Probability Question

Let's say you're playing blackjack against the house. Each game you play, you have a 51% chance of winning. The house has a 49% chance of winning. Each game you must bet $10 (you lose $10 every time you lose a game and win $10 every time you win a game) and your starting bankroll is $10,000. The house has infinite money, and you must keep playing forever unless you bust. That is, you can't ever quit the game unless you go busto.

What's the probability that you will go busto playing this game?

miniscule

100%?

^brain fart of an answer by the way... i need to think more

cool question

100% ez

Obviously 100%. A better question might be, how many hands maximize your EV in such a case? Obviously playing just 1 hand is +EV, and 2 hands even more so, but playing unlimited hands will make you go bust. I don't know what the answer is here.

100% came to mind first, but i'm not sure about it.

You need to lose 100 times more than you win to go broke. It's extremely unlikely this will happen in the first 100 or 200 times. However, when you play an infinite amount of times, you will have won 51% as the infinity cancels out all variance. But in between those 2 extremes there may be a moment where it does happen. Im not good enough at limit poker math...

playing infinite hands u will always go busto, but also to maximize your ev you would also never stop playing

the answer has to be never or always

that said im leaning more towards never

you would have to run really really bad to go busto but its still possible fuck

i think it has to be 100% doesn't it? that seems so fucking wrong though, awesome question

isn't it effectively 0? you have a slightly better than coinflip chance of winning, and you need to lose 1000 times in a row to go bust. On a whim I'd say (0.49)^1000? I'm not sure if thats right or not..

make that 1-(0.49^1000)

the beauty of infinity is that you will at some point have a run bad enough to go busto, no matter if you start with 100000000000000000000000000000 x the bet

i'd say an incredibly small nonzero number. your expectation is always positive, so the longer you play the further you will be away from busto, requiring that much worse of a downswing to busto you. id say like 1/10^500 or something.

its a mindfuck because all things either can or cannot happen and although in the course of infinity hands we are maximizing our value we are at the same time putting ourselves at the definite risk that we will go busto because it has to happen over the course of infinity just because it can.

sorta like the monkeys with a typewriter writing shakespeare theory

If you run it out long enough you will always go bust.

so this means its theoretically impossible to win at poker, investing, or anything really amirite

Based on the info you provided, you would never go busto.

On September 10 2008 18:26 CrownRoyal wrote: so this means its theoretically impossible to win at poker, investing, or anything really amirite

sick point

like .000000000000000000000000000001% chance to go bust. Heres my logic...go google percentages of hands, then go check pt. You're suppose to get pocket aces about X% of times, so take the # of times you got aces, put it over the total amount of times, and it should be within a very small margin of the odds of getting it.... If you 51% of the time, the theory of averages amd all that say's you will always win more than the house does. Unfortunately, the house has better odds than you IRL, but it is the most profitable game to play at a casino (as in you have the best odds compared to slots/roulette/craps etc...)

In conclusion, check PT to confirm the actual probability of getting cards to prove that probabilty is basically always right. With that said, if you have 51% odds, you will win more than the house and will not go busto.

i figured it out

RAKE WINS!

umm ok i think that is has to be a function it cant be a single number beacuase u increase or decrease ur total money with every hand. the base number would be the probability of loosing 51-49 1,000 in a row times that would be (x), but tyhats just a theory cuz im REEEEEALLY bad at math especially anything higher than trigonometry. ask Daut he should know

my point is if the probability of going broke varies like losing 1000 time in a row is lets say 1% the number will increase to go broke once u lose 1 time making it lets say 1.001% but if u win the hand it would decrease making it 0.999999%

hmm so taking this is mind then teh answer would be probability of losing = [49% (cuz of teh last hand u will play when u have 10 bucks left) > probability of losing 49-51 1000 times in a row > infinite]

ship teh prize?

hmm..i think this is a markov chain question

so simplified (49% > a > infinite) > 0

where a = prob of losing 1k 49-51 flips in a row


edited cuz forgot to put 0

lol now im hyped i wanna know if im right some1 tell me the answer wiiiii

if there is any chance, no matter how small, over an infinite number of trials you will go busto.

well yes because u have to keep playing always, your purpose is not to win money is just to bet 10 dollars over an infinite if your purpose was to make it profitable you would take out the original invesment after u have profited a decent amount.

Its the same thing we do with the martin gale if we had an infinite BR we could keep doubling our bet and eventually we will win even if our flip is 1 - 99 in favor of the house, hence they house puts a cap or maximum bet

Applying the concept of percent to this game will yield counterintuitive results since there are infinite possible game outcomes. The problem is that there are infinitely many possible games in which you bust and infinitely many in which you never bust (it is trivial to construct a scenario in which you never bust - simply hit your 51% every time). To try to divide this infinite set into parts that can be meaningfully compared to one another is problematic.

