From: Hal Finney (hal@finney.org)
Date: Mon Oct 28 2002 - 22:34:25 MST
Let:
H = event that a coin is heads
S = event that a coin is selected
We have immediately:
p(H) = p(~H) = 1/2
p(S) = 1/4
p(~S) = 3/4
Now what we want is p(~H | ~S), the probability that a coin is not heads
given that it is not selected. I am going to instead calculate
P(H | ~S), the probability of an unselected coin being heads, and we can
then subtract from 1 to get the desired probability.
Bayes theorem tells us:
p(H & ~S) = p(H | ~S) * p(~S)
= p(~S | H) * p(H)
I will argue that p(S | H), the problem that a coin is selected given
that it is heads, is very close to 1/2. That is the main step. From this
it follows that p(~S | H), the probability that a coin is not selected
given that it is heads is 1 - 1/2 = 1/2. And this will immediately lead
to the desired result using the two Bayes theorem equations.
We know that the most probable outcome is that there will be exactly
10 coins as heads. When we select 5 of them, the chance that a given
head-side-up coin is selected is 1/2.
With slightly less probability, we might have 9 or 11 coins be heads.
The chances of these two outcomes are equal. In the 9 coin case, the
chance off a coin being selected is slightly more than 1/2 (specifically
5/9), and in the 11 coin case, it is slightly less than 1/2 (specifically
5/11). These two cases average to very close to a 1/2 probability.
Likewise we can pair up the case of 8 and 12 coins falling heads.
These too will average out to about a 1/2 probability. And in this
way we can consider all of the possible cases and average them out
to the same 1/2 probability. Now, the approximation becomes worse in
the more extreme cases, but these cases are less common and so the bad
approximation won't matter that much.
For more accuracy you can do as Lee did and count each case explicitly.
But it is clear that 1/2 is a good probability for the chance that a
coin that fell heads will be selected.
Therefore we can substitute in the last equation above:
p(H & ~S) = p(~S | H) * p(H) = 1/2 * 1/2 = 1/4
And then we have, rearranging one of the equations above:
p(H | ~S) = p(H & ~S) / p(~S) = 1/4 / 3/4 = 1/3
And therefore the desired quantity, p(~H | ~S) is one minus this, or
2/3.
This is close to Emlyn's simulations, within about a percent.
The discrepancy is no doubt due to the approximation above that the
chance of a head being selected is 1/2. Substituting the actual figure
by looking at the 21 possible numbers of heads that could fall (0-20
possible heads) would give a more accurate figure.
Hal
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