Why does the conditional probability of having chosen a biased coin given an unusual result decrease with a higher number of coins

Question

Let's say $1$ out of three coins is biased and lands on tails with probability $p>1/2$

We choose a coin randomly from the three coins and throw that coin $10$ times. Given that it lands on tails $8$ out of $10$ times the probability of having chosen the biased coin is $$P(B|8,2)=\frac{P(8,2|B)P(B)}{P(8,2|B)P(B)+P(8,2|B^c)P(B^c)}=\frac{{10\choose8}p^8(1-p)^2\cdot{1\over3}}{{10\choose8}p^8(1-p)^2\cdot{1\over3}+{10\choose8}({1\over2})^{10}\cdot{2\over3}}$$

And this is equal to $0.7746$ if $p=0.8$ or $0.5791$ if $p=0.6$ etc

I would expect that value to increase if we had more than $3$ coins and only one of them was biased with $p=0.8$ but it's not the case.

Answer 1

If you had more than 3 coins, in the fraction $$ P(B \mid 8, 2) = \frac{P(8,2|B)P(B)}{P(8,2|B)P(B)+P(8,2|B^c)P(B^c)} = \frac{P(8,2|B)}{P(8,2|B)+P(8,2|B^c)\frac{P(B^c)}{P(B)}} $$ the only numbers that change are $P(B)$ and $P(B^c)$. Namely, $P(B)$ goes down, and $P(B^c)$ goes up. That means that the denominator goes up, which means your resulting probability goes down.

In Bayesian terms: your evidence $P(8, 2 \mid B)$ has not changed, your prior $P(B)$ has gone down, so your posterior $P(B \mid 8, 2)$ has also gone down.

Answer 2

Here is an appeal to intuition. Think of these two scenarios:

• You have ten coins, one of which is biased. You pick one of them and toss it ten times, getting 8 tails.
• You have ten coins, one of which is biased. You eliminate seven of them using some completely certain method. Then you pick up one of the remaining three and toss it ten times, getting tails 8 times.

Which of these scenarios makes you more likely to hold the biased coin, you think? Note that the second case is exactly the same as just starting with three coins.