# HU riverplay OOP

• Bronze
Joined: 09.09.2007
Could someone explain this to me?
This is from Cornholio's advanced riverplay article

If you assume your opponent will check behind all worse hands in case you check, but bet for value all better hands, it would be sufficient if the opponent called a worse hand with a probability of 1/(bets in the pot). If the opponent has a better hand, you would lose one bet, and you won't get a bet from a worse hand.

For instance, if the pot is 9 BB big before your bet on the river and the opponent calls better hands in 60% of the cases, calls worse hands in 20% of the cases and folds worse hands in 20% of the cases, you still have to bet.

suppose a sample size of 10 hands:
6 hands he calls my river bet with a better hand --> I lose 6 BB
2 hands he calls my river bet with a worse hand --> I win 2 BB
2 hands he folds to my river bet with a worse hand --> neutral

sum: I lose 4 BB

where does the pot size come into play?
• 11 replies
• Bronze
Joined: 18.03.2008
you win the pot when he folds which is 9bb.........
• Bronze
Joined: 09.09.2007
60% he calls better hands
20% he calls worse hands
20% he folds a worse hand
The sum is 100% so he's never folding a better hand in this example.
• Black
Joined: 30.07.2008
Hi!

I think article assumes you either play b/f or c/c where this is very logical.In case of c/c you will lose 6BB over 10 hands.

But it says that he will always check worse hands behind and we are not a favourite against his (calling+raising)range I see c/f as clearly the best here
• Bronze
Joined: 06.05.2008
@ihufa - don't you also win the pot when he calls your bet with a worse hand ???? In which case you will win the pot 40% of the time!!

This would seem to imply that you invest 10BB - 1 for each hand and lose 6 when he calls with a better hand but win 2 x 9 when he folds and 2 x 11 when he calls with the worse hand.

10 - 6 + (2 x 9) + (2 x 11) = 44

Investing 10 to win 44 seems ok to me or is my maths or my asumptions of how this gets worked out wrong?
• Bronze
Joined: 09.09.2007

I see now that in the example b/f gives Hero a better result than c/f.

But I still don't get this:
it would be sufficient if the opponent called a worse hand with a probability of 1/(bets in the pot)
• Black
Joined: 30.07.2008
Originally posted by DarthBobo

I see now that in the example b/f gives Hero a better result than c/f.

But I still don't get this:
it would be sufficient if the opponent called a worse hand with a probability of 1/(bets in the pot)
You probably meant ...better than c/c
I still think c/f is the preffered one. In 60% you will lose 9BB and in 40% you win 9BBs.Over 10 hands that would be -18BB which is clearly better than c/c and even b/f
• Bronze
Joined: 09.09.2007
is there a way to get cornholio in this thread?

after all he wrote the article
• Black
Joined: 07.04.2008
If you assume your opponent always checks worse hands and bets better hands and your equity <50% against his calling range then c/f is always the best option...

The analysis suggested there and your calculations here only compare b/f versus c/c. His example (actually, his assumption) is just wrong with what he wants to show; which is that b/f might be better then both c/c & c/f even if your equity against your opponents calling range is <50%.

Try assuming that in one of the ten situations you have (more specifically: 1 of the 4 times your opponent has a worse hand) he would bet if checked to.

Then you'll see that EV[b/f] > EV[c/c] > EV[c/f] then.
• Bronze
Joined: 09.09.2007
Originally posted by DukeFreedom
Try assuming that in one of the ten situations you have (more specifically: 1 of the 4 times your opponent has a worse hand) he would bet if checked to.

Then you'll see that EV[b/f] > EV[c/c] > EV[c/f] then.
So take my spreadsheet and change the value of column D, hand 7 to "bets".
E7 then become +1 instead of 0.
The sum of column E becomes -5.

This does not teach us anything new because b/f is still better cause the sum of column I is still larger than the sum of column E. It was -4 versus -6 and now it is -4 versus -5.

So... I still don't get it!

How does cornholio come to "a probability of 1/(bets in the pot)". Nobody knows? None of the replies have mentionned pot size.
• Black
Joined: 07.04.2008
If it showed nothing new, I don't think I'd have posted it.

Originally posted by DukeFreedom
Then you'll see that EV[b/f] > EV[c/c] > EV[c/f] then.

Lemme first just quote what Cornholio wrote

If you assume your opponent will check behind all worse hands in case you check, but bet for value all better hands, it would be sufficient if the opponent called a worse hand with a probability of 1/(bets in the pot). If the opponent has a better hand, you would lose one bet, and you won't get a bet from a worse hand.
He worded this very poorly and actually it's even wrong when this is what he really ment, because as I said before: if our equity <50% and we assume your opponent always checks behind worse hands when checked to then c/f is always the best move.

If you assume your opponent will play optimally in case you check, it would be sufficient if the opponent called a worse hand with a probability of 1/(bets in the pot).
Now note that playing optimally is *NOT* the same as always checking behind worse hands. To see this quickly, take your example and assume that your opponent will bluff 1/11 times when he bets (when you check to him). In that case you can't call (actually: you'll be indifferent between calling or folding), because 10/11 times you'll loose 1 BB and 1/11 times you'll win the pot of 10 BB, giving a net EV of 0, the same as c/f.

In fact, "playing optimally" is very precisely defined for your opponent according to game theory, but I really don't want to get into explaining the exact theory behind this... Basically it just says that your opponent will bluff with such a frequency that you'll be indifferent between c/c his bet or c/f.

Hence this "optimal playing strategy" for your opponent dictates that both c/c and c/f will have an EV of 0 for you, as with the example I gave above.

However, b/f will have an EV of > 0 if your opponent calls worse hands with a probability P > 1/B, where B = bets in pot. This follows, because:

with probability (1-P) we lose 1
with probability P we win B+1

Now the net EV of b/f will be P*(B+1) - (1-P)*1= P*B - 1.

Now P*B - 1 will always be > 0 if P > 1/B.

Hence b/f will be better then c/c and c/f if we assume our opponent calls worse hands with probability > 1/(bets in pot) and would play optimally if we check.
• Bronze
Joined: 01.01.2009
I though this was fairly straight forward?

You arent value betting because from the bets that go on the river you are -ev, but you are more protecting the equity you have in the pot already which you loose by by check/folding. It is better then check calling by the assumtion tht he ONLY bets better hands as you would have to fold if this assumption would be deemed correct.