Intuition behind fair vs biased result for Gambler's Ruin problem

Question

The probability of success for the Gambler's ruin problem, where gambler A starts with \$ $i$, is given by $$P_i= \begin{cases} \frac{1-\left(\frac{p}{1-p}\right)^i}{1-\left(\frac{p}{1-p}\right)^N}, & \text{p ≠ 0.5}\\ \frac{i}{N} & \text{p=0.5} \end{cases} $$ Plugging in values N = 100, i = 10, this results in $P_i = 0.1$ for a fair game and $P_i = 0.00917625$ for a biased game with $p = 0.49$. Similar large differences in probability are seen across several values of $i, N$ and $p<0.5$, such as can be found here. I understand the derivation of the results here, but am having a difficult time grasping intuitively why this difference is so large. Could someone please explain, intuitively, why a slight bias can make such a big difference?

Answer 1

Because the game is long, the chance the player loses in the biased case becomes very large. The average number of plays is $Ni=1000$. We can use the normal approximation to say the expected value is then $-10$ so the player ought to lose. The standard deviation of the number of wins is $\sqrt{1000 \cdot 0.49 \cdot 0.51} \approx 15.8$ so winning so fast is over $6\sigma$ high. It takes very little bias to be sure the favored party will win as long as you play enough times. Ask the casinos, who play craps with a house edge of only $0.5\%$ and still make money.