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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?

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.

HandwovenConsensus OP , (edited )

I don’t know if that applies to this scenario. In this game, the player is always in the lead until they aren’t, but I don’t see how that works in their favor.

ada ,
@ada@lemmy.blahaj.zone avatar

Once the player loses, the chain ends, and the house wins. So as long as the house can afford to keep pushing the player in to trying again, they’re going to create more opportunities for the player to return their winnings to the house.

HandwovenConsensus OP ,

Right, and as the chain continues, the probability of the player maintaining their streak becomes infinitesimal. But the potential payout scales at the same rate.

If the player goes for 3 rounds, they only have a 1/8 chance of winning… but they’ll get 8 times their initial bet. So it’s technically a fair game, right?

rowinxavier ,

If everyone has the same amount of starting capital it is a fair game assuming both can opt out at any time.

That said, the house appears to not be able to opt out (they definitely can, you just don’t think about that part), and the house has more capital. For them each time someone plays a round there are only 3 possible outcomes. Half are the player loses, then a quarter are the player wins and plays another round, and lastly a quarter are the player wins and ends the game. The only case where the player wins is option 3, in all other cases, so 75%, the house wins because the next round has another chance to make the player lose directly at a 50/50 chance or play another round.

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)

HandwovenConsensus OP ,

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.

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