Assume that a gambler plays a fair game where he can win or lose 1 dollar in each round . His initial stock is 200 dollar. He decides a priory to stop gambling at the moment when he either has 500 dollars or 0 dollars in his stock. Time is counted by the number of rounds played.
i) show that the probability that he will never stops gambling is zero
ii ) Compute the probability that at the time when he stops gambling he has 500 dollars , and the probability that he has zero dollars
Assume that a gambler plays a fair game where he can win or lose 1 dollar in each round . His initial stock is 200 dollar. He decides a priory to stop gambling at the moment when he either has 500 dollars or 0 dollars in his stock. Time is counted by the number of rounds played.
i) show that the probability that he will never stops gambling is zero
ii ) Compute the probability that at the time when he stops gambling he has 500 dollars , and the probability that he has zero dollars
Although I agree with whosenext, these questions can be answered even without any previous knowledge about random walks and risk of ruin, but just with some logic.
Spoiler:
1) Here you basically have to show that you surely reach one of the two thresholds (+300 or -200 assuming that you start at 0). You can reason as follow.
- Define r as the probability of ever reaching +1.
- The probability of ever reaching n (n>1) is r^n, since in order to reach n, you must reach n-1 and you have a probability of r to reach n from n-1 and so on. In other words, you reach +1 with probability r; once in +1 with probability r you reach +2 (so r^2 to reach +2 from 0), and again with r we reach +3 (r^3 from 0) and so on.
- But we have:
r = 1/2 + 1/2 * r^2
since we start at 0: with p=1/2 with reach immediately +1 while with p=1/2 we reach -1 and that means that we have r^2 to reach +1 from there.
The equation above shows two identical solutions, both at r=1. This means that, sooner or later, we certainly will reach +1. In turn, this means that we also will reach any n, since r^n = 1.
The same logic can be applied for the -300 threshold of course. So we will reach either one of the threshold with certainty.
2) Since the game is fair, any combination of bets will have 0 EV. On the other hand, as we shown before, you either win or lose (no chance you'll play forever). Say that P_W is the probability of winning 300 and P_L the probability of losing 200. We must have:
P_W * 300 - P_L * 200 = 0 (since the game is fair)
P_W + P_L = 1
