# Decision making with expected value

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?

The expected returns are $50$ and $100$ per game, and the number of games is irrelevant. It is your intuition that has failed.

The distributions of returns may be different between the two games, and you might actually wish to maximize the probability that you have some money (say, more than $25$ per game). In that case you'd have to use a Binomial distribution to find the probability of getting a final result greater than some cutoff, and here likely game A will be superior, even though its AVERAGE will be less than in game B.

Football coaches face this problem all that time. Suppose in a passing play the team gets a gain of 20 yards (50% of the time) or 0 yards (50% of the time), so the average is of course 10 years. Suppose in a running play the team gets a gain of 3 yards (33% of the time), 2 yards (33% of the time) or 1 yard (33% of the time), so the average is 2 yards. Clearly passing is "better" than running (at least on average).

But suppose the team is on the 1.5-yard line and has just one play to make the touchdown. Should the team pass or run? Clear, it should RUN, since all that matters is the probability of gaining more than 1.5 yards.

Your intuition is guided by scenarios such as that.

Of course, too, you can incorporate the number of trials, by looking at the distribution of gains according to the two scenarios.

• To elaborate a bit, people tend to be risk-averse. This is especially pointed when one outcome is guaranteed, e.g. an option between being given \$1 or getting \$X or \$0 based on the flip of a coin. Most people need X to be noticeably higher than$2\$ before they'd take the latter. This is made more extreme for higher amounts via the declining marginal utility of money and even more extreme if framed in terms of losses. – Derek Elkins left SE Jul 13 '18 at 23:09