Required overlay for flush draw on paired board

[also]

I was an observer of the following hand, and afterwards I criticized the hero's play. Well, he was right, I was wrong.

Fairly loose, passive home game. 5 players limp, so 7 see a flop of T62, two hearts. The flop is checked around, and the turn is (T62)T, two hearts and two spades. The big blind bets, a loose player calls, and our hero calls with a nut flush draw (he didn't say, but I'm assuming spades), getting 5.5:1 on his money.

Our hero's thinking is that the checked flop made a ten unlikely, so a good part of the time the board pair is irrelevant to his drawing chances. I pointed out that the big blind is more likely than any other player to hold a weak ten and check it on the flop (he is also more likely than the others to flop a trashy two pair with that board and try to check-raise), so 5.5:1 on hero's probable 7-out shot when he might be drawing dead is insufficient. Let's have a looksie.


Hero is getting 5.5:1 on the turn to call. Let's assume the bettor always bets the river, whether he is bluffing or not, and that our hero calls on the river if he hits his flush and folds if he misses.


If the bettor is already full, then hero's call EV is:

9/45 * -2 + 35/44 * -1 = -1.20

If the bettor has a ten, then hero's call EV is:

7/45 * 6.5 + 36/45 * -1 + 2/45 * -2 = .122

If the bettor is bluffing, then hero's call EV is:

9/46 * 6.5 + 35/46 * -1 = .511


Now, let's try to figure out the chance that our villain has trips vs. a boat/quads on the turn, given that he is not bluffing. He was in the blind, so let's assume all hands are equally likely. On a board of TxyT, he can hold:

xx: 3 combos
yy: 3 combos
TT: 1 combos
Tx: 6 combos
Ty: 6 combos
----------------------------
total boats/quads: 19 combos

If he has just trips (Tz), then there are 38 possible kickers (everything except the 7 cards we've seen and the 3 y's, 3 x's, and 1 ten). There are 2 * 38 = 76 trips combos he could hold.

Given that he has at least trips, then, our villain holds a boat there 19/(19+76) = .2 of the time.


In the cases where our villain is *not* bluffing, hero shows a profit of:

(.2 * -1.20) + ((1 - .2) * 1.22) = 0.736

If we mix in the bluffing case, the expectation can only improve. So, in conclusion, it's a super solid call. A big part of the reason I thought it was a bad call was that during the hand, the 6 outs odds (~6.5:1) popped into my head rather than the 7 outs odds. (Yeah, I'm mister super pro. I'm mister pot odds expert. Hahaha.) (A 6.5:1 draw is definitely *not* a call in this spot, BTW.)


This analysis ignores a few details. For instance, I assume no one behind our hero *ever* raises. This is somewhat reasonable given the fact that the players in position didn't bet the flop and that the pot is now protected, but "never" is a big word. At least occasionally someone in position will have flopped a set or ridiculous two pair and decided to slowplay. I also assume that, in the "bluffing" case, neither of our opponents holds a random pocket pair, which would reduce hero's outs from 9 to 8.

[also]

Wow, my first post to 2+2 and I suck ass.

Notice how I go:

[ QUOTE ]

If the bettor is already full, then hero's call EV is:

9/45 * -2 + 35/44 * -1 = -1.20

If the bettor has a ten, then hero's call EV is:

7/45 * 6.5 + 36/45 * -1 + 2/45 * -2 = .122

[/ QUOTE ]

and then a little later

[ QUOTE ]
(.2 * -1.20) + ((1 - .2) * 1.22) = 0.736


[/ QUOTE ]

See how that decimal place moved over one spot to the left (.122 -> 1.22) for the drawing against trips EV? This should be:

.2 * -1.20 + (1 - .2) * .122 = -0.14

(Actually, there is also a second error: the 35/44 at the top should be 36/45, but the overall EV rounds to -0.14 anyway.)

So, in fact, the call is losing if the villain is never bluffing and always has trips or better.

That means we need to mix in the value of the case where the villain is overstating his hand, which, unfortunately, is a rather player-specific calculation. For instance, villain might be running a pure bluff or betting a pair of 6s. With a pure bluff, he might give up his bluff on the river or try again if a flush doesn't hit (or even if a flush does hit); if he typically abandons his bluff and checks down the river, then hero's A-high has a lot more value (in fact, villain might bluff so much relative to his legit hands that hero always has a profitable call on the end with A-high), but hero loses value because he isn't bet into when he makes his flush. If villain is often betting a pair of 6s there (if he bets with 6s there 100% of the time and bets with trips+ there 100% of the time, which is probably the case with most players, then he is clearly a big favorite to hold just 6s), then hero should call on the end if he catches an A and should consider the three As as partial outs. All these "or"s and "%"s are pretty player-dependent, so I'm not going to bother to try to calculate anything.

Okay, but now that we know that 5.5:1 is insufficient to call with a nut flush draw against a blind opponent who we put on trips or better, I'm curious what the actual required overlay is.

