Two Heads After N TailsDate: 05/20/99 at 10:45:32 From: jack wilson Subject: Probability of two heads after n tails Hi Doc, I'm sure this is probably a simple question but I can't seem to get past the initial phase. I believe it is similar to the picking the same ball problem, but I'm interested in a portion of the non-winning set. The question is: what is the probability of flipping two heads after n number of tails? I know there is a 50/50 chance of flipping a head, and that there is a .25 probability of two consecutive heads. I can also assume from the answer in the picking the same ball problem that the probability of two consecutive heads after n flips is: 1 - (.75)^(n-1) but the non-winning set contains both heads and tails, so how do I get the probability after n number of tails? Since both are equally likely, can I assume that the number of tails is n/2? Thanks, Jack Date: 05/20/99 at 13:29:49 From: Doctor Mitteldorf Subject: Re: Probability of two heads after n tails Dear Jack, It's easy to get confused in probability problems, because our language corresponds to exact mathematical calculations only if we're oh-so-careful about how we phrase things. If what you mean is "The first n flips came out tails. What's the probability now that we will get 2 heads in a row?", then the answer is 1/4. It doesn't matter how many tails just happened - that's all done and in the past. You have a fresh start for your two shots at getting heads. If what you mean is "What is the probability of first getting n tails in a row and then 2 heads in a row?" then the answer is (1/2)^n * 1/4. The (1/2)^n is for the n tails, and the 1/4 is for the two heads. The answer that you read about 1 - (.75)^n doesn't have any application to this situation. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ Date: 05/20/99 at 14:13:25 From: Doctor Anthony Subject: Re: Probability of two heads after n tails The question is slightly ambiguous. It could mean that a mixture of n heads and tails takes place and then 2 heads, OR exactly n tails with single heads spread amongst them and then 2 heads. Since you say n tails I shall assume you mean the second situation. We could consider the following diagramatic representation: *T*T*T.......*THH where there are n tails followed by 2 heads, and we can insert single H's into gaps between the T's. The gaps are denoted by *'s There are in fact n *'s and we must consider all the possibilities of inserting 1, 2, 3, ....., n H's into the gaps. If there are no H's inserted between the T's the probability of the sequence is (1/2)^n x (1/2)^2 The final (1/2)^2 is the probability of the final two H's. = (1/2)^(n+2) If there is only 1 H to be inserted we can choose the gap in C(n,1) ways, and the probability of such a sequence will be C(n,1) x (1/2)^(n+1) x (1/2)^2 again, the final (1/2)^2 is the probability of the final two H's. So the actual probability = C(n,1) x (1/2)^(n+3). If we insert 2 H's in amongst the T's, the relevant probability is C(n,2) x (1/2)^(n+4) We can continue in this way until we have inserted n H's into the gaps between the T's. The total probability is then given by: (1/2)^(n+2)[1 + C(n,1)(1/2) + C(n,2)(1/2)^2 + ..... + C(n,n)(1/2)^n] The series in brackets is the binomial expansion of (1+x)^n with x = 1/2. It follows that the required probability is (1/2)^(n+2).[3/2]^n This is the probability of EXACTLY n tails before we get two consecutive heads. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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