1
$\begingroup$

Ok, so I have a problem that is equivalent to this game:

Two players each choose $k$ random integers between 1 and $n$ independently of each other. What is the probability that they both choose the same numbers?

Now I have a table of measured data for the probability of each number, but there are two catches.

  1. The numbers are chosen at uneven distributions

  2. The data I'm given is the percent of games each number appears in. Since up to 10 numbers can be selected each game, the total percent is greater than 100.

I have no idea where to go from here or how to work with this.

Example of data: percent of games each number was chosen 1-10% 2-5% 3-6% ...

$\endgroup$
  • $\begingroup$ You cannot add probabilities of events that are compatible (i.e. not disjoint). $\endgroup$ – Jean Marie May 21 '17 at 21:08
  • $\begingroup$ The sum of all those percentages shouldn't necessarily be less than 100% because you aren't adding up probabilities of disjoint events, like JeanMarie said. Specifically, it's very likely that one number will appear in multiple games since each person chooses $k$ numbers per game. $\endgroup$ – tilper May 21 '17 at 21:15
0
$\begingroup$

You'd have to sum over all possible selections the probability that both players made that selection:

$$\sum_{n\choose k} (p_{n\choose k})^2 $$

Where $p_{n\choose k}$ is the probability for a specific $k$-of-$n$-selection. That would equal $\prod_{i\in k}^n p_i\prod_{i\not\in k}^n (1-p_i)$

I assume that the $p_i$ are determined correctly, but the formula will give a result regardlessly.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.