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## Tuesday, November 14, 2017

### Problem of the Week 11-14-17: Tossing a Coin [Probability]

Check out this Problem of the Week.
Be sure to let us know how you solved it in the comments below or on social media!

Solution below.

Solution

Video

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.

1. 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.

This 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.

1. You are correct, please excuse our mistake! We will be correcting the solution shortly

2. This comment has been removed by the author.

2. Maybe an alternative more easy explanation:

Call 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.

Cheers!