Does the size of the bet matter in this case

I got in to an argument with a few people about this kind of scenario:

Situation A: A guy with 100$ stack plays heads and tails against a guy with 10$ stack, the bet is 1$. The game is beeing played until one of them looses all their money.

Situation B: A guy with 100$ stack plays heads and tails against a guy with 10$ stack, the bet is 10$. The game is beeing played until one of them looses all their money.

Does the guy with 100$ stack stand better chances of winning on situation B. Or are his chances as good in point situations.

point = both if you are wondering lol

his "chances" are just as good in both situations, but due to variance, it is more likely that he'll go broke in situation B.

if his chances are just as good in both situations, how can he go broke more likely in situation B due to variance. That asnwer doesn't make any sense in my mind.

because in A, he is only wagering 1/10 of his stack. Since he is Even Money, the chance of him losing all of that when only wagering 10% is low.

In B, he is wagering 100% of his stack. This means he only needs to lose his first hand to go broke, where as A has to lose his first 10 coinflips to go broke.

Basically, to go broke, A has to lose 10 times more than he wins, where as B only has to lose 1 more time than he wins.

Edit-- in this post I'm talking about the shortstack, not the $100 stack

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because in A, he is only wagering 1/10 of his stack. Since he is Even Money, the chance of him losing all of that when only wagering 10% is low.

In B, he is wagering 100% of his stack. This means he only needs to lose his first hand to go broke, where as A has to lose his first 10 coinflips to go broke.

Basically, to go broke, A has to lose 10 times in a row AT LEAST, where as B can go broke in 1 flip, 2 flips, 3 flips, etc depending on how he wins.

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In theory thats exactly what im thinking, but I don't know how to proof it. Is the formula too complicated in this kind of situation? Or does that even change the chances of bigstack winning those situations ? In situation B, the lowstack is ofcourse more at risk from the start BUT does it change his chances compared to situation A.

i edited my statement as it was incorrect, I'm not sure what I was thinking.

To restate: Situation A needs to lose 10 more times than he wins to go broke, Situation B needs to lose only 1 more time than he wins to go broke. Thus, it is more likely that Situation B goes broke, even though their odds are the same.

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To restate: Situation A needs to lose 10 more times than he wins to go broke, Situation B needs to lose only 1 more time than he wins to go broke. Thus, it is more likely that Situation B goes broke, even though their odds are the same.

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This is what I don't get. If its more likely that the lowstack in situation B goes broke, how can his odds be the same as in situation A? In both situations they are playing until other one is broke.

Just think about it. Shortstack A needs to lose 10 more times than he wins, B needs to lose 1 more time. Which is more likely to happen if you flip a coin...you get 10 more tails than heads, or you get 1 more tails than heads? Which is likely to happen SOONER?

try flipping a coin and find out.

I also slightly misread your question, but my reasoning still stands. The $100 stack has a better chance in Situation B, for the reasons provided.

[BruceZ]

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I got in to an argument with a few people about this kind of scenario:

Situation A: A guy with 100$ stack plays heads and tails against a guy with 10$ stack, the bet is 1$. The game is beeing played until one of them looses all their money.

Situation B: A guy with 100$ stack plays heads and tails against a guy with 10$ stack, the bet is 10$. The game is beeing played until one of them looses all their money.

Does the guy with 100$ stack stand better chances of winning on situation B. Or are his chances as good in point situations.

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If the coin is fair, then in both cases, the player with the $100 stack has a 10-1 advantage to win the freeze out over the player with the $10 stack. The normal freeze out calculation applies. To see this, notice that when the $100 stack wins, he wins $10, but when the $10 stack wins, he wins $100, or 10 times more. If you run this a billion times, if the coin is fair, each player must win the same amount of money on average, so the $10 player who wins 10 times more must win 10 times less often than the $100 player. So the $100 player has a 10-1 advantage to win the freeze out. This does not change with the size of the bet, and it is the same for both situations.

This is why in a poker tournament, a player's odds of winning the tournament are the size of his stack to the size of his opponent's stack(s). This does not change when the stakes change, and it doesn't depend on the size of each bet, or even if all the bets are the same size, which they are not. Heads-up at the end of a no-limit tournament, a player can bet his whole stack, or only a small fraction of it, but regardless, his odds of winning are always the ratio of his stack size to his opponent's stack size.

Now if the coin were not fair, then the player favored by the coin would win more often if he makes smaller bets. For that equation, see this post (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=0&Number=2197150&page=&vc=1), which does not apply for a fair coin (p = q).

[BruceZ]

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if his chances are just as good in both situations, how can he go broke more likely in situation B due to variance. That asnwer doesn't make any sense in my mind.

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Correct, that's nonsense. See my post in this thread.

[pzhon]

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The $100 stack has a better chance in Situation B, for the reasons provided.

