0

If I put my opponent on a draw at the turn how much do I have bet to take away their pot odds.

Say I have top pair and an unconnected under and I put my opponent on two overs with an open ended straight draw. How much do I need to bet to take away their pot odds?

1

Your opponent has 2 overs and an open ended straight draw then they have 14 outs
Unknown cards is (52 - 4 - 2 - 2) = 44
You take away the two cards in their hand as you have put them on 2 cards
At what size bet does hand odds equal pot odd
(44 - 14) / 14 = (bet + pot) / bet
44/14 - 1 = 1 + pot/bet
44/14 - 2 = pot/bet
bet = pot / (44/14 - 2) = pot * 0.875

You would need to an almost pot sized bet to chase them off the draw.

On the other side if you are on the draw you can bet 1/2 the pot and still be getting value. You could represent top pair or an over pair and possibly chase them off. If you hit the river you could again bet 1/2 the pot and probably get paid off. But don't raise the turn. If they lead out on the turn with less than a pot size bet then just call and give them a chance to lead out on the river.

In general the equation is
bet = pot / (44/outs - 2)
At 22 outs that is infinity
If you put your opponent on 22 outs then you cannot price them out
But I think 21 is the max - two overs, open ended straight, and a flush draw

here they are
the first column is number of outs and the second is bet / pot

3   0.0789
4   0.1111
5   0.1471
6   0.1875
7   0.2333
8   0.2857
9   0.3462
10  0.4167
11  0.5000
12  0.6000
13  0.7222
14  0.8750
15  1.0714
16  1.3333
17  1.7000
18  2.2500
19  3.1667
20  5.0000
21  10.5000

If someone thinks they see a mistake then please comment

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.