Imagine you have a gambling game with a house edge of 1%. The game is very simple, you flip a coin and you win or lose. Probability of either is 50%. If you win, you get x1.98 your bet.

Now what would it do to the house edge, if the player is allowed to flip the coin again if he loses? For example, a player bets $10, flips a coin, loses, and gets a chance to flip again (for free). What would the house edge be in this scenario if such rule would apply?

Chance of success now 75%, so 1 - (.75 x 1.98)?

.25 - .75(.98) = -48.5%

Same answer as DMMx69.

Do I get to retry if I win the first toss?

No, only retry if you lose the first one.

So the house edge goes from 1% to -48.5% ?

Yes. They will lose 48.5 cents on every dollar wagered over time.

And that time will be very, very short.

If a player gets a free re-flip after a loss, the house edge flips in favor of the player.

Original game (50/50 win/loss, x1.98 payout):

Expected value = $9.90 per $10 bet → 1% house edge

With free re-flip on loss:

75% chance to win $19.80

25% chance to lose $10

Expected value = $12.35 → player edge of 23.5%

So, the house edge becomes -23.5%, heavily favoring the player.

Try it yourself — flipacoinonline.com

@jassonadder careful to keep the random variable consistent when calculating EV. In gambling contexts we usually make "net profit" the random variable, but one could also define it as "final stack".

The player's expected net profit is .75*9.8−.25*10 = $4.85
The player's expected final stack is .75*19.8−.25*0 = $14.85

Half your calculation assumed the RV was final stack while the other half assumed it was net profit, so your result isn't an EV of anything.

Good thread — nice thought experiment. Quick, careful math (define the random variable consistently):

Assume a $10 bet and a win pays ×1.98 (net profit on a win = $19.80 − $10 = +$9.80). A loss is −$10.

Original single flip (no retry)

Win probability = 0.5 → net +9.80

Lose probability = 0.5 → net −10.00

Expected value (net profit) = 0.5×9.80 + 0.5×(−10.00)

Calculate: 0.5×9.80 = 4.90; 0.5×(−10.00) = −5.00

EV = 4.90 − 5.00 = −0.10 per $10 bet → −$0.10 / $10 = −1.0% (house edge 1%).

(You can think of it as the house expecting to keep $0.10 of every $10 wagered.)


With a free re-flip only if you lose the first toss

Possible outcomes and probabilities:

Win on 1st toss: probability 0.5 → net +9.80.

Lose 1st, win 2nd: probability 0.5×0.5 = 0.25 → net +9.80 (because the re-flip is free).

Lose both: probability 0.5×0.5 = 0.25 → net −10.00.

EV (net profit) = 0.5×9.80 + 0.25×9.80 + 0.25×(−10.00)

Compute step by step:

0.5×9.80 = 4.90

0.25×9.80 = 2.45

0.25×(−10.00) = −2.50

Sum = 4.90 + 2.45 − 2.50 = 4.85

So EV = +$4.85 per $10 bet → +48.5% for the player (i.e. house edge = −48.5%). The house would lose on average $4.85 every $10 wagered.

Paging the mods.

Jassonadder and senthilkumarkkr1 are both spambots using an LLM. Notice how both posts include a stupid link, among other clues (such as grunching and ancient thread to begin with).

Showelay is another, from another thread. Among other obvious tells, notice how Showelay sneaks the "poki games" link into their quote of the OP.