# Understanding a specific example of Gambler's Ruin

In Wikipedia, if Player 1 has $$n_1$$ pennies, and Player 2 has $$n_2$$ pennies, and they play the game where they flip a fair coin, if it comes up heads, Player 1 gives a penny to Player 2, and if its tails, he receives a penny from the other player, and they play until someone has no more pennies left, the probability of Player 1 winning is $$\frac{n_1}{n_1 + n_2}$$.

I confirmed this by doing the whole process for Gambler's ruin, but I am not sure how it's obvious intuitively. It's written in Wikipedia as if it's something obvious, but I don't see how you can derive this without essentially doing the computation for showing the Gambler's Ruin problem.

It would be great to understand the intuition.

Thank you!

• It's certainly a plausible answer - you need something symmetric when interchanging the players, and why shouldn't my chance of winning be simply the fraction of the total coins I have - but I don't think the Wikipedia article is written as if it's obvious. It's just stating a fact. Mar 28, 2022 at 8:35

Player $$1$$ has an equal probability of $$0.5$$ to either move up by one penny, or down by one penny, $$P(n) = 0.5[P(n+1) +P(n-1)$$],
or $$P(n+1) - P(n) = P(n) - P(n-1)$$,
From the point we are at the start, ruin will occur after $$n_1$$ steps in one direction and $$n_2$$ steps in the other in an ambit of $$(n_1+n_2)$$ steps
And intuitively, it is obvious that the more coins $$A$$ has, the more probable that he wins, thus P(A wins) = $$\dfrac{n_1}{n_1+n_2}$$