If P is the pot size on the turn before hero calls, then

EV_call =
4/5 * (7/45 * (P+1) + 36/45 * -1 + 2/45 * -2) +
1/5 * (9/45 * -2 + 36/45 * -1)
= 1/225 (4 * (7 * (P+1) - 36 - 2 * 2) + 1 * (-9 * 2 - 36)

Setting this equal to zero,

EV_call = 0 = 1/225 (4 * (7 * (P+1) - 36 - 2 * 2) + 1 * (-9 * 2 - 36)
0 = 4 * (7P + 7 - 36 - 4) + (-18 - 36)
P = 186/28 = 6.64

So, hero has to be getting around 6.7:1 or better to call there against a trips+ blind opponent, assuming exactly one bet on the river if hero hits (against tricky opponents who routinely go for a check-raise here if they're full and a flush hits, you need an even bigger overlay). This seems about right: 7 outs, minus about 1/5 for the times we're drawing dead, gives us 5.6 outs, which is 7.0:1, but then we get a little bump because the implied odds/reverse implied odds situation is favorable to us (7/9 * 4/5 = 62% of the time, the bet going in on the river is going to us).

If the villain is not blind, then this 6.7 number doesn't really apply, since the chance that he has a boat versus trips on the turn is dependent on his starting hands.

[TaintedRogue]

Also,
Thanks for the formula! A bigger thanks if you'll help me understand it.
9/45 * -2 +35/44 *-1 = -1.20
I know what the 9/45 represents, it's the 9 outs to the flush, divided by the 45 unknown cards. I don't see how you come up with 45 unknown cards b4 the river though, as it's 46.
So, I get:
9/46 * -2 + 37/46 * -1
.1957 * -2 + .8043 * -1
-1.8043 + -.1957 = -2.00

What am I doing wrong so far?

[TaintedRogue]

Well, what I am doing wrong for starters is, we are assuming the opponent has a house, so one of his cards cannot be a spade. The T. But either the 6 or 2 has to be a spade, so he may have 2 spades. I don't see anyone limping in with anything other than T6 suited, which is the only plausable hand I see making a house, so he can't have a spade. So I would have to assume that neither card is a spade and make it 9/44.

[TaintedRogue]

This is what I see:
If the button has a house, then he limped with T6 suited, 66 or 22. Only plausable hands. So, it's:

9/44 * -2 + 35/44 * -1 = -1.7955 + -.2045 = -2.00

However, it can't be that simple, as T6 suited would be without a spade and either 66 or 22 has to be without a spade.

If the button has trip T's we have to discount 2 of the spades; the button's kicker and either the 6 or 2, as one of them has to be spade. And, we know the button has a T, so it's 45 unknown cards instead of 46, so I understand your equation there.
I also understand your bluffing equation.

[TaintedRogue]

There are only 2 hands of T6 suited left after the turn and 3 each of 22 and 66 and 1 TT. We'll assume the 6 on the board is a spade, so there are 2 hands of 22 in which one of the 2s could be a spade.
Of all the hands the button could possible have (9) if he has a house, 2 of them could be with a spade.
So, when the button's hand does hold a spade, which would be 22.22% of the time, the equation would be:

1st assuming the button holds a spade:
(8/44 * -2 + 36/44 * -1) * .2222 = -.4444

Assuming the button doesn't hold a spade:
(9/44 * -2 + 35/44 * -1) * .7778 = -1.5556

and I get the same total -EV: -2.00; so, I guess it is that simple.

[TaintedRogue]

And................I missed the fact that you're talking about the Big Blind and not the Button and I was shuffling in the dark once again...................

[also]

[ QUOTE ]
9/45 * -2 +35/44 *-1 = -1.20
I know what the 9/45 represents, it's the 9 outs to the flush, divided by the 45 unknown cards. I don't see how you come up with 45 unknown cards b4 the river though, as it's 46.

[/ QUOTE ]

Actually, the "35/44" was a typo: I corrected it in my second post above to read "9/45 * -2 + 36/45 * -1". Before we look at the full house case, though, let's look at the "trips only" case, which is simpler.


Versus trips

EV_trips = 7/45 * 6.5 + 36/45 * -1 + 2/45 * -2

Here there are 6 cards we've actually seen, plus we know our opponent has a T, which cannot be a spade since the T[img]/images/graemlins/spade.gif[/img] came on the turn. So, there are 7 seen cards, 45 unseen cards, and the 9 spades that aren't in hero's hand or on the board are all in the unseen pile. So, (45-9)/45 river cards fail to complete hero's flush, 2/45 complete his flush but give the villain a boat (i.e., the one that matches the villain's kicker and the one that double-pairs the board) (this isn't quite right, but I'll get to that in a second), and 7/45 give hero a winning flush.

That's the rationale. However, two things are not correct in this formula. (Yeah, I suck further.)

The main problem is that the villain's kicker may or may not be a spade, so we actually have to merge two cases:

EV_trips_spade_kicker = 7/44 * 6.5 + 36/44 * -1 + 1/44 * -2
EV_trips_non_spade_kicker = 7/44 * 6.5 + 35/44 * -1 + 2/44 * -2
EV_trips = 8/45 * EV_trips_spade_kicker + 37/45 * EV_trips_non_spade_kicker

(8/45 rather than 9/45 because we have assumed the villain isn't full for this case so his kicker can't be the spade that matches the heart on board.) This new formula changes the EV versus trips value from 0.12 to 0.15.

