Which biased coin was more likely to have been flipped?

Question

Consider the following question:

We have two biased coins. Coin $A$ returns heads with probability $3/5$ and coin $B$ returns heads with probability $4/5$. A coin was chosen at random and flipped $10$ times, returning $7$ heads. Was it more likely that we flipped coin $A$ or $B$?

The answer is coin $A$. The probability that coin $A$ behaved as described is $$\left(\frac{3}{5}\right)^7\left(\frac{2}{5}\right)^3\binom{10}{7}\approx.215$$ while the probability than coin $B$ behaved as such is $$\left(\frac{4}{5}\right)^7\left(\frac{1}{5}\right)^3\binom{10}{7}\approx.201$$ so it was more likely to be coin $A$.

I generalized this question as follows:

Coin $A$ returns heads with rational probability $P_A$ and coin B with rational probability $P_B$. The coin is flipped $N$ times and returns $N(P_A+P_B)/2$ heads ($N$ is chosen so that this is an integer). Was it more likely that we flipped coin $A$ or $B$?

I conjecture that: the answer is always the coin whose probability of returning heads is closest to $1/2$. If the distances are equal, both coins are equally likely.

How do I prove this? I was able to reduce this problem to a messy inequality; I was wondering if there is a much more elegant explanation.

Answer 1

While I wrote the question, I found a relatively simple way to approach the problem. We can adopt the method used to solve the first specific question.

First, it's not hard to show the second part of the claim. If $P_A$ and $P_B$ are equidistant from $1/2$, the probabilities for coins $A$ and $B$ will have identical expressions.

Now, WLOG, say coin $A$'s probability of returning heads is $P_A=a/c$ and $B$'s is $P_B=b/c$. If one of them, say $a/c$ WLOG, is less than $1/2$, we can replace it with $1-a/c>1/2$ and not change the problem, using the statement above. So now $a/c,b/c>1/2$, WLOG $a>b$. Also WLOG $a+b$ is even.

So we flip the coin $c$ times, and find $(a+b)/2=h$ heads. The probability coin $A$ behaved as described is $$\left(\frac{a}{c}\right)^h\left(\frac{c-a}{c}\right)^{c-h}\binom{c}{h}$$ and the probability coin $B$ behaved as such is $$\left(\frac{b}{c}\right)^h\left(\frac{c-b}{c}\right)^{c-h}\binom{c}{h}$$ We want to compare these two probabilities to show that the first the less than the second (since $a>b$). So we can remove identical terms from both sides and compare $$a^h(c-a)^{c-h}<^?b^h(c-b)^{c-h}$$ $$\left(1+\frac{a-b}{b}\right)^h=\left(\frac{a}{b}\right)^h<^?\left(\frac{c-b}{c-a}\right)^{c-h}=\left(1+\frac{a-b}{c-a}\right)^{c-h}$$ Now here, I am close but stuck. We note that since $a/c,b/c>1/2$, that implies that $a,b>c/2$. So $b>c/2>c-a$, so we can attempt to increase the left hand side by replacing the denominator $b$ with something smaller, $c-a$: $$\left(1+\frac{a-b}{c-a}\right)^h<^?\left(1+\frac{a-b}{c-a}\right)^{c-h}$$ Unfortunately this is too strong. By similar reasoning, $h$, being the average of $a$ and $b$, is larger than $c/2$, so the exponent on the LHS is greater than the exponent on the RHS, making this new expression false.

I suppose this problem has been reduced to showing that a certain inequality: $$\left(1+\frac{a-b}{b}\right)^h<^?\left(1+\frac{a-b}{c-a}\right)^{c-h}$$ is true under the appropriate constraints: $a,b>c/2$ and $h=(a+b)/2$.

Answer 2

Here's one way to do it: $$ \frac{\Pr(A \mid {\text{observed data)}}}{\Pr(B\mid{\text{observed data}})} = \frac{\Pr(A)}{\Pr(B)}\cdot\frac{\Pr(\text{observed data}\mid A)}{\Pr(\text{observed data}\mid B)}. $$ The third fraction you see above is what you've already told us how to compute. The second fraction (i.e. the first one after "$=$") is what you haven't told us at all (obviously it is equal to $1$ if the two coins are initially equally probable). The first gives you what you're looking for.