Gambling With Rich Friends

Probability Level 4

Yudhisthira is a habitual gambler and gambles at every opportunity he can find. One day, he enters in a casino with 1 rupee in his pocket and starts betting. At each bet, he is likely to win 1 rupee with probability $0.1$ and likely to lose 1 rupee with probability $0.9$ . Yudhisthira also has a super-supportive rich friend, Krishna, who always provides with 1 rupee to keep him betting whenever Yudhisthira's total earning becomes zero.

What is the probability that Yudhisthira will eventually earn 1 million rupees from the gambling?

Note: If $q_t$ denotes the amount of money that Yudhisthira's has after $t$ bets, we have $q_{t+1}= \max(1,q_t+G_t),$ where $G_t$ is an independent random variable where $G_t=+1$ with probability $0.1$ and $G_t=-1$ with probability $0.9$ .

10^{-6}

10^{-10^6}

1 0 None of these

1 solution

Abhishek Sinha
Oct 19, 2016

We will consider a subset of cases where Yudhishira earns 1 million rupees, and show that this tends to 1, hence the probability must be 1. We will simplify the scenario by assuming that Yudhishira continue playing even after he has made 1 million rupees.

The probability that Yudhisthira gains 1 Million rupees in his first million bets is $0.1^{10^6}$ . Let's set $q = 0.1^{10^6}$ .
The probability that Yudhisthira gains 1 Million rupees in his second million bets is also at least $q$ , which happens if he starts only with 1 rupee.
Similarly, the probability that Yudhisthira gains 1 Million rupees in his Xth million bets is also at least $q$ .

Since these events are independent of each other, the probability that he does not make one million in any of the first $n$ million bets is $(1-q)^n$ . As $n \rightarrow \infty$ , since $q > 0$ , we see that $(1-q)^n \rightarrow 0$ . Hence, the probability that he makes one million will approach 1.

Yudhisthira will gain 1 Million rupees eventually.

Summary:

We know that $q = q = 0.1^{10^6}$ . By finding the complementary probablity, $\displaystyle \lim_{n\to\infty} (1-q)^n = 0$ , so the answer is 1 minus this value, which is simply 1.

Great approach of showing that the probability eventually goes to 1, but focusing on such specific outcomes that allow for distinct counting!

While I knew this result, I was surprised to see such a simple explanation for it.

Calvin Lin Staff - 4 years, 7 months ago

On a more advanced level, the probability of this kind of ``tail-events" may be determined using Kolmogorov's zero-one law , after arguing that the event happens with at least a strictly positive probability (however small).

Abhishek Sinha - 4 years, 7 months ago

Right. I love the 0-1 law.

Calvin Lin Staff - 4 years, 7 months ago

I'd rephrase it: instead of considering the amount won, just consider the number of wins. At the moment, you need to show that the probability of winning 1 million rupees if Yudhisthira starts with more rupees is also at least $q$ . Intuitively it's certainly true, but you still need to prove it. It's easier to just say "the probability that Yudhisthira wins all the second million games is $q$ ", and later say that winning any consecutive million games means Yudhisthira gains 1 million rupees at some point.

Ivan Koswara - 4 years, 6 months ago

Gambling With Rich Friends

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