No Formulas or Advanced Math

You have three dollars. Your opponent has infinite wealth. You flip an unfair coin for a dollar a pop. Thc coin is weighted two to one in your favor. What is the probability that you will ever go broke if you play forever?

It's a certainty you will go broke in an infinite contest.

I would say 100%.

Well if you're going to play forever you can't ever go broke [img]/images/smile.gif[/img] I suspect what you mean (though one is never quite sure with you) is what is the chance that you will go broke sometime before you play forever.


Suppose instead of 3 dollars we only had 1 dollar. Then there would be a 1/3 chance that we go broke on the first flip. There would be a 2/3 chance that we win the first flip and double our bankroll to 2 dollars. If that happens, the chance of going broke from that point on would be whatever it was when we had 1 dollar squared since now we have to lose 1 dollar twice in order to go broke. So if p is the chance we will lose 1 dollar, we have


p = 1/3 + 2/3(p)^2


Or multiplying both sides by 3 and rearranging:


2(p)^2 -3p + 1 = 0


factoring gives


(2p - 1)(p - 1) = 0


So p = 1/2 or p = 0.


We reject p=0 since we already said there was a 1/3 chance of losing it on the first flip.


So there is a 1/2 chance of losing a 1 dollar bankroll. The chance of losing a 3 dollar bankroll then is (1/2)^3 = 1/8 since we have to lose the 1 dollar bankroll 3 times.

100%

*(2p - 1)(p - 1) = 0


So p = 1/2 or p = 0. *


Wrong ! ... otherwise it looks GOOD !


You even got the semantics right.


But what about * No Formulas or Advanced Math * ;-)

100% - you'll never win forever

Why would you necessarily go broke in the long run? I would agree with all of you if this was a -EV game, but it isn't. If the answer is 100%, every blackjack card counter, +EV VP player, and sports better should give up and go home. There is a certain risk of ruin involved in this (and a high one with such a small bankroll), but I just can't see how you will inevitably go broke 100% of the time. All comments (read as: flames) are welcome.


-- Homer

Yeah, I saw that after I posted and had to think about how to dismiss p=1.


There cannot be a probability of 1 for going broke after N flips for any N no matter how large. This is because there is always a non-zero probability that we do not go broke after any N flips. For example, it is always possible to win all N times with probability (2/3)^N. In fact, as long as our number of losses does not exceed our number of wins by 3, we will not go broke, and this will have a non-zero probability for any number of flips. So we can't have p=1 or else the probability of going broke plus not going broke would be greater than 1 which is nonsense. It would also imply that our opponent with a negative expectation would have a positive expectation of 3 dollars after N flips which we know can't be right.


So you can make as many flips as you please, and the probability of losing our last dollar will never exceed 1/2, and the probability of losing 3 dollars is 1/8.

But if you have to play _forever_ sooner or later you'll have a swing that is big enough to eat your entire bankroll... I think...


Sincerely, Andreas