1
$\begingroup$

An urn contains $N_1$ red balls and $N_2$ green balls. Each ball has an associated weight. Each ball is drawn (without replacement) with a probability proportional to how much its weight contributes to the urn.

What is the probability that when we draw $x$ balls, we get $x$ red balls and $0$ green balls?

Example: 2 red balls of weight 0.3 and 0.4, and 2 green ball of weight 0.2 and 0.1.

In first attempt, the probability to get each ball is the following: Pr(red1)=0.3, Pr(red2)=0.4, Pr(green1)=0.2, Pr(green2)=0.1.

Improve this question
$\endgroup$
5
  • $\begingroup$ Keep going: after drawing the first red ball, what is the new probability of drawing a red ball? Multiply and repeat. $\endgroup$
    – whuber ♦
    Mar 29, 2022 at 15:30
  • $\begingroup$ Do you mean using the Law of total probability? I think that that could be hard since each ball have a different weight and I am drawing up to 15 balls. $\endgroup$
    – Cohensius
    Mar 29, 2022 at 15:35
  • $\begingroup$ Use the law of conditional probability. But are you saying the weights vary within each color of ball? In that case, there is no hope for it: you have to compute an exponentially large polynomial. But the formula is straightforward to write down, FWIW. $\endgroup$
    – whuber ♦
    Mar 29, 2022 at 15:44
  • $\begingroup$ @whuber, yes, its multivariate, each ball (card at the opponent hand actually) has a different weight. exponentially large polynomial -> this is exactly my problem since the opponent can have up to ~10 cards, that's to much to calculate. $\endgroup$
    – Cohensius
    Mar 29, 2022 at 17:27
  • 1
    $\begingroup$ It's too much with pencil and paper, but a computer will make short work of it. Some useful information appears in the closely related question at stats.stackexchange.com/questions/20590. $\endgroup$
    – whuber ♦
    Mar 29, 2022 at 18:15

0

Reset to default

Your Answer

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