Rolling Dice and Probability
Date: 11/07/2007 at 07:26:09 From: Harri Subject: Probability, rolling dice Having thrown a die 10 times, what is the probability of rolling all six numbers? Order does not matter.
Date: 11/07/2007 at 10:03:55 From: Doctor Tom Subject: Re: Probability of rolling numbers 1-6 in ten throws. Hello Harri, This is a tough problem, and you can find details if you look up "coupon collector's problem". (Waiting to get all 6 numbers is like a coupon collector who is trying to get a complete set where each new coupon is equally likely to be any of them.) There are easy tricks to find the expected (average) time to get all six numbers and that turns out to be about 14.7 rolls. In other words, if you repeat the experiment, rolling until you get all six, and write down the number of rolls it takes you to do so each time, you'll find that the average of those numbers is about 14.7. To solve your problem exactly for small numbers of rolls, like 10, we can proceed as follows. Let p0, p1, p2, p3, p4, p5 and p6 be the probabilities that at a certain stage in the process you've seen 0, 1, 2, ..., 6 different faces. Before you start, p0=1 and p1=p2=...=p6=0. After one roll, p0=0 (and will be zero from now on), p1 = 1, p2=p3=...=p6=0. In general, there is a nice formula for figuring out the new p0, ..., p6 from the old ones, one roll later. Suppose they are p0, ..., p6 at one stage, and you want to compute p0', p1', p2', ..., p6' one roll later. Here are the formulas: p0' = 0 p1' = p1/6 p2' = 5 p1/6 + 2 p2/6 p3' = 4 p2/6 + 3 p3/6 p4' = 3 p3/6 + 4 p4/6 p5' = 2 p4/6 + 5 p5/6 p6' = p5/6 + p6 Let's look at one of these carefully; the others can be argued in the same way. Let's look at how we obtain p3'. Well, either we had three faces before and didn't get a new one, and this will happen 3 times out of 6, which accounts for the term 3 p3/6. OR, we had two faces and got a new one. If we had two, then 4 of the six rolls from there would yield a new face, which is where the term 4 p2/6 comes from. There are no other ways to arrive at three faces. Now just do the arithmetic. Here are the first few steps. You'll have to continue, step by step, to stage 10 if you want the answer for 10 rolls. The numbers in parentheses represent (p0, p1, p2, ..., p6). The "stage i" means the situation after i rolls. stage 0: (1, 0, 0, 0, 0, 0, 0) stage 1: (0, 1, 0, 0, 0, 0, 0) stage 2: (0, 1/6, 5/6, 0, 0, 0, 0) stage 3: (0, 1/36, 15/36, 20/36, 0, 0, 0) stage 4: (0, 1/216, 35/216, 120/216, 60/216, 0, 0) ... For practical purposes, you don't need exact fractions; you can do each stage with a calculator, and keep the decimal expansion. This technique of keeping track of the probability of being in various "states" as time progresses is called a "Markov chain", which you can also look up. There is a very easy way to work out the numbers from my formulas by repeated matrix multiplication. Look up "Markov chain" for more information. As time goes on, the probabilities will shift to the right, toward p6, but there is always some small probability that smaller numbers will be present. Thus at every stage after 5 rolls, the odds of having completed the set of faces increases since the probability of being in stage 5 is always positive, but after a certain point it will get smaller and smaller. - Doctor Tom, The Math Forum http://mathforum.org/dr.math/
Date: 01/29/2008 at 19:00:08 From: Doctor Anthony Subject: Re: Probability of rolling numbers 1-6 in ten throws. Hi Harri - I would add a technique that can be used in this type of problem. The number of ways that 10 numbered balls could be distributed in 6 numbered boxes such that no box is empty is a suitable model. The number of ways is given by the coefficient of x^10/10! in the expansion of [e^x - 1]^6 and is denoted by T(10,6). The number of unrestricted ways of distributing the balls is 6^10 so that the required probability is T(10,6)/6^10 We have [e^x - 1]^6 = e^(6x) - C(6,1)e^(5x) + C(6,2)e^(4x) - C(6,3)e^(3x) + C(6,4)e^(2x) - C(6,5)e(x) + 1 We now pick out the terms in x^10/10! = (x^10/10!)[6^10 - 6(5^10) + 15(4^10) - 20(3^10) + 15(2^10) - 6] = (x^10/10!) and so the coefficient of x^10/10! is 16435440. If you throw a die 10 times the total number of possible outcomes, with 6 possibilities at each throw, is 6^10. So the required probability is 16435440/6^10 = 0.271812128 - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum