help me out with a math/odds question

[badonkadonk]

this should be simple for you but i'm having problems

there is $25 in a pot. your opponent bets another $25 which would end up putting you all-in. to call what % of the time do you have to think your hand is good?

is it

a) 50% of the time
or
b) 33% of the time

or something completely different. dumbed down answers appreciated. thanks.

[Lost Wages]

There is $50 in the pot at the time of your decision. It costs you $25 to call. You have to win X% of the time to break even. If you win X% of the time then time then you lose 1-X% of the time (disregarding ties).

($50)*(X) - ($25)(1-X) = 0
$50X - $25 + $25X = 0
$75X = $25
X = $25/$75 = 1/3 = 33%

Lost Wages

[Crooked Paul]

Lost Wages is entirely right. But maybe a bit more explanation is in order.

Basically, you always want the relationship of the portion of money you put into the pot to the total value of the pot (after your call) to be less than your chance to make the best hand.

In your example, you're being asked to call 1/3 or 33% of the pot. (Your $25 call to the $50 in the pot plus your $25 call = 25/(50+25) = 25/75 = 1/3 = 33%)

So if you think your hand is already best and will remain so more than 1/3 the time, or that you might be beat but your hand should improve to the best hand more than 1/3 of the time, then you should call.

Basically, you're saying, if I have to put in 33% of the pot here, I need to win this pot 33% of the time not to lose money. If my chance to win this pot is greater than 33%, then if I keep making this call over time, I expect to make money.


Crooked