We came across this brilliant puzzle on Plus Magazine, and they've kindly allowed us to share it.

Post solutions below!

2

We came across this brilliant puzzle on Plus Magazine, and they've kindly allowed us to share it.

Two people have asked you out for a date and you decide who to go with by flipping a coin. The only coin you have, however, lands heads with a probability of $0.75$ and tails with a probability of $0.25$. How can you a use a series of flips to decide who to go with so that both people have the same chance of being picked?

Post solutions below!

Puzzle

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3 comments
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This is nice, though it only has a $\frac{3}{8}$ probability of making a decision in two coin flips, so it will take an expected number of $2\times \frac{8}{3}=\frac{16}{3}$ flips to make a decision. Is there some way of reducing this expected number?

jules
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What if you add to the conditions that a decision must be reached in a finite number of flips?

Payne
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This is an interesting point. With probability 1, the decision will be reached in a finite number of flips with this scheme.

If you want a decision reached within a fixed number $N$ of flips, then it is probably impossible to do it exactly, but an extremely inefficient method of doing as well as possible would be to work out the probability of every possible sequence of $N$ flips (all $2^N$ of them), and find the subset with has a total probability of as close to $\frac{1}{2}$ as possible. There are probably efficient algorithms which get to within a given percentage of the optimum solution.

jules
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Hi all, I have a question. In this case, it is impossible to guarantee that the decision will be reached in $N$ flips (that's because the sum of the probabilities of some the individual outcomes can never be 1/2). Are there a probability $p\neq 1/2$ of head and $N$ so that we can guarantee that the decision can be reached in $N$ flips?

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I guess this is the kind of task for which statistics should not be used. If you pick up who you date based on random choices, then you need to date more :) ! Then again, in the end maybe it's all random with that respect too...

Nice solution by @muraii. $16/3=5.33$ flips on average. I think that one can define a way to ensure that at most in 5 flips you get it. It might be possible to define even a smaller number of flips and ensure one get's it...