Fourth St Decision

[David Sklansky]

There are two of you left in the pot. The board is JT64 different suits. You have AK. The other player bets into you. You can't make him fold. You know he has JT, QJ, KJ, or AJ. He will play all those hands. Some are more likely only because of cards out.

If an ace, king or queen, hit on the river he will check and call no matter what he has. Assuming that the river bet is the same size as the turn bet, what odds do you need on the turn to make it right to call with your AK? (Both estimaters and calculators should feel free to take a stab at this question. But please identify which one you are.)

[skp]

Edited note: Useless analysis as I forget about the Queen outs for Broadway:

Guesstimate:

Odds of improving are at 13:2 i.e. 6 outs out of 45 unknown cards given that you know he has a Jack.

But the odds of improving to the best hand are only at about 10:1 or 11:1 given that if you improve, he will still have you beat about 40% of the time.

When he checks the river and when you improve, you should bet given your 60%/40% advantage (asuming that he will never checkraise with two pairs). So, you have some implied odds working in your favour. So, I'll say that the pot should have 9 or 10 bigs for you to call you on the turn.

[brad]

5 second guess 4.5:1

[brad]

Queen gives you nuts here. (straight)

[skp]

thx

[skp]

I figure it about 4: 1

Your ways to improve are 10/45

When you improve, you will win about 80% of the time.

But your river bet will be profitable 80% of the time.

Put it all in a blender and I would ballpark it at 4:1. A little worse perhaps.

I probably made another bonehead blunder there
somewhere :-)

[Johnny G]

If you hit an A on the river...you beat QJ and KJ. If you hit a K on the river...you beat AJ and QJ. You need specifically a Q to beat JT, therefore if I felt confident that my opponent was holding specifically JT, I probably wouldn't make the draw on any less than great odds.

So what we're talking about is 4 solid outs to the nut hand (if you catch a river Q), 3 outs on A that should win the hand 2/5 of the time, 3 outs on the K that should win 2/5 of the time.

This is how I do it for a very rough "sitting at the poker table" sort of calculation, and please, correct me if I am wrong...I would count 4 solid outs with the Q's, and reduce the K's and the A's to about 1 out each (it's actually about 1.2 outs each but I always like juicy odds). So I would probably count myself for 6 outs for the purposes of pot odds. (I forget most of my college statistics so I know that this isn't quite right because you probably have to multiply the probabilities or something??)

Therefore, my odds are 40-6 of improving to the winning hand which reduces to not quite a 7 to 1 shot. I would therefore like to have 7 bb's in the pot on the turn to call the bet but would likely settle for 6 bb's since he will check and call if an ace, king or queen falls on the river.

I know this is a rough calculation so please show me a better way!

Thanks!!

[Ulysses]

Estimate:

The 4 Queen outs are always good.

Kings and Aces are only good against 2 of his hands, so instead of 6 outs that's 3. But it's a little more than three because he'se a little bit more likely to have QJ than the other hands, so it's probably more like 3.2 outs.

10 A/K/Q out of 45 cards. So, 10 times, I put in 2 bets.
35 other cards, so 35 times I put in 1 bet.

That's a total of 55 bets I put in. I win about 7.2 times, so I need to make about 7.6 bets each time. The times I win, I put 2 in the turn and river, so that means 5.6 need to be in the pot for me to call the turn.

I'll call w/ 6 BB in the pot.

[Johnny G]

yeah, i like your idea...you can take a look at how my mind works in my posting above...

[VeryTnA]

Guess...
I think you stay in this only for the nut straight.
With that thought in mind you need at least 14+ BB. If the other player is holding a Q (which you previously decided) you only have 3 real outs.
If you "beleive" you are behind and some of your outs (AK)are counterfit is folding a terrible decision?

[Ian]

[MHoydilla]

5.5 BB just a guess no calc. figured hands avg. about 7 outs each with 1 bb on river i would call with 5.5 bb in pot

[mikelow]

The number of outs is either 4, 6, 7, or 7, depending on which hand he holds.

This is an estimate. He will have JT half as likely as the others (not quite accurate).

You average 6.29 outs and you have to call twice on a 6:1 shot. So my estimate is 10:1.

[Ulysses]

The number of outs is either 4, 6, 7, or 7, depending on which hand he holds.

It's 4,10,7, or 7, right?

That's not why I responded, though. Why do you think JT is half as likely as the others? My thought was that since there's a T on the board and I have an AK, JT/JK/JA were all equally likely and JQ is a little more likely. Am I thinking about that incorrectly?

