Suppose that you place a single bet on red in roulette. Then the probability of not losing money is \(18/38=\) 0.4736842. If you place two consecutive bets, the probability of not losing money is

(18/38)^2+2*(18/38)*(20/38)
## [1] 0.7229917

If you place three bets:

(18/38)^3+3*(18/38)^2*(20/38)
## [1] 0.4605628

We can write a general formula for the probability of not losing money after \(n\) bets by reasoning as follows: in order to come out ahead over \(n\) bets, one must win at least \(n/2\) times. Therefore, the probability of not losing money is \[ \sum_{x = \lceil n/2 \rceil}^n \binom{n}{x}(18/38)^x(20/38)^{n-x}. \] While this is difficult to compute by hand, we can use R’s built in Binomial Cdf calculator by equating the above expression with \[ P(X\geq \lceil n/2 \rceil) \] where \(X\) is Binomial with success probability \(p=18/38\). Since the cdf of a random variable is defined in terms of the probability of being less than or equal to some value, we must rewrite the desired probability as \[ P(X\geq \lceil n/2 \rceil)=1-P(X \leq \lceil n/2\rceil -1). \] If \(n=50\), this comes out as

1-pbinom(24,50,18/38)
## [1] 0.4078581

The Central Limit Theorem provides a quick alternative to this computation by allowing us to estimate the above probability using normal approximation. To motivate this idea, the plot below shows the distibution (pmf) of \(W\) when \(n=50\):

plot(seq(-50,50,by=2),dbinom(0:50,50,18/38),xlab="W",ylab="Probability")

Basically a normal curve! The Central Limit Theorem is the mathematical fact which guarantees that for large \(n\), the distribution of any sum or average of independent and identically distributed random variables can be well approximated by a normal distribution with the “appropriate” expected value and standard deviation. For example, in our roulette example, the expected value of our winnings \(W_n\) after \(n\) plays is is \[ E(W_n) = -n/19 \] and the standard deviation is \[ D(W_n) = \sqrt{\frac{360n}{361}}. \] Now we use the normal cdf to estimate \(P(W\geq 0)\) by

1-pnorm(0,-50/19,sqrt(50*360/361))
## [1] 0.3546941

This isn’t too far off from the exact Binomial answer, but we can do slightly better if we use a continuity correction:

1-pnorm(-1/2,-50/19,sqrt(50*360/361))
## [1] 0.3813759

The graph below compares the continuity correction probabilities of not losing money as a function of \(n\) with the actual values

x=seq(50,1000,by=50)
plot(x,1-pnorm(-1/2,-x/19,sqrt(x*360/361)),ylim=c(0,0.4),ylab="Probability")
points(x,1-pbinom(floor(x/2-1),x,18/38),pch=2)
legend("topright",legend=c("Normal","Binomial"),pch=c(1,2))

Notice that the estimate gets better and better as \(n\) increases. Notice also that you are very unlikely to come out even or ahead after 1000 rounds of roulette so you better enjoy the free drinks!