There are m white and n red marbles in a closed box.
Two players take turns on drawing randomly a marble, without returning it back.
The 1st player drawing a white marble wins.
Evaluate the probability that the first player wins.
Check your formula for (m,n)= (2,3) and (2,4).
The m=2 case is interesting enough to be its own problem, and probably D2 by itself (rather than the problem as stated which seems D3 or D4).
Lets draw all the marbles out even after we know who won. We'll also pair them up (P1 draw, P2 draw) with the possibility of a singleton for odd n.
For any sequence of draws we'll construct a mirrored sequence by swapping those pairs. Example with n=5: (W, W), (R, W), (W, R), W [a 1st player win] becomes (W, W), (W, R), (R, W), W [a 2nd player win].
The only time we don't get a match between a Player 1 win and a Player 2 win is when both Red marbles are in the same pair, which Player 1 always wins.
For all n+2 marbles there are T=(n+2)C2 = (n+2)(n+1)/2 total possible sequences of draws.
The special sequence of draws when both reds are in the same pair happens S=floor(n/2)+1 times out of those.
Then the probability of Player 2 winning is half of the times the special sequence does not occur: (T-S)/(2T) = 1/2 - S/(2T) = 1/2 - (floor(n/2)+1) / ((n+2)(n+1))
Then the probability of Player 1 winning is the complement: 1/2 + (floor(n/2)+1) / ((n+2)(n+1))
Analytic Solution for m=2
