Guessing how many balls picked based on number of white balls only

Question

I have a pool for white and black balls. For this example let say 20% is white and remaining 80% is black.

Now, at random, someone picks different numbers of balls following some distribution of picks, e.g. 1 pick 10% 2 pick 20% 3 pick 60 4 10%.

If at the end, I am only allowed to see how many white balls the person picked for each try, how can i tell what is the likelihood that person picked X number of balls from the pool?

How should I go about thinking about this ?

Example. : Again an infinite pool of 20% white and 80% black balls. Person B can pick different numbers of balls randomly for each trial following the pick probability of 1 pick 10% 2 pick 20% 3 pick 60% 4 pick 10%. Person A can only ask about how many white balls B picked for each trial. (sequence does not matter )

Let say B picked:

1 white 1 black

0 white 2 black

1 white 2 black

1 white 1 black

2 white 2 black

1 white 2 black

1 white 0 black

3 white 0 black

0 white 1 black

0 white 1 black

Person A has a list of white balls (1, 0, 1, 1, 2, 1, 1, 3, 0, 0) B picked for the 10 trials. How can I go about thinking about the likelihood of how many balls B picked from the pool for each trial?

Or I can ask a different question too, how can I estimate the pick probability of B, i.e. without knowing how B is picking the balls, how can I guess the pick probability of 1 pick 10% 2 pick 20% 3 pick 60% 4 pick 10%?

Answer 1

This is a simple application of Bayes rule. The probability that the person picked $n$ balls conditional on observing $k$ picked white balls is just the probability that the person picked $n$ balls AND $k$ of those were white, divided by the probability that $k$ white balls were picked.

To put this into action, suppose the number of observed white balls is $k$. If the person had picked $j$ balls then the conditional probability that $k$ of them are white is given by the binomial distribution (for $j\ge k$): $$ {j \choose k} p^k (1-p)^{j-k}$$ where $p$ is the proportion of white balls in the pool. Let $a_j$ be the probability that the person picked $j$ balls. Then the probability that the person picked $j$ balls and $k$ of them were white is $$a_j {j \choose k} p^k (1-p)^{j-k}$$ again assuming $j\ge k$. The overall probability that $k$ white balls were picked is obtained by summing expressions like this over all possible values of $j$: $$\sum_{j: j\ge k} a_j {j \choose k} p^k (1-p)^{j-k}$$ Therefore, applying Bayes rule, the probability that $n$ balls were picked conditional on $k$ of them were white is the ratio $$a_n {n \choose k} p^k (1-p)^{n-k} \over \sum_{j: j\ge k} a_j {j \choose k} p^k (1-p)^{j-k}$$