Binomial and Geometric Probability Distributions
Date: 10/20/2005 at 00:02:08 From: Josť Subject: the probability of rolling _exactly_ one 6 in n rolls What is the probability of rolling exactly one 6 in n rolls? I know how to find the expected value, and I know how to compute it if it's not n, but I don't know how to compute it for n. I think it would have something to do with a probability distribution like 1-(5/6)^n.
Date: 10/20/2005 at 08:36:55 From: Doctor Wilko Subject: Re: the probability of rolling _exactly_ one 6 in n rolls Hi Jose, Thanks for writing to Dr. Math! It looks like to answer your question, you'd use the binomial probability distribution. For a dice roll, the probability of getting exactly 1 success (in this case a six) out of n trials is calculated by, nC1 * (1/6)^1 * (5/6)^(n-1) Let's think about a concrete example. If your question was, "What's the probability of rolling exactly one six in four rolls?", you could have a six in any of the four spots and any other non-six's in any of the remaining spots, i.e., _6_ _2_ _1_ _5_, or _2_ _6_ _3_ _3_, or _3_ _3_ _6_ _5_, or _3_ _2_ _2_ _6_ The probability would then be, 4C1 * (1/6)^1 * (5/6)^3 ~ 0.3858 | | | 6 can be | | in any of | | the four | Prob. of three spots | non-six's | | Prob. of getting one 6 About a 38.5% probability of getting exactly one six in four rolls. This link from our archives will explain more about the binomial distribution: Binomial Probability Formula http://mathforum.org/library/drmath/view/66627.html Just for fun, what if the question had been, "What's the probability of getting a six for the _first time_ on the nth trial?" Notice how this wording is different from the first question. It's still exactly one six in four rolls, but this time the six is restricted to the last roll. This time the probability would be calculated as, (5/6)^(n-1) * (1/6) Looking at a concrete example again, "What's the probability of getting a six for the _first time_ on the fourth roll of the die?" You could have, _4_ _3_ _3_ _6_ | \ / | \ / Last spot reserved for 'success', | in this case a 6 | (n-1) non-six's (4-1) = 3 non-six's So the probability would be calculated as, (5/6)^(3) * (1/6) ~ 0.0965 About 9.5% of getting a six for the first time on the fourth roll of a die. This is referred to as the geometric probability distribution. Note the difference in probabilities. It looks like since the six can be in any of the four spots in the first example (binomial), the probability is higher (4 times in fact!) than when the six is restricted to be in the last spot (geometric). Does this help? Please write back if you have further questions. - Doctor Wilko, The Math Forum http://mathforum.org/dr.math/
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