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Solution
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This video solution is incomplete, please refer to the solution document above for a full correct version of the solution to this problem, which includes conditional probability.
Haven't you forgotten the conditional probability piece? We are saying that Danny has already flipped a Heads, so we don't need need the initial H in each element of the sequence; this would make the answer off by a factor of 1/2. Another way to think about it would be to define W:Danny wins with a Heads flipped first, and A:Danny's first flip is Heads. Then we want P(W|A) = P(W&A)/P(A) = (1/3)/(1/2) = 2/3.
ReplyDeleteThis also makes sense because the game should be fair knowing nothing else; each should win with probability 1/2 due to symmetry. So the fact that Danny has flipped a Heads already would only increase his chance of winning (not decrease to 1/3).
Just some initial thoughts.
You are correct, please excuse our mistake! We will be correcting the solution shortly
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DeleteMaybe an alternative more easy explanation:
ReplyDeleteCall p the probability that Danny wins after he throws head. Conditioning on the next throw (head and tail, resp.) gives;
p = 1/2 + 1/2 * q
where q is the probability that Danny wins after Mike has thrown a tail. Note that at this stage, due to symmetry, Mike has exactly the same probability p to win as Danny did initially. Therefore, it holds that q = 1 - p. Filling q = 1 - p in the conditioning above and solving for p gives p = 2/3.
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