The second problem is that this formula assumes that hero always pays off on the end when he makes a flush and the villain makes a boat. While paying off is unavoidable when villain hits his kicker, hero should arguably be able to get away from his flush when the river double-pairs the board. However, in practice, that depends on the exact board, which I don't remember. It was either T[img]/images/graemlins/heart.gif[/img]6[img]/images/graemlins/spade.gif[/img]2[img]/images/graemlins/heart.gif[/img] T[img]/images/graemlins/spade.gif[/img] or T[img]/images/graemlins/heart.gif[/img]6[img]/images/graemlins/heart.gif[/img]2[img]/images/graemlins/spade.gif[/img] T[img]/images/graemlins/spade.gif[/img]. If the river comes (T[img]/images/graemlins/heart.gif[/img]6[img]/images/graemlins/spade.gif[/img]2[img]/images/graemlins/heart.gif[/img] T[img]/images/graemlins/spade.gif[/img]) 2[img]/images/graemlins/spade.gif[/img] on the first board, hero is probably paying off a bet in practice cuz he reads villain for a T or 6. If the river comes (T[img]/images/graemlins/heart.gif[/img]6[img]/images/graemlins/heart.gif[/img]2[img]/images/graemlins/spade.gif[/img] T[img]/images/graemlins/spade.gif[/img]) 6[img]/images/graemlins/spade.gif[/img] on the second board, though, then hero should be able to get away from it (unless he is also willing to bluff catch on the end with A-high). So, on the first board the formulas would be as above, but on the second board, they would be:

EV_trips_spade_kicker = 7/44 * 6.5 + 37/44 * -1
EV_trips_non_spade_kicker = 7/44 * 6.5 + 36/44 * -1 + 1/44 * -2


Versus boat/quads

Again we are going to run into complications if we try to account for every case exactly. If the board is T[img]/images/graemlins/heart.gif[/img]6[img]/images/graemlins/spade.gif[/img]2[img]/images/graemlins/heart.gif[/img] T[img]/images/graemlins/spade.gif[/img] and villain holds T[img]/images/graemlins/club.gif[/img]2[img]/images/graemlins/diamond.gif[/img], then hero will probably pay off another bet if the river is the (T[img]/images/graemlins/heart.gif[/img]6[img]/images/graemlins/spade.gif[/img]2[img]/images/graemlins/heart.gif[/img] T[img]/images/graemlins/spade.gif[/img]) 2[img]/images/graemlins/spade.gif[/img], but this can't happen if villain holds the T[img]/images/graemlins/club.gif[/img]2[img]/images/graemlins/spade.gif[/img] boat. If you (TaintedRogue) want to rework the formulas to account for the cases where villain does and does not hold a spade, please do, but I'm just about out the door; it should be straightforward but tedious (categorize villain holdings by whether they contain a spade, paying to attention to which river spades hero will and will not pay off, for both of the above boards).

Barring the exact per-holding case analysis, I think "8.8/44 * -2 + 35.2/44 * -1" should be pretty close for the T[img]/images/graemlins/heart.gif[/img]6[img]/images/graemlins/spade.gif[/img]2[img]/images/graemlins/heart.gif[/img] T[img]/images/graemlins/spade.gif[/img] board: Villain holds a spade 4/19 times (2[img]/images/graemlins/spade.gif[/img]2[img]/images/graemlins/club.gif[/img], 2[img]/images/graemlins/spade.gif[/img]2[img]/images/graemlins/diamond.gif[/img], T[img]/images/graemlins/club.gif[/img]2[img]/images/graemlins/spade.gif[/img], and T[img]/images/graemlins/diamond.gif[/img]2[img]/images/graemlins/spade.gif[/img]), which is about .2. We know both villain's cards here, so 44 cards are unseen. (I probably should have used /44 instead of /45 for the older, less-accurate version of this formula, too; in practice such changes don't alter the results much, though).

For the T[img]/images/graemlins/heart.gif[/img]6[img]/images/graemlins/heart.gif[/img]2[img]/images/graemlins/spade.gif[/img] T[img]/images/graemlins/spade.gif[/img] board, "8/44 * -2 + 36/44 * -1" should be right, because the only spade that villain can hold is the 6[img]/images/graemlins/spade.gif[/img], and hero was already folding if that card came on the river, so it's a non-out.

[Barry]

You guys are way overthinking this, IMHO. Look at the board. There is very little chance that the BB is full here. BB has to have T6, T2, 66 or 22 to be full. Possible, but not likely. 4:1 to make the hand, pot is laying 5.5:1. That's good enough for me.

Change the 6 or 2 to a broadway card, then you just might have a point. Even then, if he was closing or close to closing the action, a call still would be OK.

[TaintedRogue]

EV_trips_spade_kicker = 7/44 * 6.5 + 36/44 * -1 + 1/44 * -2

shouldn't it be 6/44? He cannot catch the spade in his opponent's hand either.