# Thread: Poker math question (algebra)

1. ## Poker math question (algebra)

Our opponent bets $70 into a$100 pot on the river. How often to we need to win to make the call break even?

The answer given was $170(X) -$70(1-X) = 0, so X = 0.292 or 29.2%

2. ## Re: Poker math question (algebra)

Never mind, I think I've got it.

70 to call so pot will be 240.

70/240 = 0.2916

29.2%

3. ## Re: Poker math question (algebra)

Anyone know how I'd get the algebraic answer though, as in line 2 of my opening post?

4. ## Re: Poker math question (algebra)

Are you sure you have copied the problem correctly? It should be "us" who bet $70, not "our opponent". What they are saying is that if you win "X" of the time, you will lose 1- X of the time. When you win you win$170 dollars. When you lose, you lose \$70 so in the long run you will win 170X dollars and lose 70(1- X) dollars. "To break even" means you neither lose nor win any money: 170X- 70(1- X)= 0. That is the same as 170X- 70+ 70X= 240X- 70= 0 so 240X= 70 and X= 70/240= 7/24= .2916666 or about 29.2%.

5. ## Re: Poker math question (algebra)

Thanks for replying. The problem was definitely posed saying "Our opponent bets 70 into 100 on the river"

That aside though, how do I learn to work this stuff out myself? What mathematical topic is it?