Imagine a game in which a player has two choices:
A: 50% chance of winning $100 and 50% chance of winning nothing
B: 0.1% change of winning $100,000 and 99.999% chance of winning nothing
The expected values are calculated as
E(A) = 0.5 *100 = 50
E(B) = 0.001 * 100,000 = 100
So if a player was allowed to play this game many times (e.g. 1000 times), choosing B every time would be the “rational” decision to make, as it would be expected to pay off more.
However, if a player is only allowed to play this game for a few times, chances are choosing B will give the player nothing. And choosing A, there is a high chance that the player will take at least $100 home.
Now the expected values of A and B remains the same for each round no matter how many times the player is allowed to play, because each round is independent. So, the overall expected value of B is still higher than the expected value of A in any situation.
This contradicts the intuition described above, in which different choices should be made when the game is played different number of times.
Also, the numbers in the example could be tweaked easily to make choosing B much more absurd, while having a way higher expected value than A (e.g. 0.0001% chance to win $1,000,000,000,000). In this situation, it is clearly not realistic to expect winning by choosing B for just a few times, hence choosing by the expected value would be a wrong decision.
My question would be: is there a way to incorporate the number of times something happens to the expected value, so that the player can make his decision with this new expected value?