Expected value vs your path

Play a game that wins on average and see how often a single player still goes broke.

Chance of winning90%

Win amount$1

Loss amount$5

Starting money$10

Plays per player100

Expected value per play+$0.40

Average says you end with$50

Players who went broke14%

Ended below their start14%

Each faint line is one player. The dark line is your path. Dashed: where everyone started.

On paper this game pays +$0.40 per play. Yet of 200 players starting with $10, 14% went broke and 14% ended below where they started after 100 plays. The median player finished with $50.

The law of large numbers: one player, 5,000 plays

Average result per play so far. The dashed line is the expected value it slowly converges to.

Try this

Recreate the episode's game: 90% chance to win $1, lose $5 otherwise. The expected value is +$0.40.
Cut the plays to 10. How many players end below where they started, despite the positive edge?
Raise the plays to 500. Does the share of losing players shrink as the average takes over?
Keep the game the same but start with $5. How much more often do players go broke before the math can work?

The math

E[X] = p \cdot w - (1-p) \cdot L \qquad \bar{X}_n \to E[X]

$E[X]$: expected value of one play
$p$: chance of winning
$w,\ L$: win amount and loss amount
$\bar{X}_n$: average result after n plays

Watch the episode behind it

3:26Why a positive expected value doesn't guarantee profit

Ask the Loop about this Next: The central limit theorem