If you wish to argue either 100% or 0% is the answer, then I submit that by the same argument you can conclude that either 0% of all integers or 100% of all integers contain the digit 3.

it can never be greater than 49% cuz thats the probability of losing the hand that would declare you busto it can never be 0% because theres always a chance to run extremly bad and go busto. but the point remains if the purpose is to play infinite u will indeed go busto if the purpose is to make a profit then u have probability of losing 1k flips in a row of going busto.... AHH my heard hurts to much thinking

That's not to say that an argument that arrives at 0% busto is wrong. I think 0% is actually the correct answer here. It's simply a consequence of trying to apply concepts used for regular sets on infinite ones. If you want to try to use percentages when talking about a set such as the integers, the result you obtain is actually that 100% of all integers contain the digit 3.

On September 10 2008 18:26 CrownRoyal wrote: so this means its theoretically possibly impossible to win at poker, investing, or anything really amirite

On September 10 2008 18:26 CrownRoyal wrote: so this means its theoretically impossible to win at poker, investing, or anything really amirite

if you don't win someone else does, the house in this case.

On September 10 2008 19:29 Jorge wrote: it can never be greater than 49% cuz thats the probability of losing the hand that would declare you busto it can never be 0% because theres always a chance to run extremly bad and go busto. but the point remains if the purpose is to play infinite u will indeed go busto if the purpose is to make a profit then u have probability of losing 1k flips in a row of going busto.... AHH my heard hurts to much thinking

No it's the other way around. The probability of going bust is AT LEAST and definitely greater than 49% since that's the chance we go busto on the first hand. Since we can also go bust on the 3rd hand etc we see that it is definitely represented by an increasing series with decreasing extra terms in the sequence of partial sums. The question is whether it has a limit. I've been working on it and I'm fucked if I know. It's a great question though.

edit: that's a limit <100%

wait what? u can never go bust on the first 999 hands cuz " Each game you must bet $10 "

edit: its the chance of loosing all ur moneys not the hand sir

oh whoops sorry I missed that, I was assuming starting from $10. The principle of the theory stays the same for $10,000 I imagine but it just gets a whole lot uglier.

yo it's a standard risk of ruin calculation. the winrate is easy, it's $0.10/hand, the bankroll is $10k and I'm too tired to work out how to work out the standard deviation and dig up the equation; but that's all the variables I believe...

u dont have to work out anything it's 100 percent.

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?

On September 10 2008 18:49 CrownRoyal wrote: i figured it out RAKE WINS!

im with crown lol

My English isn't refined for this stuff but if you have maths background hopefully you can dechiper what I am trying to type and verofy / fix it correctly.
The outcomes follow binomial spread (or whatever it translates to in English ) since each deal is independant with a predefined probability for both winning / losing. In order to bust you have to lose 1000 more than winning, aka in n attempts you win n/2-500 and lose n/2+500.

The probability in "n" attempts is ( n! / ((n/2-500)!) * (n/2+500)!) ) * (0.51^(n/2-500)) * (0.49^(n/2+500))

I hope I typed in all neccessary ( ). I don't have a scanner atm to make it simplier.

this

On September 10 2008 20:56 killThemDonks wrote: Show nested quote + 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

and this

On September 10 2008 21:16 Oly wrote: 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.

are correct

On September 10 2008 23:52 Catul wrote: Show nested quote + 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.

it's so clear to me now

probability < infinity
you have to play until you bust, so you will bust, % dont matter as long as they are not 0, or 100%.
Taking large sample sizes to proof a point doesnt make sense because you never get close to infinity.
Even if your a 99 % favorite every time you still will go broke vs. infinity

On September 11 2008 07:21 zodion wrote: probability < infinity you have to play until you bust, so you will bust, % dont matter as long as they are not 0, or 100%. Taking large sample sizes to proof a point doesnt make sense because you never get close to infinity. Even if your a 99 % favorite every time you still will go broke vs. infinity

i cant believe this. i know what u are saying but i dont think just cos it goes for infinity, means u will always bust

On September 10 2008 23:52 Catul wrote: Show nested quote + 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.

Well, this analysis has so many holes it's not even worth mentioning them all (just one - you can go busto in more then one way with more than 1000 flips), and you're pulling numbers out of your ass. It's more complicated than this.

