Probability Question - Page 3

lol why the fuck are you guys trying to do math

even if there is a .0000000000000000001% of it happening it will happen given the time frame of infinity

All you people who are so confident it's 100% and laughing at others trying to do math have no idea how hard you are owning yourselves.

On September 10 2008 20:34 iamalex wrote: u dont have to work out anything it's 100 percent.

nonononononono, just because there is always a chance of going bust does not necessarily mean that chance tends to 100% as the number of goes tends to infinity even though it will always exist as a nonzero chance.

On September 10 2008 20:49 Pacifist wrote: All you people who are so confident it's 100% and laughing at others trying to do math have no idea how hard you are owning yourselves.

this is simply a random walk question that can be solved using markov chains

to solve it one must assess the probability of going bust from different states. define your state to be the amount of money you have remaining (for simplicity assume you have 1000$ and each round you wager 1$), so:

state 0: BUST
state 1: 1$ remaining
state 2: 2$ remaining
....etc

So in this case we start at state 1000 and each round (independently) we have probability (pr) of winning as 0.51 and losing as 0.49.
The question really is, on average, which state will we be in, given that there is no limit on how much we can win.
i think the answer is non trivial and...i just lost all motivation to do this now....maybe later...but if you are interested in working it out check "gambler's ruin" or "asymmetric random walks" on google

http://en.wikipedia.org/wiki/Gambler's_ruin

The reason is because the longer you go without busting, the larger your roll on average. It certainly is 100% if the size of bad luck necessary to go bust stays the same, but in this system busting becomes over any given interval increasingly LESS likely the further we extend, but busting overall becomes more likely. The question is where the curve representing our probability of busting finds it's least upper bound.

do they serve drinks at this table?

No, but you are immune to dehydration.

On this page: http://www.mathandpoker.com/index.php/?cat=23

Is this graph:



So on this graph our bankroll would be 1000, and our risk or ruin <<<<<<<<1%. There's all the equations there of how it's worked out too if you want the exact number.

OKKKK If you win more than you lose, how the hell can you lose all your money???

Impossible, you continually win, by a small margin.

If you are going to win 51% of hands, once your at 10,000 hands you are suppose to have won about 5,100 and lost 4,900... Thusly you have won 200 hands more than you've lost,x10 per hand won = $2,000.

On September 10 2008 21:21 eightfourO wrote: OKKKK If you win more than you lose, how the hell can you lose all your money??? Impossible, you continually win, by a small margin.

On average, yes; and occasionally mistress variance comes and fucks you and sometimes she will do it really really really hard. Ever played poker?

what? even if there is a .0000001% chance of going busto over the course of infinity it has to occur

what am i doing wrong here

yo crown, there IS a chance of about 0.000000001% of going busto and it IS over infinity, that's the point. AT INFINITY (or mathematically speaking, the limit as it tends to infinity) that is the chance of going busto. It's the limit of an infinite series, similar to the way 1+1/2+1/4+... never reaches 2 but one might say that at infinity it equals two.

I think the question is worded wrong, especially since we can't play infinitely if we go busto.

, which is pretty close to 0.
Now I'm gonna read the thread.

I thought we were betting $1, make that obviously, which is still almost zero.

On September 10 2008 20:43 CrownRoyal wrote: lol why the fuck are you guys trying to do math

Because you're wrong and we're right :D

I can post a proof, but I'll try to convince you intuitively first.

First of all, the probability has to be stricly > 0. If you lose the 1000 first flips (which will happen with probability .49^1000, which is > 0), you're busto. So it can happen and the solution can't be exactly 0%.

Overall you'll be winning money and your bankroll will be growing. The more it grows, the less the chance you can go busto. As the time goes to infinity, so does your expected bankroll. The probability it can reach 0 goes down exponentially with the size of your bankroll.

On September 10 2008 20:43 CrownRoyal wrote: lol why the fuck are you guys trying to do math

LOL indeed

It can of course not be 0%, simple, chances are .49^100 that you'll be broke after 100 hands.

Probability that you have have lost 100 times more than won goes up as you play more hands, but average roll goes up as well, so probability at a certain point may go up or down, i can't tell.
But looking at it now, a small finite probability seems reasonable. Math guys?