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Is there a house advantage in a "double-or-nothing" coin flip game?

The idea is that if the coin flip goes in the player’s favor, they win double their bet. After winning, they can either collect their winnings, or risk them all on another coin flip to have a chance at doubling them. The initial bet is fixed at, let’s say $1.

Mathematically, this seems like a fair game. The expected value of each individual round is zero for both house and player.

Intuitively, though, I can’t shake the notion that the player will tend to keep flipping until they lose. In theory, it isn’t the wrong decision to keep flipping since the expected value of the flip doesn’t change, but it feels like it is.

Any insight?

OsrsNeedsF2P ,

If you have 100$, and you bet 1$ at a time, infinitely, you will lose.

More generally (simplified to assume you’re always betting the same amount):

P(ruin after X bets) = (edit: I removed my formula because it was wrong…but I’m sure you could mathematically prove a formula)

solrize ,

If both players have infinite bankrolls, but only one of them is allowed to stop the game once they are ahead, the one with the option of stopping has an advantage. They can play until they are in the lead, then stop. The reason this doesn’t work in real life is that real bankrolls aren’t infinite.

See also: en.wikipedia.org/wiki/Gambler's_ruin

rimu ,
@rimu@piefed.social avatar

And, in real life, the house has a much larger bankroll.

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