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View Full Version : All-In every hand, how big of a stack do you need...

fatdave
08-19-2005, 01:04 PM
Heads up no-limit hold 'em: You and your opponent are going to go all in every hand, no matter what. How many times larger than your opponents stack must your stack be to make their odds of winning it all 1%?

Originally I thought it may be 99-1, but if they double up on the first hand, that will change... then it will especially change if they double up again on the second hand.

If you change it so that they will only go all in with Ax, Kx, any broadway, any pair, any suited connector or any connector 56o and above, while you still call with any two cards, how big must your stack be to make their odds still 1%?

elitegimp
08-20-2005, 02:38 AM
this is probably extremely wrong, but if your opponent has a 50% chance of doubling up then thew odds of him doubling up 7 times in a row is 1/128... so if you can cover him those 7 times then it would be around 1%. So I'm saying you should have 128 times his stack. If I weren't too drunk, I would make a table showing why this is my answer.

BruceZ
08-20-2005, 03:39 AM
[ QUOTE ]
this is probably extremely wrong, but if your opponent has a 50% chance of doubling up then thew odds of him doubling up 7 times in a row is 1/128... so if you can cover him those 7 times then it would be around 1%. So I'm saying you should have 128 times his stack. If I weren't too drunk, I would make a table showing why this is my answer.

[/ QUOTE ]

You should have 99 times his stack. See explanation here (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&amp;Number=2999066&amp;page=&amp;view=&amp;s b=5&amp;o=&amp;vc=1).

Note that he must win the first 6 in a row to not go bust, and that has probability 1/64. At that point he will have 64% of the chips, and a 64% chance of winning. This checks out since 1/64 * 64% = 1%.