In the long run...you'll never win!

[Felipe]

<ul type="square"> I read this article about Gambler's Ruin (http://en.wikipedia.org/wiki/Gambler%27s_ruin) last night while browsing the web. Where it begins "Coin Flipping" is where I begin.

In short, it says that if there are two players playing a game with even odds(1:1), and they could play an infinite amount of games, one of them would eventually lose all of his money.

"Even with equal odds, the longer one gambles, the greater the chance that the player starting out with the most pennies wins."

The important clause in the article is, I think, that one of the players must have more money than the other. I assume that since they are playing an infinite amount of games, the possibility of one player busting out exists. It is greater than 0.

It follows that the player that starts with fewest pennies is most likely to fail."

But how can this work practically? I mean, does it really matter than the one player with more money will eventually win all the money in the long run? It seems counter-intuitive and wrong, but perhaps it has to do with playing infinite number of games, which is realistically impossible.

What does this really mean? Does this have significance for a player that gets better than even odds many times in a session? Like a good poker player?

But what if we gave both players in equal amounts of money? Does the "gambler's ruin" hold true in this case? Or does it all go to shits?

Frankly, I'm having trouble agreeing with the article. Playing with finite money and finite events, it doesn't seem right (or relevant?) that on even odds (1:1) the player starting with the most money will eventually win it all "in the long run," and the player with the least will lose it all.
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Sounds pretty dumb to me unless someone has a short BR and busts out due to variance. How can a 50/50 gamble mean that anyone is going to eventually lose otherwise???

Since it's a "coin" flip I guess the bet is only the value of the coin? Naturally if the bet size varies it changes everything. I just wanted to make sure.

If player A starts with a very small amount more than B, it seems to me a run of wins by B can change the stack sizes.

As the starting stack sizes become more lop-sided, it seems the larger stack has to have an advantage. Barring extremely long runs by the short stack.

Hell, what do I know about finite problems? I'm still trying to figure out all the simple HE math.

[elitegimp]

think of a simple case -- you and I will each bet $1 on a coin flip. Heads you win, tails I win... but there's a catch. You start with $1, I start with $100. There's a 50% chance you bust out on the first wager!

It would also apply if they started out even. If the game is infinitely long and each bet is the same there is a non-zero probability that one or other player would have a long enough losing streak to bust out.

With uneven stacks it is by no means a certainty that the short stack would bust out first. The probability that the short stack does bust out first is related to the relative size of the starting stacks

interesting thread.

[soko]

Let's just be glad it's impossible to play any game infinity times...

Speaking of time, what a waste of it reading this was.

[soko]

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think of a simple case -- you and I will each bet $1 on a coin flip. Heads you win, tails I win... but there's a catch. You start with $1, I start with $100. There's a 50% chance you bust out on the first wager!

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Sure, and there is also a 1 in 2^100 (or expanded, a 1 in 1267650600228229401496703205376) chance that the I will win 100 times in a row and you will bust out.

[golfboy7]

Your question is answered by your previous statement.

basically this 'theory' lies on the basis that the larger stack can better absorb variance than the smaller stack.

This is also why we, as poker players, play with large bankrolls.

I have built my roll to 3k and keep it there making frequent withdrawls. Yet I'm still playing 2/4.

Reason? Variance. I don't want to bust out.

[Felipe]

I think I agree with what most people said. There is a good reason why poker players manage their bankrolls, and keep themselves well rolled for a particular stake (limit). I also think that if there are infinite games, then one person will bust, because his/her probability of busting is greater than 0.

[Felipe]

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Sounds pretty dumb to me unless someone has a short BR and busts out due to variance. How can a 50/50 gamble mean that anyone is going to eventually lose otherwise???

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Did you attempt to read the article?

i'd say for sure... but we aren't playing 50% odds. some of us are higher, some of us are lower.... but playing blackjack where you're playing 49.5% odds if you're good, you will eventually lose all your $$$$$$.

it is true though that this is a good reason to have a 300BB bankroll (5/10 loose passive limit).

i set up some excel spreadsheets with 2BB hour win rate and 12BB/hour variability and it's amazing the different graphs you can get over 100 hours. i have it so i hit F9 and it just updates, but the swings can be frightful or occasionally you get a line that goes straight up. but also occasionally you can get a line that goes almost straight down.

Essentially, looking specifically at this statement

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In short, it says that if there are two players playing a game with even odds(1:1), and they could play an infinite amount of games, one of them would eventually lose all of his money.


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This is certainly true (if not obvious), assuming they have finite amount of money.

That is because with infinite # of games (in the long run), there's guaranteed a chance of hitting a long enough bad streak (doesn't have to be infinite) for either player to lose all of their money.

You absolutely don't need an unequal starting amount of money for this to happen.

Btw, your title of "In the long run...you'll never win!" is obviously false, cause if one person loses, the other person obviously wins.

