Puzzle time - bias evidence

[jon-nyc]

Years ago Alice bought a special coin from a magic shop which was supposed to flip heads with probability 60%, or was it tails? She's forgotten which.

Alice flips the coin 10 times and it comes up heads every time, so she decides it was probably biased toward heads.

Unbeknown to Alice, Bob bought (another copy of) the same coin, and also forgot which way it was biased. With lots of time on his hands, he flips the coin 100 times and gets 55 heads. Like Alice, he decides to treat it as heads-biased.

Of Alice and Bob, which one actually had the best evidence that the coin was biased toward heads?

[Klaus]

It's very close! Not what intuition suggests.

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I think this is a pretty straightforward application of Bayes' rule.

Alice:

P(HB | 10H) = P(10H|HB) * P(HB)/P(10H)
= (0.6^10 * 0.5)/( (0.6^10 * 0.5 + 0.4^10 * 0.5)
= 0.9829540725450702

Bob:

The probability of 55H out of 100 flips is 0.0478111801 for head-biased and 0.00082911901 for tail-biased via Binomial distribution formula.

Hence:
P(HB | 55H/100) = P(55H/100|HB) * P(HB)/P(55H/100)
= 0.0478111801 * 0.5 /(0.0478111801 * 0.5 + 0.00082911901 * 0.5)
= 0.9829540725453817

This is so close in fact that the difference is maybe just due to rounding errors and they are in fact the same. To find out for sure, one would have to calculate the difference between the two terms symbolically, i.e., via algebraic manipulation. But I can't be bothered to do that now and I'm happy enough with "the chances are practically identical".

[Klaus]

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WolframAlpha thinks the difference between the two probabilities is 0. I'm not quite sure whether it arrived at this result numerically or analytically, though.

[Horace]

***Not safe for work content? More like too much work content.***

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Maybe there's an elegant solution, but otherwise it's a straight math problem. The probability of 10 for 10 is simple, but 45 for 100 less so. I think the binomial theorem could be used, or integrating the probability distribution from 0 to 45. Anyway, that would be too much work for me. I wonder if there's something elegant.

[jon-nyc]

I have an elegant analytic solution:

click to show

Suppose you know that your coin has probability heads equal to either p or 1-p. You flip the coin n times and get h heads and t tails, with h+t=n and a difference d=h-t.

Using the standard Bayesian formula and simplifying, you can show (setting tail probability as q=1-p):

P(coin bias is p given observing h heads in n coin tosses)

= (p^d)/((p^d) + (q^d))

This is a function of the difference in the observed numbers of heads and tails, not the overall number of coin tosses n. So in the example the probabilities must be equal, since 10-0=55-45. Done.