⚖️Expected Value

Expected value answers the question "what is this chance situation worth, per go, in the long run?"

It is an average of the payouts, but not the plain average — the likely payouts have to count for more than the rare ones. So every payout is weighted by its own probability, and the weighted payouts are added up:

E = p1 x v1 + p2 x v2 + p3 x v3 + ...

A game paying 10 gold with probability 1/5, 8 gold with probability 1/10, and nothing otherwise:

E = 1/5 x 10 + 1/10 x 8 + 7/10 x 0 = 2 + 0.8 + 0 = 2.80 gold.

The plain average of 10, 8 and 0 is 6, which is more than twice as much — because it pretends the 70% chance of nothing is no more likely than the 20% chance of ten gold.

Because the probabilities add to 1, the expected value always lands somewhere between the smallest payout and the largest.

The expected value of a single roll of a fair die is 3.5, and no die has ever shown 3.5. Expected value is not a prediction of one play; it is the long-run average of many.

Roll a die 1000 times and the mean of the results will sit very close to 3.5. Roll it once and you will get a whole number, guaranteed not to be the expected value.

That is why expected value is a tool for deciding whether to keep doing something, not for guessing what happens next. It answers "if I did this every day for a year, where would I end up?" — and for that question it is exactly right.

Most games cost something to play. The stake is certain, so it comes off in full:

expected net gain = expected winnings - stake

A game is fair when the expected net gain is zero: expected winnings exactly equal the stake, and over a long run neither side ends up ahead. Above zero the game favours the player; below zero it favours the stallholder.

Real stalls run at a negative expected value for the player, and that is how they stay open. A big top prize does not change this. A prize of 1000 gold that comes up once in 5000 plays contributes 1000/5000 = 0.20 gold to the expected value, and no more.

To find the prize that makes a game fair, set expected winnings equal to the stake and solve. If a prize turns up with probability 1/8, then to add 3 gold to the expected value it has to be worth 3 x 8 = 24 gold.

Expected values add up over repeated plays, so scaling to many plays is a single multiplication:

expected total over n plays = n x (expected net gain per play)

If each play has an expected net of -0.40 gold, then 200 plays have an expected net of -80 gold. That figure is far more trustworthy than the per-play one, because the wobble grows more slowly than the total does — this is the same law of large numbers that steadies an experimental probability.

Do one play properly first. Everything about the game is identical every time, so the per-play figure is all the work there is; the rest is multiplication.