# Number of expected winners in a game involving three kinds of marbles

In a bag there are some red marbles, some blue marbles, and some green marbles.

There is at least one marble of each color in the bag.

Richard, Bob and George ignore the exact numbers of marbles of each color in the bag, and they decide to play a game.

Richard bets to get, in $N$ trials with replacement, at least one red marble, Bob to get at least one blue marble, and George to get at least one green marble.

However, we additionally know that, in $N$ trials, the probability for Richard to win is exactly equal to the probability for Bob to lose.

Denoting with $X\in\{1,2,3\}$ the number of boys which can win (*), what is the expected value of the random variable "In $N$ trials, $X$ boys win?"

(*) $X\neq 0$, because we know that there is at least one ball of each kind in the bag.

I tried to use the linearity of the expectation value, illustrated e.g. in this post, but I did not arrive anywhere. Moreover, I suspect that the additional information does not play any role, but I don't know how to prove it.

I apologize in case the question might be not perfectly posed. However, thanks a lot for your suggestions!

• The conclusion that $N$ must be $2$ made headlines in the New York Times in the 1990s. – Empy2 Aug 15 '18 at 11:10

The expected value, as you mentioned, is the sum of individual winning probabilities by linearity. Let there are $R$ red, $B$ blue, and $G$ green marbles in the bag. The probability of getting at least $1$ red marbles in $N$ i.i.d. trials is

$$1 - \left(\frac {B + G} {R + B + G}\right)^N$$

and similar for the other two colors. The condition given implies that

$$1 - \left(\frac {B + G} {R + B + G}\right)^N = \left(\frac {R + G} {R + B + G}\right)^N$$

Summing the three winning probabilities gives you the expectation: $$3 - \frac {(B + G)^N + (R + G)^N + (R + B)^N} {(R + B + G)^N} = 2 - \frac {(R + B)^N} {(R + B + G)^N}$$

where we used the condition to simplify the expectation.

• But, then, the conclusion is that the expected number of winners is less than 2, i.e. one boy? Sorry if the question is silly! – user559615 Aug 15 '18 at 11:16
• On average only one of Richard and Bob will win, so X's average is at most 2. The average over many games doesn't have to be a whole number. – Empy2 Aug 15 '18 at 11:21
• I see now. But... what about George? He's also included in the $2-\epsilon$, $0<\epsilon<1$, right? – user559615 Aug 15 '18 at 11:24
• From the first equation, $(B+G)^N+(R+G)^N=(R+B+G)^N$. Why can you conclude $N=2$? – Empy2 Aug 15 '18 at 11:26
• Another question: This conclusion, allows you to say that the probability that the three boys win is null? I would say no, but I am not sure! Thanks! – user559615 Aug 15 '18 at 11:27