Jtdylktuy

A random tangent I have been thinking about, please be gentle I am not a mathematician:

Imagine a coin toss where Heads means we win a coin and Tails means we lose a coin. We start with 1 coin and cannot go negative (i.e. if we flip a Tails on the first try, we lose and it's game over).

The coin is weighted so that it flips Tails with a probability of a. If a = 0, the probability of bankruptcy is 0%. At some value between 0 and 1, the probability of bankruptcy approaches 100%. For what value of a does this probability cross 50%?

So I drew a graph of the outcomes for three flips of a coin with a = 0.4:

       1

(0.4) /  (0.6)

     0   2

  (0.4) /  (0.6)

       1   3

      /  / 

     0   2   4



First Flip: P(0) = 40%, P(2) = 60%

Second Flip: P(0) = 40%, P(1) = 24%, P(3) = 36%

Third Flip: P(0) = 49.6%, P(2) = 28.8%, P(4) = 21.6%

The first thing I noticed is the probability of bankruptcy on evened-numbered flips (e.g. flip #2, flip #4, ...) does not increase. So you have a 40% of busting on the first hand, the probability remains unchanged on the second flip.

I attempted to try and solve this problem in a closed form solution shown below (sorry it's a picture, I don't know LaTeX...).

For a = 1/2, I got a P(0) = 2/3. Obviously this is wrong because for a equally weighted coin, we would expect that playing infinitely long, you would succumb to gambler's ruin pretty quick.

So again restating the original question: How weighted does the coin have to be in our favor that after playing for an arbitrarily long amount of time, our odds of bankruptcy approaches 50%? Is this a problem that can be solved in closed form? Also if you can point out the fallacies in my attempt, I greatly appreciate it.

Thanks in advance!

Here's a combinatorics way of doing the problem. Suppose $p$ is the probability of losing on any given flip.

Then the total probability of ruin is given by
$$sum_{n=0}^{infty}C_np^{n+1}(1-p)^n$$
where $C_n$ counts the number of sequences of wins and losses on flips such that you lose your last coin on exactly turn $2n+1$.

This sort of restricted random walk shows up everywhere in combinatorics and the count $C_n$ is known as the Catalan numbers, with formulas such as $C_n=frac{1}{2n+1}binom{2n}{n}=binom{2n}{n}-binom{2n}{n+1}$.

In particular (see this link for a proof), we know the generating function for the Catalan numbers, that is to say
$$g_C(x)=sum_{n=0}^{infty}C_nx^n=frac{1-sqrt{1-4x}}{2x}text{.}$$

Now we can just simplify the probability of ruin as
$$sum_{n=0}^{infty}C_np^{n+1}(1-p)^n=pg_Cbigl(p(1-p)bigr)=pfrac{1-sqrt{1-4p(1-p)}}{2p(1-p)}=frac{1-left|2p-1right|}{2(1-p)}=begin{cases}1&pgeqfrac{1}{2}\frac{p}{1-p}&pleqfrac{1}{2}end{cases}text{,}$$
where gambler's ruin manifests in the absolute value.

If specifically solving for $Pr($ruin$)=frac{1}{2}$, $p=frac{1}{3}$.