# Going from 30 to 100 in a coin flip game.

The question is this:

Say you're playing a coin flip game, where you start with 30 dollars. If you flip heads, you win 1. If you get tails, you lose 1. You keep playing until you either run out of money or reach 100. What is the probability that you will reach 100. The coin is fair.

I have seen many questions about expected number steps, expected earnings and so on. However I am having trouble with this question because the game need not even end for certain (I think). In such situation, how would we either get a probability, or explain why it can’t be arrived at, if that is the case?

• The probability that you double your money $n$ times is $(1/2)^n$, and the probability that you reach 3 from 1 is $1/3$. I'm not sure how to prove it, but I would guess the probability that you reach $a$ times your starting money is $1/a$, or 3/10 in your case. Dec 5 '21 at 16:22
• @Angelica - that is assuming a fair coin, and is a result first proved by Christiaan Huygens now called "Gambler's Ruin" Dec 5 '21 at 16:24
• The probability of the game eventually ending is $1$ either with a fair or an unfair coin. But there is no upper bound on the number of steps Dec 5 '21 at 16:26

Since the coin is assumed to be fair, the expectation value of the game is zero. Now the expectation value is given by $$E = P*70 + (1-P)*-30$$, where $$P$$ is the probability of winning the game. Setting $$E$$ equal to zero, we can solve for $$P$$ with the result: $$P = 0.3$$.
• @Aniruddh I suggest that you examine the case where you start with 2 coins. You lose when you have zero coins, you win when you have 4 coins. After 2 turns the probability that you have 2 coins is 1/2. After 4 turns it is 1/4 etc. This shows that the probability that the game has not finished after $N$ moves decreases exponentially. Jan 10 at 5:31