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# Martingale revoked

## Question

In a game of chance a player wins his stake with a favorable outcome (W) and loses it otherwise (L).

A famous (Martingale) gambling strategy is to double the stake after every loss.

A win is thus assured, even after a (finite) string of losses.

The series of outcomes W, LW, LLW, LLLW, LLLLW, ... all win the original stake.

Nevertheless, the strategy loses, and it's a classic puzzle to show why.

Let's change the game a bit and eliminate that approach.

The player doubles his stake with every W, but must leave the game after only a single L.

Note that is the only penalty for L; the player keeps his accumulated winnings.

The player has no strategy: he plays until he must quit.

If his initial stake is \$1, and W and L are equally likely outcomes, what are his expected winnings?

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If the player loses his stake after a loss, then he will end up with \$0 no matter what.

To illustrate:

Round 1: Total = \$1, Stake = \$1, Loss = \$0, Win = \$2

Round 2: Total = \$2, Stake = \$2, Loss = \$0, Win = \$4

Round 3: Total = \$4, Stake = \$4, Loss = \$0, Win = \$8

.

.

.

To summarize, by doubling his stake each round, he keeps betting all his money. Since he will eventually lose, he will eventually lose all the money he won plus the dollar he started with.

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@benjer, the player doesn't lose the accumulated winnings after losing a game. It just forces him to stop playing.

In the version of the Martingale problem that I'm familiar with, the answer to why the Martingale strategy fails is that eventually the gambler will lose enough money that his bank account won't be able to cover a bet to try to recoup his losses, and at that point the game ends with the player bankrupt. (It should be a safe assumption that the gambler doesn't have an infinitely large bank account.)

And for this problem, the answer depends on how much money the house has available to pay the player.

The expected winnings from game #N would be (the probability that the player makes it to and wins game #N) x (the amount of money he makes by winning game #N). For game #1, the player has a 1/2 chance of making it to and winning that game, and the amount of money made would be \$1 (doubling the initial wager), so game #1 is worth \$0.5. For game #2, the player has a 1/4 chance of making it to and winning that game, and the amount of money made would be \$2 (because he would have \$2 after having won the first game and would bet it all on the second game), so game #2 is worth \$0.5. It should be easy to see that each possible game adds \$0.5 to the expected winnings. So if the player could potentially play an infinite number of games, and each game has an average payoff of \$0.5 (with increasing payoff for each game balancing out the decreased probability of winning that game), then superprismatic's answer is correct.

If the house has a finite amount of money, then the game must end when the house is bankrupt. If the house has H dollars, then it will be bankrupt after the player wins N games where 2^(N-1) [the winnings from game N... remember that the possible winnings from game #1 is \$1, from game #2 is \$2, from game #3 is \$4, etc] + 2^(N-1)-1 [the winnings from all the games preceding game N: if the player won a certain amount of money in a game by betting all of his previous winnings plus his initial \$1, then the amount of winnings prior to that round must have been the amount won in that round minus one dollar] = H.

2^(N-1) + 2^(N-1) - 1 = H

2^N = H+1

N = log(base 2)(H+1)

Since there are only N games possible, and each game is worth \$0.5, the expected winnings would be

[log(base 2)(H+1)] / 2

Even if the house has a little over a billion dollars to spare, the expected winnings would be \$15.

Edit: corrected the formula for the amount of money won from game #N

Edited by plasmid

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@benjer, the player doesn't lose the accumulated winnings after losing a game. It just forces him to stop playing.

But he has no accumulated winnings. (Or if you prefer, the amount added to each subsequent bet matches that of the accumulated winnings, nullifying them.)

Bet (-): \$1 -> Win (+): \$1 -> Accumulated winnings (+): \$1 -> Net \$1

Bet (-): \$2 -> Win (+): \$2 -> Accumulated winnings (+): \$3 -> Net \$3

Bet (-): \$4 -> Win (+): \$4 -> Accumulated winnings (+): \$7 -> Net \$7

Bet (-): \$8 -> Win (+): \$8 -> Accumulated winnings (+): \$15 -> Net \$15

...

Bet (-): \$16 -> Lose -> Accumulated winnings (+): \$15 -> Net -\$1

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psykomakia, i think you keep your most recent bet after a loss as well.

in other words there's no penalty for losing other than having to stop playing.

let's make heads a win and tails a loss. so we first need to know how many heads we are likely to flip before seeing tails.

which is pretty simple, we are likely to see 1 heads.

which gives an expected winning of 1 dollar.

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