Rebrained!

You can probably answer this quickly just by thinking about the symmetry of the problem. In the patient approach, you have an equal chance of gaining or of losing $1 at each turn. You are the same number of steps from going bust as you are from doubling your money. So, just as with the bold gambler, your chances of doubling your money will be 50/50.

Before answering this problem it’s useful to read and understand the previous post http://rebrained.com/?p=237.

Let’s call P the probability that we win this game (by reaching $20 before we go bust) and 1-P the probability that we’ll lose this game (by going bust before ever reaching $20). We’ll also call x our starting amount ($10 in this case) and y our target amount ($20 in this case).

Here’s the key insight: if this game kept going even after you’d reached the the target, there would be two ways to eventually go bust – never reaching the target, or reaching the target and then going bust.

Remember from the prior post that from any starting position x, the chances of eventually going bust if we don’t stop at a target amount are ((1-p)/p)^x.

In other words the probability of eventually going bust from our starting position if we didn’t stop after reaching our target (which we get from the above formula) is equal to the probability of eventually going bust without ever reaching our target (which is effectively 1-P, the probability of losing the game), plus the probability of reaching the target (which is P, the probability that we win the game), and then keeping going from the target and eventually going bust from there (which using the above formula we know is ((1-p)/p)^y.

So we can write:
((1-p)/p)^x = (1-P) + P * ((1-p)/p)^y

With some algebra you get the following:
P = (1 – ((1-p)/p)^x)) / (1 – ((1-p)/p)^y))

Something strange happens here when we plug in p=1/2 into the above formula – we get an indeterminate 0/0. To understand what is going on and how to solve this, read L’Hopital’s rule in wikipedia. Using this rule you’ll find that the answer for the p=1/2 case is that P = x/y, or 1/2.

So in the case of the fair coin, as per our argument from symmetry, it makes no difference whether you’re a bold or patient gambler. But what if the coin is weighted?

If p=2/3, the bold gambler has a has a 2/3 change of winning. Using the above formula, the patient gambler’s chances of winning are almost certain – something like 99.9%.

In the case of p=1/3, the bold gambler has a 1/3 chance of winning the game. The patient gambler’s chances of winning are vanishingly small – something like 0.1%.

What about the roulette case? From the above, we can probably suspect that where p is less than 1/2, the patient gambler’s chances of winning plummet much more quickly than the bold gambler’s chances.

In fact, while the bold gambler’s chances of winning are 18/38 – or something like 47% – the patient gambler’s chances of winning are only about 26%.