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.