways to distribute 4 balls into 10 boxes. The general formula for distributing k balls into n boxes is: (n+k-1 choose k) See my paper: http://www.geometer.org/pascal.pdf for a complete description of why. Anyway, that makes: (13 choose 4) ways to get a 96 (12 choose 3) ways to get a 97 (11 choose 2) ways to get a 98 (10 choose 1) ways to get a 99 and (9 choose 0) ways to get a 100 (13 choose 4) = 715 (12 choose 3) = 210 (11 choose 2) = 55 (10 choose 1) = 10 (9 choose 0) = 1 Luckily, the last three numbers match our previous calculations, making us much more confident of the answer! The total is 991 ways to get more than 95. Thus the probability is 991 out of the 10^10 possible rolls: 991/10^10 = 9.91 x 10^{-8} which is a little bit less than 10 times in every 100 million attempts, or about 1 in 10 million. You said that sometimes you get numbers in the 80's and that's not too surprising. The counts go up pretty fast, although things get REALLY ugly if you ask for the number of ways to obtain scores below 90, since then you can't use any distribution of balls in the 10 boxes since you can't put more than 10 in a box. Since it's relatively easy to get to 90, I did that by just continuing as above, and there are a total of 320982 ways to get a score of 90 or more. 320982/10^10 = 3.20982 x 10^{-5} so that will happen about 3.2 times in every 100,000 rolls. Still not too common! - Doctor Tom, The Math Forum http://mathforum.org/dr.math/ Date: 06/25/2008 at 16:17:14 From: Vera Subject: Thank you (The odds of a number greater than 95 from 10 10-sided dice) Thank you so much for providing me with such a well explained, thorough answer! I'm glad I bookmarked your site! |
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