# Does doubling up after a loss really work?

Further and deeper exploration of paradoxes and challenges of intuition and logic can be found in my recently published book, *Probability, Choice and Reason.*

The basis of the martingale betting system is a strategy in which the gambler doubles the bet, such as a coin toss, after every loss, so that the first win would recover all previous losses plus a profit equal to the original stake. The martingale strategy has been applied to roulette in particular, where the probability of hitting either red or black is near to 50 per cent.

Take the case of a gambler who wagers £2 on Heads, at even money, so profits by £2 if the coin lands Heads and loses £2 if it lands Tails. If he loses, he doubles the stake on the next bet, to £4, and wins £4 if it lands Heads, minus £2 lost on the first bet, securing a net profit over both bets of £2 (£4 – £2). If it lands Tails again, however, he is £6 down, so he doubles the stake in the next bet to £8. If it lands Heads he wins £8, minus £6 lost on the first two nets, securing a net profit over the three bets of £2 (£8 – £6). This can be generalized for any number of bets. Whenever he wins, the gambler secures a net profit over all bets of £2.

The strategy is essentially, therefore, one of chasing losses. In the above example, the loss after n losing rounds is equal to 2+2²+2³+…+ 2^{n}

So the strategy is to bet in the next round 2+2²+2³+…+ 2^{n }+2

In this way, the profit whenever the coin lands Heads is 2.

For a gambler with infinite wealth, and hence an infinite number of coin tosses to eventually generate heads, the martingale betting strategy has been interpreted as a sure win.

However, the gambler’s expected value remains zero (or less than zero) because the small probability of a very large loss exactly balances out the expected gain. In a casino, the expected value is in fact negative, due to the house edge. There is also conventionally a house limit on bet size.

The martingale strategy fails, therefore, whenever there is a limit on earnings or on bets or bet size, as is the case in the real world. It is only with infinite or boundless wealth, bet size and time that it could be argued that the martingale becomes a winning strategy.

*Appendix*

Probability of losing three fair coin tosses = 1/8

Probability of losing n times = 1/2^{n}

Total loss with starting stake of 2, with 3 losses of coin toss = 2 + 4 + 8 = 14.

So martingale strategy suggests a bet of 14 + 2 = 16.

Loss after n losing rounds = 2 + 2^{2 }+ … + 2^{n}

So martingale bet = (2 + 2^{2} + … + 2^{n}) + 2 = 2^{n+1}

This strategy always wins a net 2.

This strategy, of always betting to win more than lost so far, works in principle, regardless of the odds, or whether they are fair. If each bet has a 1 in 10 chance of success, for example, the probability of 12 successive losses is about 30%, but the martingale strategy is to bet to win more on the 13^{th} coin toss than the sum of losses to that point.

This holds so long as there is no finite stopping point at which the next martingale bet is not available (such as a maximum bet limit) or can’t be afforded.

So, let us assume that everyone has some number of losses such that they don’t have enough money to pay a stake large enough for the next round that it would cover the sum of the losses to that point. Call this run of losses n.

n differs across people and could be very high or very low.

Probability of losing n times = 1/2^{n}

Using a martingale +2 strategy, the player wins 2 if able to play on, and then wins.

So, the player wins 2 with a probability of (1-1/2^{n})

Total losses after n losing bets = (2 + 2^{2} + … + 2^{n}) = (2^{n+1} – 2)

Expected gain is equal to the probability of not folding times the gain plus the probability of folding times the loss.

Expectation = (1 – 1/2^{n}) . 2 – 1/2^{n} (2^{n+1} – 2)

= 2 – 2/2^{n} – 2 + 2/2^{n} = 0.

So the expected gain in a fair game for any finite number of bets is zero using the martingale system, but it is positive if the system can be played to infinity. The increment per round need not be 2, but could be any number, x. The net gain to a winning bet is this number, x.

The intuitive explanation for the zero expectation is that the player (take the simplest case of an increment per round of 2) wins a modest gain (2) with a very good probability (1 – 1/2^{n}) but with a small probability (1/2^{n}) makes a disastrous loss (2^{n+1} – 2).

More generally, for an increment of x:

Expectation = (1 – 1/x^{n}) . x – 1/x^{n} (x^{n+1} – x)

= x – x/x^{n} – x + x/x^{n} = 0.

The mathematical paradox remains. In the case where on the nth round, the bet is 2^{n}, the martingale expectation = ½ x 2 + ¼ x 2^{2} + 1/8 x 2^{3} + … = 1 + 1 + 1 + 1 … ∞

Yet the actual expectation, when the odds are fair, in all realistic cases = 0.

If the odds are tilted against the bettor, so that for example the bettor wins less if a fair coin lands Heads than he loses if it lands Tails, the expected gain in a finite series of coin tosses is less than zero, but the same principle applies.

*Exercise*

Show that the expected value of martingale strategy in a fair game of heads/tails is zero. Show how this can be reconciled with the fact that whenever the player wins, the net overall profit to the player is positive.

**References and Links**

Martingale (betting system). Wikipedia. https://en.m.wikipedia.org/wiki/Martingale_(betting_system)