# What size bet to take away pot odds?

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?

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