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# The Gambler's Fallacy

Module by: Justin Romberg. E-mail the author

Summary: Of course the casinos have thought of that. They want you to play that way ...

## The Gambler's Fallacy

Just about everybody who has walked into a casino at some point in their lives has had the following can't-lose idea. You approach the roulette table (or whatever your game of choice) and bet $1 on black. If you win, you repeat. If you lose, you double your bet to$2 to chase the dollar you just lost. If you win, you lower your bet back to $1, but if you lose again, you double to$4, etc. The idea is that you have to win at some point, and when you do you are guaranteed to have won a dollar.

This turns out to be a very bad way to gamble pretty much no matter what the particulars are of the game you are playing. To see this, let us calculate the probability that we make nn dollars before we suffer a catastrophic loss; that is, we lose a hand and do not have enough money in front of us to double the previous bet. For example, if we bring $127$127 to the table, we can suffer a losing streak of no longer than 6 games before we win 127 hands. We will declare “success” if we make nn dollars before we lose nn dollars. To make things simpler, we will assume that nn are “powers of two minus one”; that is n=2u-1n=2u-1 for some integer u1u1. This just means that we have enough money to cover u-1u-1 losses in a row, but if we have uu losses, we walk away empty-handed.

We will call each sequence of games which ends in a win a trial. We will use the following simple model:

• each game is an independent event, and
• the probability we win each game is pp, which is fixed.

If there are kk games in a trial, that means that there were k-1k-1 losses followed by a win. Thus the length of each trial is a geometric random variable:

Prob a trial lasts exactly k games = ( 1 - p ) k - 1 p , Prob a trial lasts exactly k games = ( 1 - p ) k - 1 p ,
(1)

and

Prob a trial lasts u games = k = 1 u ( 1 - p ) k - 1 p = p k = 0 u - 1 ( 1 - p ) k = 1 - ( 1 - p ) u . Prob a trial lasts u games = k = 1 u ( 1 - p ) k - 1 p = p k = 0 u - 1 ( 1 - p ) k = 1 - ( 1 - p ) u .
(2)

If the goal is to make nn dollars, then we sit at the table for nn of these trials. Since each game is independent, the trials are also independent of one another. Thus the probability that all of the trials have length at most uu is

Prob all n trials have length u = 1 - ( 1 - p ) u n . Prob all n trials have length u = 1 - ( 1 - p ) u n .
(3)

Taking u=log2(n+1)u=log2(n+1) (our assumption on nn means that uu will be an integer), we have the expression

Prob success = 1 - ( 1 - p ) log 2 ( n + 1 ) n . Prob success = 1 - ( 1 - p ) log 2 ( n + 1 ) n .
(4)

First let's consider the case where the game is fair, that is p=1/2p=1/2. The expression above simplifies to

Prob success = 1 - 2 - log 2 ( n + 1 ) n = 1 - 1 n + 1 n e - 1 0 . 3679 as n . Prob success = 1 - 2 - log 2 ( n + 1 ) n = 1 - 1 n + 1 n e - 1 0 . 3679 as n .
(5)

Note that this game gets worse for you the larger nn is; clearly, the chance you make $1$1 before losing $1$1 is 0.50.5, but that's actually where things peak for you. In fact, nn does not need to be too large for e-1e-1 to be a very close approximation to Equation 5. For example, if n=15n=15 (risk $15 to make$15, and we cannot have a trial last longer than u=4u=4 games), we already have Probsuccess.3798Probsuccess.3798. For n=127n=127, the difference is less than two parts in a thousand.

Moral: If you bring nn dollars to an fair game, the chance that you will double your money before losing it all is less than 37%.

The situation becomes darker when the game is even slightly unfair. For a general value of p<1/2p<1/2, and u=log2(n+1)u=log2(n+1), we have

( 1 - p ) log 2 ( n + 1 ) = ( 1 - p ) log 1 - p ( n + 1 ) log 2 ( 1 - p ) = 1 n + 1 log 2 ( 1 / ( 1 - p ) ) ( 1 - p ) log 2 ( n + 1 ) = ( 1 - p ) log 1 - p ( n + 1 ) log 2 ( 1 - p ) = 1 n + 1 log 2 ( 1 / ( 1 - p ) )
(6)

and so

Prob success = 1 - 1 n + 1 log 2 ( 1 / ( 1 - p ) ) n . Prob success = 1 - 1 n + 1 log 2 ( 1 / ( 1 - p ) ) n .
(7)

This expression is accurate and straightforward to calculate, but it is a bit unwieldy. We will finish by finding simpler expressions that bound Equation 7 from above and below.

To simplify Equation 7, set r=log2(1/(1-p))r=log2(1/(1-p)) — notice that if p1/2p1/2, then r1r1 — and then take the (natural) logarithm of both sides:

log Prob success = n log 1 - 1 n + 1 r - n ( n + 1 ) r , - α n ( 1 - r ) for n N , log Prob success = n log 1 - 1 n + 1 r - n ( n + 1 ) r , - α n ( 1 - r ) for n N ,
(8)

for any αα that obeys

α N N + 1 r . α N N + 1 r .
(9)

Since r<1r<1, we can safely take α=N/(N+1)α=N/(N+1). Then

Prob success e - α n ( 1 - r ) . Prob success e - α n ( 1 - r ) .
(10)

Using similar arguments, we can also get the lower bound

e - n ( 1 - r ) Prob success e - n ( 1 - r ) Prob success
(11)

The upper bound clearly goes to zero as nn increases. But in practice, the convergence can be quite slow if rr is close to 1 (i.e. pp is close to 1/21/2). The lower bounds gets tight very quickly; for values of pp close to but less than 1/21/2, the lower bound is accurate to within one part in 100 for nn on the order of 15 (see Table 1 below).

To fix some numbers, set p=18/38p=18/38 — this is equivalent to “betting on black” at roulette. Table 1 contains evaluations of Equation 7 for various value of nn (again, we are bringing nn dollars to the table and not leaving until we have doubled it or lost it all).

 n Prob success Prob success lower bound e-n1-re-n1-r upper bound e-αn1-re-αn1-r from Equation 7 from Equation 11 from Equation 10 with N=63N=63 $1 0.4737 0.3679 –$3 0.3779 0.3380 – $7 0.3318 0.3151 –$15 0.3019 0.2947 – $31 0.2786 0.2755 –$63 0.2583 0.2570 0.2625 $127 0.2396 0.2390 0.2444$255 0.2218 0.2216 0.2269 $511 0.2047 0.2047 0.2098$1023 0.1883 0.1882 0.1932 $2047 0.1724 0.1724 0.1772$4095 0.1571 0.1571 0.1618

Moral: If you come to the Roulette table with \$1023, you have a less than 19% chance of doubling your money before you lose it all.

Moral: In all cases, a better strategy is just putting the entire amount on one roll.

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