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Looking over my daughter's homework, I had a question that seems best to be asked here. It's from Everyday Mathematics (Volume 1, Grade 5, p 42.)

Tim flipped a coin 10 times. It landed heads up 7 times and tails up 3 times. Tim said, "I'll flip it 4 more times to get the same number of heads and tails." Is he right? Explain why or why not.

My gut tells me the new 4 flips will be 50/50 yielding 2 heads and 2 tails and won't take into account the previous flips. However, maybe they all land on heads giving a 50/50 overall distribution. Am I letting my knowledge of the Monty Hall Problem confuse me?

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None of my close coin friends even remember what they did when they were previously flipped, let alone decide to make up for any excess of heads that they produced in the past. – André Nicolas Oct 7 '11 at 1:15
up vote 6 down vote accepted

We're supposed to understand that Tim thinks the coin must come up heads in the next four throws, because otherwise the total outcome of the 14 throws wouldn't be 50/50. This is a fallacy, of course.

However, your gut is guilty of the same fallacy when it thinks that the next 4 throws have to be 50/50. Possibly your gut is confusing the expected number of heads with the possible actual numbers of heads. But the problem does not seem to be about expected results -- Tim's problem is that he thinks probabilty can ever guarantee him any result.

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If there are 4 more heads than tails accumulated, and the coin is flipped several more times, then the chance of the excess falling to $4-k$ is the same as the chance of it rising to $4+k$ heads (for every possible $k$). There is no statistical tendency favoring a reduction of the excess by additional flips of the coin.

There does exist a tendency to push the ratio (total heads)/(total tails) toward 1 with more coin flips. Maintaining a 7:4 ratio for another 110 flips would be extremely unlikely. This is not a result of the coin remembering its past history but of the nature of ratios compared to differences. To maintain a 7:4 ratio or stay close to it one would need to get about 70 heads and 40 tails. But outcomes with equal or nearly equal numbers of heads and tails are more likely so that ratios further from 1 are disfavored for a large enough set of coin flips.

(Assuming a fair coin)

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Your gut is right. My interpretation is that Tim is thinking that the coin will somehow magically restore itself back to having the same number of heads and tails, since it's a fair coin. But of course you know that coins don't do that.

There's nothing that forces the coin to give two heads and two tails in the next four flips, though -- the flips are independent -- but certainly four heads in the next four flips is not the most likely outcome.

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Tim's thinking is especially naive, because why didn't he think the same thing after six flips, when the results were (say) 5 heads and 1 tails? If his theory is that the coin will conspire to even itself out, then he's already disproved that theory. (unless 14 is an especially magic number for him....) – Greg Martin Oct 7 '11 at 4:29

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