Essentially, it's because of Variance and like what Felipe said
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I also think that if there are infinite games, then one person will bust, because his/her probability of busting is greater than 0.

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Essentially, looking specifically at this statement

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In short, it says that if there are two players playing a game with even odds(1:1), and they could play an infinite amount of games, one of them would eventually lose all of his money.


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This is certainly true (if not obvious), assuming they have finite amount of money.



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Ah, I’m pretty sure this isn’t right. This is beyond my limited math knowledge but I had a conversation about a similar problem with a much more knowledgeable friend.

Think about the coin problem this way. Imagine a string of events where that the coin perfectly alternated heads and tails for an infinite number of flips. Well, in this instance, neither player would go broke, ever. Sure it’s ridiculously unlikely, but the key is that it is logically possible and so the event of one of the gamblers going bust is not guaranteed. Does this make sense at all?

There's a way to calcualte this that I don't know; involving a remus(sp?) sum.

[Felipe]

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Ah, I’m pretty sure this isn’t right. This is beyond my limited math knowledge but I had a conversation about a similar problem with a much more knowledgeable friend.

Think about the coin problem this way. Imagine a string of events where that the coin perfectly alternated heads and tails for an infinite number of flips. Well, in this instance, neither player would go broke, ever. Sure it’s ridiculously unlikely, but the key is that it is logically possible and so the event of one of the gamblers going bust is not guaranteed. Does this make sense at all?

There's a way to calcualte this that I don't know; involving a remus(sp?) sum.

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I also think that if there are infinite games, then one person will eventually bust, because his/her probability of busting is greater than 0.


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You say that its rediculously unlikely, but not impossible.

The more I look at it, the more it's making sense to me. An infinite number of games is necessary for the 'theory' to hold, however. Lets say I have a 99.99999999999999999999999999999999999999999999999 99% chance to win a bet. If I play an infinite amount of games I will still lose sometime - guaranteed! I won't lose my whole roll, but I will lose one 'bet'.

edit: Ya. I need to work on that title. Too bad it's too late though.

[wadea]

I think in statistics this is referred to as an "absorbing barriers" problem. If you win 100 times in a row you continue playing to infinity and eventually things will regress to the mean in a 50/50 game. However, the first time you lose 100 in a row you are finished playing and you are the loser. To change this, you need to create an upper absorbing barrier. For example, I think you would be more likely to win than to lose if you played the same game except each time you get to 150, you would start the same game over with 100.

Maybe someone who actually knows statistics can verify this or correct it.

-w.a.

[Exitonly]

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The more I look at it, the more it's making sense to me. An infinite number of games is necessary for the 'theory' to hold, however. Lets say I have a 99.99999999999999999999999999999999999999999999999 99% chance to win a bet. If I play an infinite amount of games I will still lose sometime - guaranteed! I won't lose my whole roll, but I will lose one 'bet'.

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Well if you're only 99.999% then over an infinite amount of games, vs an opponent with a bankroll infinitely larger than yours, you would eventually hit a streak where you would lose your whole bankroll.

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Ah, I’m pretty sure this isn’t right. This is beyond my limited math knowledge but I had a conversation about a similar problem with a much more knowledgeable friend.

Think about the coin problem this way. Imagine a string of events where that the coin perfectly alternated heads and tails for an infinite number of flips. Well, in this instance, neither player would go broke, ever. Sure it’s ridiculously unlikely, but the key is that it is logically possible and so the event of one of the gamblers going bust is not guaranteed. Does this make sense at all?


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I don't think it's correct to think about it this way.

The infinite number of coin flips in this problem produces an infinite random sequence, but you're talking about a specific non-random sequence. Any infinite random sequence must contain every possible finite sequence, this is a fundamental property of infinite random sequences. Hence it follows that the game will eventually end since there will eventually be a run depleting one players stack.

This is especially true if the shorter stack goes all in every time. Say you have two people playing poker one with $100 the other with $5. If delt two randome hands and they both go all in every time the one with $100 is more likely to win, because he covers the $5 bet or the $10 bet or the $20 every time. This to me seems obvious and intuitive in terms of poker, but the part that I think is counter intuitive is the concept of infinity, since it is imaginary. It would be imposible to do it an infinite number of times and for the sake of the exercise you could just say a trillion coin flips. The larger stack just cushions the margine of chance and the shorter stack forces a more volitle situation on the gambler since, in this example, if your broke your done.

i deny it all as a fallacy! if it were true, some sumerian would have all the gold in the world at this very moment. wait, maybe there IS a sumerian with all the gold at this very moment. hmmmm.

[Mark1808]

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i deny it all as a fallacy! if it were true, some sumerian would have all the gold in the world at this very moment. wait, maybe there IS a sumerian with all the gold at this very moment. hmmmm.

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There is - Bill Gates!