[brad]

AK
Jx

JTxx

so

AAA
KKK
QQQQ
JJ
TTT

so

3 AJ
3 KJ
4 QJ
3 JT

A) AK vs AJ +4 +3 -2 outs +7 -2
B) AK vs KJ +4 +3 -2 outs +7 -2
C) AK vs JT +4 -3 -3 outs +4 -6
D) AK vs QJ +3 +3 +3 outs +9

6*(+7 -2)
3*(+4 -6)
4*(+9 -0)

so

4*(5:1 )
3*(11:1 )
6*(6:1 )

4*(16)+3*(8)+6*(7) +1 = 64+24+42+1= 131/13= 10% or 9:1 plus a bet on end so 8:1.

a little more than 8:1

[Soh]

My poor calculation says if 6BB is in the pot, you win a little over $3 in the long run.

Soh

[Soh]

My new calculation says you win a little less than $0.07 a hand, not $3.00.

Soh

[brad]

if he has QJ then actually that is best for you as then you have 9 outs.

[mikelow]

Ok here goes:

vs. JT--four outs
vs. QJ--nine outs (one is held by your opponent)
vs. KJ--seven outs
vs. AJ--seven outs

So, I would think JT is 3/4 as likely as QJ. There are nine ways for him to hold JT and twelve ways to hold QJ. There are also nine ways for him to hold KJ and nine ways for him to hold AJ. So there are 39 possible hand combinations.

So your odds of winning are (assuming 44 unseen cards) are:

[(1/11 x 3/13) + (9/44 x 4/13) + (7/44 x 6/13)] = .157 or 1 in 6.356.

Thus you need about 5.4 to 1 odds to call the turn.

[brad]

so if you figure the bet you win on the end its 4.4?

gee i did a lot better guessing /forums/images/icons/smile.gif

[mikelow]

That would be coorect if you were going to bet on the end and
were called all the time (which was stated).

[dak]

I came up with 2.5-1 - with the assumption that the other
player would never check raise.

A summary of my calculations:
An A will hit the board and win with probability of .0524
An A will hit the board and lose with probability of .0105
A K has the same probability as an A
A Q will hit the board (always wins) with probability of .0839

You can always bet when an A or a K hits the board

So: 79/100 times lose one bet
2/100 times lose 2 bets
19/100 win 2 bets

So I came up with 19X + 38 - 83 = 0
Solving for X I get 2.3

So if there are 2.5 bets in the pot you can call

[deadbart]

But some of the times you bet on the end and get called, you will lose. Or you can only bet when the Q comes, and always win. But either way the math is more complicated.

[mikelow]

Yes you will lose when an ace or king comes and you lose to two pair. Gee, it's getting complicated!
Always beat QJ if you improve, an ace or king loses to JT, ace loses to AJ, and king loses to QJ.

The chance of improving and losing is:

[(6/44 x 3/13) + (2/44 x 6/13)] = 30/572 = .052

So about 1/4 of the time you do improve, you lose. If you are checkraised some of the time, that worsens your odds a bit, so maybe you need about six big bets in the pot.

But if you could read your opponent more accurately, my answer of 4.4 big bets stands.

[deadbart]

Given the many ridiculous mistakes made by the other posters, some of whom I'm certain are smarter than me, and others of whom I suspect of being smarter than me, I'll probably @#*% this up.

We'll start with some things the others got right:

AJ and KJ can be treated as one with twice the likelihood.
The ratio of occurences of AJ/KJ:QJ:JT is 6:4:3.
The respective number of outs is 7:9:4.

We also need to know the how many "negative outs" we have, where we hit a card that helps our hand but doesn't win us the pot: 2:0:6.

So first lets assume I always bet when I hit any of my possible outs.
Let "pot" be the size of the pot on the turn before he bets (as a programmer, I hate just using 1 letter variables; sorry if it looks weird) (it would be easier to at first to let pot be the size of the pot after he bets, but doing it this way makes it easier to see what happens if I don't bet on the river, I think/hope).

EV vs. AJ/KJ: ([2+pot]*7 - 2*2 - 35)/44 = (7*pot - 25)/44
EV vs. QJ: ([2+pot]*9 - 35)/44 = (9*pot - 17)/44
EV vs. JT: ([2+pot]*4 - 2*6 - 34)/44 = (4*pot - 38)/44

So now we take the weighted sum of the EVs:

(6*[7*pot - 25] + 4*[9*pot - 17] + 3*[4*pot - 38])/(44*13) = (90*pot - 332)/572

Set that equal to zero and solve for pot: pot = 3.68888888888...

So if the pot is 3.69 bets or better, you should call. To answer Sklansky's question specifically, you need pot odds of 4.69 to 1 or better, assuming you always bet when you improve.

