Double Barreling Quiz

I'll rough math because why not.

gutshot is 8% but Ill call it 10% out of laziness.

EV when he calls is:

Lose $2 90% of the time = -$1.80
Win $3 10% of the time = +$0.30
= -$1.50

[edit] fail. We win $5 when he calls and we win. Final answer is smaller than I said first [/edit]

So If a call costs us $1.50 and a fold wins us $3 we need him to fold once for every 2 calls to break even => 33%.

Haven't looked at other responses yet because I want to try and do this on my own first.

EV = x * $3 + (1-x)[(4/46) * $7 - (42/46) * $2]

we set EV = 0 and solve for x:

1.2174 = 3x + 1.2174x

x = 28.87% of the time he needs to fold


Final conclusion: This really doesn't seem like that much, especially if he's the type of player who's going to peel flop wide.


Is this right? I had to look back at caddie's showdown thread to double check my equation but I think its correct.

Originally Posted by kfaess

EV = x * $3 + (1-x)[(4/46) * $7 - (42/46) * $2]

4/46 times you bet $2, and then won a pot of 7, so you netted +5.
Otherwise everything is neato.