The random walk is a great idea, however fact that we stop playing when we go busto makes this bit complicated, as it affects all states probabilities, except for most optimistic 1000, for sample sizes greater than starting br, and probability of going busto may start growing, with no visible limit (I'm not even sure IF it's growing, or just going down slower)

But you know that actually the house has an edge on you at blackjack and not the opposite right ? Unless you count cards and they only use 1 deck.

sheitan ive seen 21 the movie and rainman and you're wrong they won lots of monies

catul why do we have to lose consecutively? 1000 times wtf? there are so many other ways to lose

On September 11 2008 07:33 GsOne wrote: Show nested quote + On September 10 2008 23:52 Catul wrote:   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. Well, this analysis has so many holes it's not even worth mentioning them all (just one - you can go busto in more then one way with more than 1000 flips), and you're pulling numbers out of your ass. It's more complicated than this. The random walk is a great idea, however fact that we stop playing when we go busto makes this bit complicated, as it affects all states probabilities, except for most optimistic 1000, for sample sizes greater than starting br, and probability of going busto may start growing, with no visible limit (I'm not even sure IF it's growing, or just going down slower)

lol wtf I'm trying to dumb it down and even saying I'm doing so and you're telling me this has many holes and that it's more complicated than this ?

you can go busto in more then one way with more than 1000 flips

Irrelevant, reread my post. What I said about losing the first 1000 flips only proves you have a > 0 chance to go bust since I showed one possibility.

I don't know where in my message I'm pulling numbers out of my ass. My answer above ( (49/51)^1000 ) is the exact answer, and I'd have bothered to write a proof if killThemDonks and Oly hadn't already provided links to it. The random walk isn't a "great idea", it's the fucking solution.

Now that I've read the links they gave, there is no complete proof (although it's strongly hinted at in the wikipedia article), so here's one. I've put it in spoiler because it takes quite some place. It's nothing difficult though and I've detailed every step.

+ Show Spoiler +

You start with a bankroll $x.
You play until your capital reaches N or 0 (those would be absorbing states in Markov chains lingo). We'll solve for the infinity case later.
Each time you play, you have a probability p of winning $1 and a probabilty q of losing $1, with q=1-p.
Let be the probability that you go busto before reaching N given you start with $k. We're looking to find the general form of

We have and .
If your roll is at $k and you play once, with probability p you're going to be at $k+1 and with probability q at $k-1. In the first case, the probability of eventually busting is and in the second case . We can write :
, for .

We can rewrite this as :
(because p+q=1)





By replacing the right term using the same equation until you reach it is easy to see that gives



Now we're going to sum the left term for all k from N to 0 :





For the special case p=q=1/2 this gives



For p not equal to q, we get

(the barely readable thing in ( ) is q/p)





Replacing with what we obtained for p1-p0 above :



And finally



For p=q, it's very easy to find the same way




Now, if we look at the probability of reaching N before going busto, we can do it the same, but going towards instead of and we find



and for p = q = 1/2



Note that it was not (completely) obvious that . As it is the case, we have proven the game ends with probability 1 (we reach either N or 0 at some point or another).


Now, back to our original problem. We play against the house who has infinite money. We find the probability of busting by letting N go to infinity in our expression above. 3 cases appear :

1. q > p : that means we have more chance to lose a flip than win a flip





When the house has an edge and infinity money, we busto with probability 1 (no kidding).

2. q = p : we're flipping 50/50





Even when the house has no edge, as long as it has infininte money and we don't have an edge either, we busto with probability 1 (no shock either).

3. q < p : we have an edge

There we go. This defines the probability that was asked in the original post :



where
is the probability of busting starting with a bankroll of $x and playing for $1 everytime
is the probability of winning each time
is the probability of losing each time (p+q=1)

On September 11 2008 10:00 CrownRoyal wrote: catul why do we have to lose consecutively? 1000 times wtf? there are so many other ways to lose

Doesn't anybody read what I write or something ?
I said we could lose 1000 times in a row to prove that we have at least some chance of busting. It can't be 0.

wtf is that. i don't even want to quote it. or read it.

holy crap Catul i have no idea what any of that means but i believe you

I put it in spoiler now, it was taking way too much space.

On September 11 2008 11:05 Catul wrote: Show nested quote + On September 11 2008 10:00 CrownRoyal wrote: catul why do we have to lose consecutively? 1000 times wtf? there are so many other ways to lose Doesn't anybody read what I write or something ? I said we could lose 1000 times in a row to prove that we have at least some chance of busting. It can't be 0.

oh wtf i thought you was disagreeing with me and saying that we could never go busto

and your proof is so ridiculous i don't have any ambition to look at it or understand it sorry.

seriously where do you learn to write walls of math problems like that, im pretty sure if you was just bullshitting that no one would call your bluff its so impressive

anyone who disagrees with this extremely basic markov chain solution put forth by catul has just never received the proper mathematical education to even understand the original problem anyway.

there's a good reason we don't hire people like [insert whoever still thinks you're 100% to go busto] to teach courses in higher math.

i immagine the number has to be very close 0 since we have positive expectation on every investment of 20 cents for every $10, obviously as N (N being number of intervals) approaches infinity so will our bankroll as craytoul described

i stand corrected.

so do you always go broke if the odds are 50/50?

