If you place a one dollar bet on red in roulette you have an 18/38 probability of netting $1 and a 20/38 probability of netting $-1. The expected value of a single play is $-2/38 = $-.053 and the variance is essentially 1 ($^2). If you make this bet n times your expected net winnings is -.053*n with a variance of n and a sd of sqrt(n).
First we use R to simulate n=100 plays:
box <- c(rep(1,18), rep(-1,20))
x <- replicate(10000,sum(sample(box,100,repl=T)))
prob <- sum(x>0)/10000 # to estimate the
# probability of netting a positive amount.
hist(x)
prob
Here are the theoretical values, computed with CLT, of being ahead after n plays.
n -.053n sqrt(n) z* P(W>0)=P(Z>z*)
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50 -2.65 7.07 0.37 .3539
100 -5.3 10.00 0.53 .2981
1000 -53 31.6 1.67 .0469
10,000 530 100 5.30 0
1,000,000 53,000 1000 53.00 0^2
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Here is R code to make these computations.
n <- c(50,100,1000,10000,1000000) sqrtn <- sqrt(n) expval <- -.053*n zstar <- .053*n/sqrt(n) prob <- 1-pnorm(zstar) cbind(n,expval,round(sqrtn,1),round(zstar,2),round(prob,4))Here is the link to roulette information: click here