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You are wrong.

The expected amount of money for each player is preserved under a fair bet or a sequence of fair bets. The expected amount of money for the player starting with $100 is always $100. Since he ends up with $0 or $110, he wins with probability 10/11.

Here is another way to see that you are wrong, and the scenarios are equivalent. Instead of tossing a coin for $1 in an unbroken stream, imagine that the coin is tossed until one player or the other has gained $10, and then a break is taken. This is repeated until one player has all of the money. The amount of money after the breaks follows a random walk with steps of size $10.

This is standard.

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To see this, notice that when the $100 stack wins, he wins $10, but when the $10 stack wins, he wins $100, or 10 times more. If you run this a billion times, if the coin is fair, each player must win the same amount of money on average, so the $10 player who wins 10 times more must win 10 times less often than the $100 player.

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This is flawed. Especially the last line I quoted. The $10 player does not make more money in 1 coinflip victory. If he made $100 for each winning coinflip, then this would be correct, however that is not the case.

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The expected amount of money for each player is preserved under a fair bet or a sequence of fair bets. The expected amount of money for the player starting with $100 is always $100. Since he ends up with $0 or $110, he wins with probability 10/11.

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I don't understand this part, can you clarify?

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Here is another way to see that you are wrong, and the scenarios are equivalent. Instead of tossing a coin for $1 in an unbroken stream, imagine that the coin is tossed until one player or the other has gained $10, and then a break is taken. This is repeated until one player has all of the money. The amount of money after the breaks follows a random walk with steps of size $10.

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This doesn't work because you are not taking into account when the shortstack may lose his whole stack before he makes each $10.

The 2 situations APPROACH equality, however because Situation B has the shortstack gambling his entire stack, it may not REACH equality in time, the game may end before that. This would be rectified if there were unlimited rebuys and the game were replayed a billion times, but it is not that way. The game is played once, so equality is only being approached in theory, not necessarily reached.

If you were the big stack, and you could choose to be situation A or B...you'd be on crack to choose A.

I hope I don't sound too hostile hehe, I was re-reading my posts and I realise I might. I'm happy to change my view if someone can provide good reason /images/graemlins/smile.gif

[BruceZ]

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To see this, notice that when the $100 stack wins, he wins $10, but when the $10 stack wins, he wins $100, or 10 times more. If you run this a billion times, if the coin is fair, each player must win the same amount of money on average, so the $10 player who wins 10 times more must win 10 times less often than the $100 player.

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This is flawed. Especially the last line I quoted. The $10 player does not make more money in 1 coinflip victory. If he made $100 for each winning coinflip, then this would be correct, however that is not the case.

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There is no "1 coinflip victory", what are you talking about? I'm talking about winning or losing the whole freeze out. Each time he wins the whole freeze out, he wins $100. Each time he loses, he loses $10. Those are the only two options.

This is not "flawed", it is the standard way to analyze a freeze out. I didn't just invent it, and anyone who knows anything about tournaments knows this result. Read the paragraph I added to the above post about poker tournaments if you have not already done so.

If you are talking about losing the whole freezeout, then you are not taking into account that it is more likely the shortstack may lose all his money quicker in situation B.

Bet your wholestack on a coinflip repeatedly and you will go broke quicker more often than if you bet 10% of your stack on a coinflip repeatedly.

[pzhon]

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The expected amount of money for each player is preserved under a fair bet or a sequence of fair bets. The expected amount of money for the player starting with $100 is always $100. Since he ends up with $0 or $110, he wins with probability 10/11.

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I don't understand this part, can you clarify?

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What don't you understand about it? The first line is the most fundamental property of martingales (http://mathworld.wolfram.com/Martingale.html).

You are disagreeing with the result here (http://mathworld.wolfram.com/GamblersRuin.html). However counterintuitive it may be to you, it is standard.

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Instead of tossing a coin for $1 in an unbroken stream, imagine that the coin is tossed until one player or the other has gained $10, and then a break is taken.

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This doesn't work because you are not taking into account when the shortstack may lose his whole stack before he makes each $10.

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No, there is no chance the short stack goes broke (strictly) between the breaks. At each break, both players have a multiple of $10.

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The 2 situations APPROACH equality,


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You are confused. The probabilities are numbers. Numbers do not move. The probability the large stack wins is 10/11 in both situations.

Yes, the probability is the same, the NUMBER is equal. But variance will cause instances where the bigstack will win a bunch of coinflips in a row, and also will cause the big stack to lose a bunch of coinflips in a row. In BOTH of these cases, situation B is more favourable for the big stack, because variance is more likely to bust the short stack.

Variance can kill shortstack B, it is unlikely it will kill shorstack A.

[pzhon]

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situation B is more favourable for the big stack, because variance is more likely to bust the short stack.


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Variance is responsible for any loss here.

