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I have a friend bend a coin as much as he wants out of my sight. We agree on an even money bet on the 1st flip. I randomize my choice 50-50.
It flips heads, 3rd party judge reports to us blind.
I want to do another even money bet on the 2nd flip. I'm +/=EV taking heads this time. The range could be from 0 to 1. Is my blind start and 50-50 assumption now irrelevant or am I actually building on it?
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You are actually building on your 50-50 assumption. The math is a little technical. f97tosc talked about "priors"
here:
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Suppose that we write f(P) for the prior probability distribution that the coin is such that when tossed, it will result in heads a long-term fraction P of the time.
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The function f(P) is the prior. Essentially, P is the probability of heads for the bent coin, and f(P) is your estimate of the probability that the coin is actually bent that way. For example, if you think all bends are equally likely, then the function f(P) is flat. If you think extreme bends are less likely than small bends, then the function f(P) has a hump in the middle.
Your initial 50-50 assumption is represented by choosing a prior with a certain property. For example, if you choose a prior which is symmetric about the line P = 1/2, then you would get a 50-50 initial estimate. The actual shape of the prior beyond that is up to you.
You must use the prior if you want to apply Bayes theorem and estimate the probability of heads on the second flip. This estimate will depend on the exact shape of the prior, and not simply on its symmetry. So it depends on more than just your initial 50-50 assumption. But you
can say, based only on symmetry, that the second flip is more likely to be heads than tails. As f97tosc says in his post,
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we only need to make some very mild additional technical assumptions to be able to conclude that, for example, after seeing one head, we rationally should judge it more likely than not for the next toss to be head as well
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This means that if you start with a prior function f(P) that gives you a 50-50 initial estimate, then the fact that heads is more likely than tails on the second flip does not depend on the shape of the function. Even though the exact probability of heads on the second flip does depend on the shape of the function.
In the end, though, the conclusion that heads is more likely on the second flip is based both on the first flip being heads and on your initial 50-50 assumption.
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If a different unknown coin is flipped 6 times, all heads, I'm tempted to say "hey, let's see that bloody coin!" Where does that come from, other than my theoretical 50-50 expectation?
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It is a mix of both your initial assumption and the 6 flips. You might say it is "mostly" the 6 flips. The reason you might say that is because, even if your initial assumption was biased towards heads, 6 heads is pretty hard to get. So as long as the bias in your assumption was not too extreme, you would still be surprised by 6 heads in a row.
But it also comes from another assumption you might be making. You might imagine that extreme bends are less likely than small bends. That would mean your prior function f(P) has a hump in the middle. In some sense, you
must make this assumption in order to be surprised by 6 heads. For example, a flat prior would make the probability of 6 heads exactly 1/7. In fact, it makes the probability of k heads exactly 1/7 for all k. If you initially assume that all bends are equally likely, then all numbers of heads are also equally likely, and you should not be surprised if you see 6 heads or 3 heads or 0 heads. They would all have the same chance of happening, if all bends were equally likely.