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08-08-2002, 12:01 PM
You are playing a coin flip freezout with someone for a dollar a pop. You have \$500 and he has \$100. You play until one of you is broke. What are the chances you will win his hundred without ever being behind?

08-08-2002, 12:24 PM
1:5 ??

he has fav 1 to five odds of winning

08-08-2002, 12:52 PM
Assuming a fair coin, you are a 5-1 favorite (wining 5/6 of the time). This is because if you repeated this freezout many times, you must win the same amount of money that he does. But when you win, you only win \$100 while when he wins, he wins \$500, so you must win 5 times as often as he does.

08-08-2002, 01:01 PM
There is a 2/3rds chance that he will bust the short bankroll without ever becoming the short bankroll himself.

There is of course a 5/6ths chance that he will win the freezout.

08-08-2002, 01:01 PM
That means you can't even lose a dollar. That's the same as if you only had a \$1 stack to begin with, so the odds are 100-1 against you.

08-08-2002, 01:15 PM
Maybe you mean behind as in never having less than the other player. Yes I'm sure that is probably what you mean. We will never be behind as long as we never lose 201 dollars. That will be true 201/301 of the time.

08-08-2002, 01:21 PM
He could mean never having less than \$500, in which case the \$500 can be considered \$1.. and he'd have about a 1/100 chance.

08-08-2002, 01:23 PM
That was my first interpretation, but I don't think he means that, since "freezout" means you're done when you lose all you're money. You can't reach into your pocket for more.

08-08-2002, 01:27 PM
Sorry, I misread your interpretation which is the same as my second interpretation.