This is kind of a crazy question.
I was looking up the expected value of a binomial distribution here the other day. I was looking at it in relation to basketball where it is clear if you took $n$ shots and had a probability of making $p$ of making a $3$ point shot that the equation for $E[X]=3nP$. Similarly, if you made a two-point basket with a probability of $p$ it would be $2np$ given the same number of shots.
Is there a way of relating the probability of winning by these expected values given $n$ baskets if you just shoot three-point shots vs just shooting two-point shots? I would be curious if this would be related to its expected value in some way. Like it seems to me that you want the expected value to be greater than the two-point one iff the probability of winning with three-point shots is greater than the probability of winning with two-point shots given the same number of shots. I just feel like there is some sort of theorem like that about this that I am missing. Does anyone know of a theorem that can be related to expected value and winning in this context given you shoot all three point shots or two with some said probability?
Best Answer
The probabilities of winning with 3-point shots vs 2-point shots will depend on $p_{3}$ and $p_{2}$, but also on $n$. The only related "theorem" I can think of that relates to the expected values is that according to the law of large numbers, the 3-point shots will win over the 2-point shots if $% 3p_{3}n>2p_{2}n\rightarrow p_{3}>\frac{2p_{2}}{3}$ when $n\rightarrow \infty $. Hence, when $n$ increases indefinitely the strategy with the highest expected value will win.