So I am in a casino betting on Black in an American Roulette wheel.
Scenario1: I bet for successive 1000 bets starting with a bet of 1 dollar.
If I loose, I bet the double of amount that I bet earlier and if I win, I reset the betting amount to 1 dollar. I also stop betting once I reach 80 dollars and keep my winning as 80 dollars. I also assume that I have infinite amount of bank roll. So I never run out of money to bet.
Scenario2: Same as Scenario 1 but I dont have an infinite bank roll. I have a finite bank roll of 256 dollars. And also I quit once I loose all 256 dollars.
I am trying to calculate the probability of me winning 80 dollars in 1000 successive bets and also the estimated expected value of my winnings after 1000 spins.
I wrote python code to simulate scenario 1 and empherically I never came back without winning 80 dollars. So can I say that the probability of me winning 80 dollars in 1000 spins is 100%?
Similarly, I ran simulation for scenario 2 and found that I lost 38 among 100 "1000 successive spins".
I am trying to calculate the probability mathematically and not able to do so.
Regarding expected value, for 1 spin, my expected value is -0.06 dollars. Is it right to multiply it by 1000 to calculate the expected value for 1000 spins?
Thanks in advance!!
Best Answer
The expected value of a $1$ dollar bet on black on an American roulette wheel is $\frac{18}{19}-1 = -\frac{1}{19} \approx -0.05263$ dollars (a little better than $-0.06$). If you were to bet $1$ dollar on each of $1000$ spins, your expected gain would be approximately $1000 \times -0.05263 = -52.63.$ But you are not betting $1$ dollar on each of $1000$ spins in either scenario; you are betting varying amounts of money on (usually) fewer than $1000$ spins. What's more, the amount bet depends on the previous spins.
The chance of winning seems a little easier to compute, though still not easy.
In Scenario 1, since you start with an unlimited bankroll, we cannot measure your progress by subtracting your initial bankroll from your final bankroll. So let's just say you start with zero net gain and you are allowed to let your net gain go as negative as you need in order to keep playing.
During a long run of red and green spins, your net gain may go very negative indeed. But every time black wins, your net gain becomes equal to the number of times black has won so far.
Moreover, your strategy is that if your net gain ever reaches $80$ you stop playing.
So the only way that you will not win $80$ dollars within $1000$ spins is if black wins fewer than $80$ times in $1000$ spins. The probability that this happens is very, very small, but it is not zero.
The probability is so small, in fact, I did not manage to get Wolfram Alpha to give me an answer. But the probability of black winning exactly $79$ times is approximately $2.020 \times 10^{-164},$ the probability of black winning exactly $78$ times is approximately $1.923 \times 10^{-165},$ and the probability for each smaller number of wins is progressively much smaller even than that. I think we can safely say that the probability of losing is between $2.2\times 10^{-164}$ and $2.3\times 10^{-164}.$ You could make this a little more exact by adding in some of the other $78$ losing cases.
This is not exactly zero, so your probability of winning is not exactly $1.$ But the probability is so close to $1$ that you are almost certainly not going to simulate any losing case.
In Scenario 2, you have (almost) an asymmetric random walk with two absorbing states, also known as gambler's ruin. The state is the size of your bankroll, the initial state is $256,$ and the absorbing stats are $0$ and $336.$ I say that this is almost that problem, because the usual random walk does not limit the number of steps that may be taken, whereas you limit your problem to $1000$ steps.