Comment on Is there a house advantage in a "double-or-nothing" coin flip game?
OsrsNeedsF2P@lemmy.ml 4 weeks ago
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) = (1 - (p / q)^(X)) / (1 - (p / q)^N)
where:
- p is the probability of winning a single bet (in this case, 0.5)
- q is the probability of losing a single bet (in this case, also 0.5)
- N is your initial capital (in this case, 100)
- X is the number of bets
HandwovenConsensus@lemm.ee 4 weeks ago
You’re saying that the player pays a dollar each time they decide to “double-or-nothing”? I was thinking they’d only be risking the dollar they bet to start the game.
That change in the ruleset would definitely tilt the odds in the house’s favor.