Probability of an Even Number of 6'sDate: 05/03/2003 at 05:45:36 From: Liya Andres Subject: A fair die Dear Dr. Math, A fair die is thrown n times. Show that the probability that there is an even number of sixes is 1/2 * [1+(2/3)^n] Note: 0 is considered an even number. What I find confusing about it is how to get the probability of an EVEN number of sixes. I'm also confused about where the 2/3 that is found in the equation came from. I was only able to get the probability of getting at least one six for n tosses, which I calculated as 1-(5/6)^n, and the probability of getting no sixes at all, which I calculated as (5/6)^n. But I can't figure out how to work this into showing the equation is true. Date: 05/03/2003 at 07:35:59 From: Doctor Mitteldorf Subject: Re: A fair die Liya - Here's a shortcut, where you don't have to calculate the probability of 0, 2, 4, etc. We'll use induction on n. In a proof by induction, we show that the statement is true for n = 0, then use the fact that it's true for n to show that it's also true for n + 1. Thus the proof for n = 0 implies it is true for n = 1, which implies it is true for n = 2, etc. For n = 0, the formula works out to p=1, which is the right answer since 0 sixes is the only possibility. Now suppose it is true for n = n. Then the probability of an even number of sixes is 1/2 * [1+(2/3)^n] and for an odd number of sixes it is 1 - 1/2 * [1+(2/3)^n]. We can calculate the probability of an even number of sixes for n + 1 tosses as coming about in two ways: either the result for the previous n tosses was even, and the next toss is not a six (5/6 * 1/2 * [1+(2/3)^n]), or else the result for the previous n tosses was odd, and the next toss is a six (1/6 * (1 - 1/2 * [1+(2/3)^n]). Add these two expressions and simplify, and you get 1/2 * [1+(2/3)^(n+1)] which is the original formula as it applies to (n + 1) tosses. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ Date: 05/04/2003 at 05:24:46 From: Liya Andres Subject: A fair die Thank you so much! Is there a way to show how the given formula was derived, using the concepts of probability instead of induction? -Liya Date: 05/04/2003 at 05:40:49 From: Doctor Mitteldorf Subject: Re: A fair die Dear Liya, I can't do it for a general n, but I can show you how it would be done using a particular n for an example. Let's say n = 5. The probability of rolling r sixes out of 5 rolls comes from two parts. The first part comes from the probability of (six) and the probability of (not six). These are 1/6 ad 5/6. And the factor we need is (1/6)^r * (5/6)^(5-r). This is because the (1/6) probability came up r times, while the (5/6) part came up (5-r) times. The second factor comes from the number of different orderings that are possible. For example, if r = 5, then you are talking about every roll being a 6. There is only one way to do that. But for r = 4, there is one roll that is not a 6, and it could be the first, the second, etc. There are 5 different positions for that one (not-six). For r = 3, it's a little more complicated: There are 3 sixes and 2 (not-sixes) to mix together in any order. There are 10 ways to do this, such as 66NNN, 6N6NN, 6NN6N, 6NNN6, N66NN, etc. (Can you list the other 5?) In general, the number of different orderings is the combinatorial formula C(5,r), sometimes written as 5Cr, with the 5 and the r as subscripts in the front and back. The way to calculate 5Cr is 5! C(5,r) = ------------- (5 - r)! r! So we can put this all together: The probability of getting r heads out of 5 is C(5,r)* (1/6)^r * (5/6)^(5-r). I suggest you calculate this for r = 0, r = 2, and r = 4 and add them up to see that the number you get agrees with your formula for the proability of an even number. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ Date: 05/04/2003 at 09:19:35 From: Liya Andres Subject: Thank you (A fair die) Dear Doctor Mitteldorf, Just want to let you know how much I appreciate your quick and helpful replies. Thank you! :) -Liya |
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