On September 11 2008 11:03 Catul wrote: Show nested quote + On September 11 2008 07:33 GsOne wrote:   On September 10 2008 23:52 Catul wrote:   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. Well, this analysis has so many holes it's not even worth mentioning them all (just one - you can go busto in more then one way with more than 1000 flips), and you're pulling numbers out of your ass. It's more complicated than this. The random walk is a great idea, however fact that we stop playing when we go busto makes this bit complicated, as it affects all states probabilities, except for most optimistic 1000, for sample sizes greater than starting br, and probability of going busto may start growing, with no visible limit (I'm not even sure IF it's growing, or just going down slower) lol wtf I'm trying to dumb it down and even saying I'm doing so and you're telling me this has many holes and that it's more complicated than this ?   you can go busto in more then one way with more than 1000 flips Irrelevant, reread my post. What I said about losing the first 1000 flips only proves you have a > 0 chance to go bust since I showed one possibility. I don't know where in my message I'm pulling numbers out of my ass. My answer above ( (49/51)^1000 ) is the exact answer, and I'd have bothered to write a proof if killThemDonks and Oly hadn't already provided links to it. The random walk isn't a "great idea", it's the fucking solution. Now that I've read the links they gave, there is no complete proof (although it's strongly hinted at in the wikipedia article), so here's one. I've put it in spoiler because it takes quite some place. It's nothing difficult though and I've detailed every step. + Show Spoiler + You start with a bankroll $x. You play until your capital reaches N or 0 (those would be absorbing states in Markov chains lingo). We'll solve for the infinity case later. Each time you play, you have a probability p of winning $1 and a probabilty q of losing $1, with q=1-p. Let be the probability that you go busto before reaching N given you start with $k. We're looking to find the general form of We have and . If your roll is at $k and you play once, with probability p you're going to be at $k+1 and with probability q at $k-1. In the first case, the probability of eventually busting is and in the second case . We can write : , for . We can rewrite this as : (because p+q=1) By replacing the right term using the same equation until you reach it is easy to see that gives Now we're going to sum the left term for all k from N to 0 : For the special case p=q=1/2 this gives For p not equal to q, we get (the barely readable thing in ( ) is q/p) Replacing with what we obtained for p1-p0 above : And finally For p=q, it's very easy to find the same way Now, if we look at the probability of reaching N before going busto, we can do it the same, but going towards instead of and we find and for p = q = 1/2 Note that it was not (completely) obvious that . As it is the case, we have proven the game ends with probability 1 (we reach either N or 0 at some point or another). Now, back to our original problem. We play against the house who has infinite money. We find the probability of busting by letting N go to infinity in our expression above. 3 cases appear : 1. q > p : that means we have more chance to lose a flip than win a flip When the house has an edge and infinity money, we busto with probability 1 (no kidding). 2. q = p : we're flipping 50/50 Even when the house has no edge, as long as it has infininte money and we don't have an edge either, we busto with probability 1 (no shock either). 3. q < p : we have an edge There we go. This defines the probability that was asked in the original post : where is the probability of busting starting with a bankroll of $x and playing for $1 everytime is the probability of winning each time is the probability of losing each time (p+q=1)

On September 11 2008 16:43 iamalex wrote: so do you always go broke if the odds are 50/50?

nvm figured it out

I believe the answer is the result of some power series that converges to a very small number

thanks for the proof, 'pulling number out of your ass' was about stating that the answer is (q/p)^x, and since you were talking about probability of going busto in x consecutive flips I somehow menaged to confuse q^x and (q/p)^x. Also I had no idea the answer will be so good looking.

thanks for straighting this out.

On September 10 2008 18:44 collegesucks wrote: Show nested quote + On September 10 2008 18:26 CrownRoyal wrote: so this means its theoretically impossible to win at poker, investing, or anything really amirite sick point

You can quit poker.

So in simple words it is a profitable game where you can go broke but on the long term your chances to go broke decrease ?

On September 12 2008 01:31 KwarK wrote: Show nested quote + On September 10 2008 18:44 collegesucks wrote:   On September 10 2008 18:26 CrownRoyal wrote: so this means its theoretically impossible to win at poker, investing, or anything really amirite sick point You can quit poker.

no i can't. fuck my life and fuck this game... T__T