One urn 2 Colours get majority using Bayes theorem

Question

There is one urn containing 10 red and blue balls. 4/10 are red and 6/10 are blue. I have to simulate how a set of rational players would decide what the dominant color of balls is in an urn, if they one after another randomly draw a ball out of the urnl look at the color of the ball and without showing the ball to the other players put it back in the urn. The player then guesses whether the urn is majority blue or majority red and publicly announces this guess to the other players. To simulate the game i have to let 20 player draw a ball.

The players know that there are 10 balls in the urn, and that the ratio of balls is 6:4

How can i calculate the conditional probability for the player that the urn is majority blue or majority read, given the player's draw and given what the player has heard so far.

So far i have:

Let A=Red ball drawn, B=Blue Ball drawn then,

Prior probability P(A)=4/10

Marginal probability P(B)=6/10

But now i don't know how to get P(A|B) and P(B|A).

I already searched the web but only found questions with two or more urns. I hope that someone can help me calculate the conditional probability.

Answer 1

I assume that the prior belief for the initial stage would be five red, five blue. As each person declares their belief after seeing one ball it should indicate a shift in the prior for the next person. The conjugate prior for the bernoulli trail is a beta distribution. This question shouldn't be closed as the edited question makes sense. I am guessing there will be a "burn in" period before the priors stabilise to dominate however I have not tried to establish it. My gut says you'll need to simulate it before you can put an analytical bound on it but maybe there's a trick I am missing. Assuming perfect call by the people and truly no asymmetric information then the ability to form a true belief may not be possible if the first two draws are from the q side. It's an interesting problem. If you are not sure how to simulate it then draw out a table, I am guessing you need to get 3-4 draws with all permutations before stabilisation.