# Maximum Call Amount By Expected Value (e.g. in no limit hold'em)

Recently I started learning the basics of poker and I came across the 2015 MIT course by Kevin Desmond. In Episode 3 he mentioned the concept of Maximum Call Amount.

In short, the formula to calculate the EV is:

``````Win% * WinAmount - Lose% * LoseAmount = EV
``````

The WinAmount would refer to the pot before your Call and LoseAmount is just yoiu Call Amount.

He then re-wrote the above formula to introduce the Maximum Call Amount:

``````Win% * (Pot + X) - Lose% * X = 0 -> X = (Win% * Pot) / (Lose% - Win%)
``````

EV being zero is our threshold to call, thus the Maximum Call Amount.

If EV falls below zero apparently that's a no-go for us.

And here's what baffles me.

Say we have a pot of \$380 and hero faces a \$100 bet, Pot odds here is roughly 17% ( `100/(380+100+100)` ).

Any Win% greater than 17% would be worth a call, according to the course.

Kevin in the course gave an example of Win% being roughly 34% (Lose% came to 66%) and X came to a +404 (rounded up).

I tried playing with the formula, and I found that:

1. if our win% were 100%, then X = (380 * 1) / (0 - 1) = -380

2. if our win% were 0%, then X = 0 (this makes sense because if we can't win at all, we'd not bet anything)

3. if our win% were 50%, the formula would be rendered invalid because denomenator would be `0` (i.e. 50 - 50).

4. if our win% were 49%, then X = (380 * 49%) / (51% - 49%) = +9310

5. if our win% were 51%, then X = (380 * 51%) / (49% - 51%) = -9690

My questions here:

• What does -380 in 1 mean, given we are to win 100%? if we guarantee to win, shouldn't we just call as much as can?
• given 50% winning chance does not make mathematical sense in this formula, 49%/51% of winning chance would give us a really large call amount. How does this make sense comparing to 1? if we are almost equally possible to win or lose, why would our maximum call be so big? the sign before the number doesn't seem to make much sense to me.