For which numbers of r does a fair game result if s = 45 is given?

Question

Let there be two urns. One urn contains r red and s black balls, the other urn s red and r black balls. One urn is selected at random and one ball is taken out. The removed ball is placed in the other urn. There, enough shuffling is done and one ball is taken at random. The winner is when the first ball taken and the second ball taken are of the same color.

For which numbers of r does a fair game result if s = 45 is given?

Answer 1

Let (U, A, B) be the ordered triple corresponding to the event "Urn U is selected, then the first ball of color A is drawn from that urn, then the second ball of color B is drawn from the other urn". There are eight such ordered triples corresponding to outcomes in this experiment:

(1, black, black)

(1, black, red)

(1, red, black)

(1, red, red)

(2, black, black)

(2, black, red)

(2, red, black)

(2, red, red)

The probability distribution for each of these outcomes is as follows:

P(1, black, black) = 1/2 * s/(r+s) * (r+1)/(r+s+1)

P(1, black, red) = 1/2 * s/(r+s) * (s)/(r+s+1)

P(1, red, black) = 1/2 * r/(r+s) * (r)/(r+s+1)

P(1, red, red) = 1/2 * r/(r+s) * (s+1)/(r+s+1)

P(2, black, black) = 1/2 * r/(r+s) * (s+1)/(r+s+1)

P(2, black, red) = 1/2 * r/(r+s) * (r)/(r+s+1)

P(2, red, black) = 1/2 * s/(r+s) * (s)/(r+s+1)

P(2, red, red) = 1/2 * s/(r+s) * (r+1)/(r+s+1)

The winning triplets are (1, black, black), (1, red, red), (2, black, black), and (2, red, red). Thus the probability of a win is 1/2 * s/(r+s) * (r+1)/(r+s+1) + 1/2 * r/(r+s) * (s+1)/(r+s+1) + 1/2 * r/(r+s) * (s+1)/(r+s+1) + 1/2 * s/(r+s) * (r+1)/(r+s+1), which simplifies to (2rs+r+s)/( (r+s)(r+s+1) ). Substituting s = 45 yields (91r+45)/( (r+45)(r+46) ).

The losing triplets are (1, black, red), (1, red, black), (2, black, red), and (2, red, black). Thus the probability of a loss is 1/2 * s/(r+s) * (s)/(r+s+1) + 1/2 * r/(r+s) * (r)/(r+s+1) + 1/2 * r/(r+s) * (r)/(r+s+1) + 1/2 * s/(r+s) * (s)/(r+s+1), which simplifies to (r²+s²) / ( (r+s)(r+s+1) ). Substituting s = 45 yields (r²+45) / ( (r+45)(r+46) ).

From the context of the problem statement, we assume that the payoff for a win is the same as the the penalty for a loss. Thus, in order for this game to be fair, the probability of a win must be equal to the probability of a loss. Equating these two yields the following equation:

(91r+45)/( (r+45)(r+46) ) = (r²+45) / ( (r+45)(r+46) )

=> 91r+45 = r^2 + 45

=> r = 91

Therefore, we require r = 91 to make this experiment a fair game. Hope this helps!