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 Six hundred coins (Posted on 2006-12-22)
I created six hundred coins. I tell you that each is red on one side, but may be red or blue on the other side. I flip each coin, and show you the resulting colors. You count 400 red and 200 blue. What is your best estimate of the number of coins that are red on both sides?

I flipped all the same coins again, and you count 350 red and 250 blue. How should you modify your estimate?

 See The Solution Submitted by Tristan Rating: 4.5000 (2 votes)

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 I might think you cheated. | Comment 3 of 12 |

All that matters is how many turn up blue.  Then the situation can be thought of as a binomial trial where the number of attempts is the number of red/blue coins.

The number of red/blue coins is clearly between 400 and 500.  We want to maximize the probability that both extreme outcomes occurred.

Call the number of blue coins n.  This is then the number of trials.  The probability of 200 or fewer blue occurring when these are all flipped is (using the notation of a TI-83 calculator) binomcdf(n,.5,200)
The probability of 250 or more blue for the same n coins is
1-binomcdf(5,.5,249)

The two situations are independent so we can multiply to find the probability that both would happen.

The most likely case is when n=453

I'd  think you cheated because even then the probability is only .000110245617

 Posted by Jer on 2006-12-22 11:42:40

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