You have an urn with 10 coins in it: 9 fair, and one that is heads only. You draw a coin at random from the urn, then flip it 5 times. What is the probability that you get a head on the 6th flip given you observed head on each of the first 5 flips?

Let $H_i$ be the event we observe head on the $i$th flip, and let $C_i$ be the event we draw the $i$th coin, $i = 1,…,10$.

Then we wish to calculate (using range syntax for brevity) $$( P(H_6 | H_1 H_2 H_3 H_4 H_5) = P(H_6 | H_{1:5}) $$)

Conditioning on which coin we drew, and exploiting the symmetry between coins 1 to 9:

So it just remains to calculate $P(C_i | H_{1:5})$. This can be done via bayes rule:

where, playing the same conditioning trick:

Thus:

Note that we can quickly self-test and verify $ \sum_{i=1}^{10} P(C_i) = 1 $.

Returning to eqn (2)

Alternatively, you can use R to calculate the probability via brute force by repeatedly sampling according to our problem and counting the number of heads observed.

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max_itrs <- 5*1e6
used_itrs <- 0
skipped_itrs <- 0
num_H <- 0

for(itr in 1:max_itrs){
  coin <- sample(1:10, 1)

  if( coin <= 9 ){
    if( !all( runif(5) <= 0.5)){
      skipped_itrs <- skipped_itrs + 1
      next
    }
    num_H <- num_H + ifelse( runif(1) <= 0.5, 1, 0)
  } else {
    num_H <- num_H + 1
  }
  used_itrs <- used_itrs + 1
}

num_H / used_itrs

# vs 73/82
num_H / used_itrs - 73/82

my sample run produced

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> num_H / used_itrs
[1] 0.8901657
> 
> # vs 73/82
> num_H / used_itrs - 73/82
[1] -7.821522e-05