Betting game with numbers Someone is challenged to play the following game: there are 36 marbles in an urn. 21 of them are red, 9 are blue and 6 are green. Each red is worth 0 rupees, each blue 100 rupees and each green 1000 rupees. By betting 1000 rupees only once, you must pick as many marbles as you want (obviously without seeing its color), one by one (you draw each marble, look at its color and either stop or continue) and without replacement and you can stop at any time, and you earn the product of the values of the marbles you have picked so far.
Assuming that the player plays optimally, what is the average percentage of the average percentage of profit for the game?
My understanding is that we must calculate the probability (not) to draw a 0 in a series of consecutive draws from 1 to 36.
For the 1st draw, the probability to get a 0 is $P_1 = \frac {21}{36}$ and the game ends, so with probability $P'_1 = \frac {15}{36}$ the player continues ("optimal" playing refers to this, otherwise if he continues, he will get 0 anyway).
Up to now, the player has earned 100 rupees with probability $Pw'_1 = \frac {15*9}{36*15}$ and 1000 rupees with probability $Pww'_1 = \frac {15*6}{36*15}$.
For the 2nd draw, the probability not to get a 0 is $P'_2 = \frac {14}{35}$ and so on.
The player will earn again the product of either 100 or 1000 of the first earning, with probability equal to the product of $P'_2$ by $\frac {8}{14}$ or $\frac {5}{14}$.
Can you please help me continue? It is getting really confusing from this point forward!
Thank you very much!
 A: As long as there is at least one blue or green marble in the urn, it makes sense to keep playing, since you have a better than 1 in 36 chance of multiplying your winnings by at least 100. So the optimal strategy is to draw 15 marbles. Your expected earnings will be the product of: (1) the probability that every one of the first 15 marbles drawn is non-red, times (2) the number of rupees you will earn if you draw all 15 non-red marbles and no red marbles. Can you compute these two quantities? The result is kind of funny, in that you wind up with a minuscule probability of winning an astronomically large amount of money.
Edit: There are
$$
{36\choose 15}=\frac{36!}{15!\, 21!}=5567902560
$$
possible sets of 15 marbles you could draw. Exactly one of them has no red marbles, so using the optimal strategy, the probability you win money is $1/5567902560$. In the event you draw 15 non-red marbles, the amount of money you win is $100^9\cdot 1000^6 = 10^{36}$. Your expected winnings are then
$$
\frac{10^{36}}{5567902560}\approx 1.796\cdot 10^{26},
$$
and your expected percent profit is
$$
\frac{\frac{10^{36}}{5567902560}-1000}{1000}\cdot 100\%\approx 17960084416419816082413526\%.
$$