You recognize that the big stack will win 10/11 of the time in both scenarios, yet you say the big stack should prefer scenario B? Exactly how much would you be willing to pay to change the situation from A, worth $100, to B, also worth $100? Do you think other people should also share your strange preference?

Your denials of standard probability theory consist of errors and nonsense.

[BruceZ]

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If you are talking about losing the whole freezeout, then you are not taking into account that it is more likely the shortstack may lose all his money quicker in situation B.

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I'm not taking it into account because it's irrelevant. Review "my" (classic) argument carefully, and tell me which step this contradicts. There are 2 steps to the argument. First, the $10 player will either win $100, or he will lose $10. Second, since the coin is fair, the EV of each player is zero. The only way their EVs can be zero is if the $10 player wins $100 with probability 1/11 and loses $10 with probability 10/11. Then his EV is 1/11*100 + 10/11*(-10) = 0. The $100 player wins $10 with probability 10/11, and loses $100 with probability 1/11, so his EV is 10/11*10 + 1/11*(-100) = 0. If they continue to play a large number freezeouts, they will have a net EV of 0. The duration of each freeze out is irrelevant. This is true for both situations A and B. If you think that one player wins more often in one situation, then you must believe that one player has a positive EV for this freezeout while the other has a negative EV, even though the coin is fair, and that is preposterous. The EV is zero in both situations, and the variance of each situation is the same too, for both players the variance is 1/11*(100)^2 + 10/11*(10)^2 = 1000.

If we are evenly matched at heads-up no-limit hold'em, and we agree to play every day for the rest of our lives, each of us will have an average win for our lifetimes of zero. It doesn't matter how much we bring to the table each day, or how long we play. That's fundamental. If I always bring $100 to the table, and you always bring $10, and we always decide to play until one of us is broke, then I must win 10/11 of the time and you must win 1/11 of the time, so that our EVs are both zero.

If you don't believe this classic argument, then you must also not believe that 2 equally skilled players playing heads-up at the end of a no-limit hold'em tournament each has odds of winning equal to the ratio of his stack to his opponent's stack. This will be true independent of the stakes, or whether they sometimes go all-in and sometimees bet one chip at a time. The argument is the same as the one I made above, and this argument is independent of bet size. This argument appears, for example, in Sklansky's Tournament Poker for Advanced Players in the chapter "Freezeout Calculations". If you don't understand this, then you would be unable to respond correctly at the end of a tournament when someone offers you a deal, because you don't know how to make a simple freezeout calculation.

As for your argument that it is more likely for the $10 player to have 1 more loss than wins in situation B than 10 more losses than wins in situation A, this is false. Note that if this were true, then the $100 player would be more likely to have 100 losses in situation A than 10 losses in situation B. Actually, in situation B, the $10 player must have 1 more loss than wins before player A has 10 more losses than wins. In situation A, while it is true that the $10 player must now have 10 more losses than wins, he must do this before the $100 player has 100 more losses than wins. So while the $10 player has to have 10 times more losses than wins in situation A, he has until the $100 player also has 10 times more losses than wins, so he has many more chances to accumulate these 10 losses, so that the probabilty of accumulating 10 losses in situation A is the same as for 1 loss in situation B.

If you still believe that the $10 player will win more often in situation A, then I challenge you to prove it by computing the probability that he will win in both situations. I have shown you the proper way to compute the probability that he will win, and that probability is 1/11 in both situations. If you disagree, then please tell me by how much you believe situation A is favored over situation B, and I will happily take situation B in a computer simulation, and allow you take situation A, and give me something better than even money odds. We'll let it run a good long time so that I can drain your entire bankroll.


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I'm happy to change my view if someone can provide good reason

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I have answered the question at the top of this thread, computed the probabilities for both situations, and provided a complete proof of the correct answer. We don't owe you an understanding. Rather than understand the proof that you have been given, you persist in making statements which neither prove or disprove anything.

Now both Pzhon and I have told you that you are wrong, and that your arguments are nonsense, yet still you persist in arguing. You are still a relative newbie, but eventually you will realize that when both of us use words like "nonsense", your most efficient route to the truth is not to assume that you are right and argue, but to read our arguments carefully with a mindset to understand why they are correct. Otherwise you will be wasting both time and forum space. I recommend starting with my explanation in this case, because if you can't understand mine, it is not likely that you will understand Pzhon's. From a purely probability standpoint, the probability that you are right is zero.

[LetYouDown]

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From a purely probability standpoint, the probability that you are right is zero.

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I continuously fluctuate between whether your wit or your knowledge I appreciate more. Today I go with wit.

[WhiteWolf]

I won't rehash the logical proofs others have given. Instead, I'll try to address where your intuition is failing you.

The chance that the big stack wins the freezeout is the same in both situations.

The chance that the big stack will win "quickly" is higher in the second situation.