The only real alternative to always betting is to only bet when you make the straight. In this case you win 3 less bets when you beat AJ/KJ, but lose 2 less when you improve to a 2nd best hand. Against QJ, you win 6 less bets. Against JT, you save 6 bets. 6*(-3+2)-4*6+3*6 = -18. So you make significantly less if you don't always bet when you improve.

Let me know what I did wrong /forums/images/icons/wink.gif.

David "DeadBart" Bartholow

[deadbart]

[Gabe]

I estimate 5 to 1.

[Aggressor]

My calculation:

Opponent's Hand / Number of Ways / Percent of Ways / Your Outs to Win / Probability / Your Outs to Bet and Lose / Probability

JT / 9 ways / 23% / 4 outs / 9% / 6 outs / 14%
JQ / 12 ways / 31% / 9 outs / 20% / 0 outs / 0%
JK / 9 ways / 23% / 7 outs / 16% / 3 outs / 7%
JA / 9 ways / 23% / 7 outs / 16% / 3 outs / 7%

Thus, you will win 15.7% of the time, and you will catch an out, bet, and lose 6.3% of the time. (By the way, you should always bet if you catch an A or a K because you are likely to have your opponent beat. For example if you catch an A, you beat JQ and JK (21 ways) and lose to JT and JA (less than 18 ways since you now can account for 2 Aces.)

Your expected value calculation looks like the following, with X = the size of the pot before your opponent bets fourth street:

15.7% x (X + 2) + 6.3% (-2) + 78% (-1) > 0

or X > 3.76.

The X + 2 comes from the fact that if you catch an out and win, you win the pre-4th street pot + 2 BB. The -2 comes from the fact that you lose 2 BB when you catch an out, bet the river, and lose. And the -1 comes from the fact that you lose 1 BB when you miss all of your potential outs (since you check the river).


So you need 3.76 BB in the pot heading into fourth street to make it a call.

[Kevin J]

If you're good enough to always check a worse hand and bet a better one on the river, I'd GUESS a little under 5. But this is a generous assumption and I'd say that the average player would need more than this to make a call profitable. How much more, depends on his ability to collect rather than lose a bet those times he makes a pair on the river. IMO-

[mikelow]

When he has AJ--there are two cards that improve and lose. Same with KJ. So, the improve and lose percentage is 5.2%.

[dak]

My terminology was incorrect and I had a small error in my
calculation. The correct odds are 3.56-1. I didn't count
the original bet in the pot so 2.5-1 should have been 3.5-1.
Also I made an error in the calculation forgetting about the
case where you draw an A or K and lose to JT.

A quick summary:
Ways for opponent to have hands:
JT 9
QJ 12
KJ 9
AJ 9

Drawing an A (or K)
3/44*21/39 = .036 Win
3/44*9/39 + 2/44*9/39 = .026 Lose
Since you win more often than you lose drawing an A or K you
can alway bet. Adding them together
.072 win an extra bet
.052 lose an extra bet
Drawing a Q
4/44*27/39+3/44*12/39 = .084 win an extra bet

Drawing anything else
34/44*9/39+35/44*12/39+35/44*9/39+35/44*9/39 = .79 Lose 1 bet

So out of 100 times
79 lose 1 bet
5 lose 2 bets
16 win the pot + your turn bet + an extra bet
If X is the pot size when you have to call one bet on the turn:
16X +16 + 16 - 89 = 0
X = 3.5625
Since I rounded .0156 up to 16 I would actually want 4-1 in this case.

[pudley4]

I got the same answer

Let x=pot size after his turn bet

If he has AJ you win x+1 (the pot plus one bet) 7/44 times; you lose 2 bets (your turn call and river bet) 2/44 times; you lose 1 bet (your turn call) 35/44 times. The EV formula is:

7/44(x+1) - 2/44(2) - 35/44(1)

If he has KJ, the formula is 7/44(x+1) - 2/44(2) - 35/44(1)
If he has QJ, the formula is 9/44(x+1) - 35/44(1)
If he has JT, the formula is 4/44(x+1) - 6/44(2) - 34/44(1)

Now multiply each formula by the probability of facing that specific hand (6/26 for AJ, KJ, JT, and 8/26 for QJ), then add all 4 results together and set it equal to 0. Solve for x and you get:

Pot odds of 4.68888888... to 1 on the turn.

[David Sklansky]

Obviously

[Rick Nebiolo]

David,

I’m a very quick estimator today. /forums/images/icons/grin.gif

You have almost four queen outs to the nuts and I’d reduce my AK outs to about four instead of six to adjust for the times my outs are no good (he is twice as likely to have a hand where an ace or king is good for you). That’s eight outs or so.

With approximately eight outs, I have 38 bad cards and 8 good ones so I’d need almost five to one. I’d need a pot containing four big bets since I always bet the river and pick up a bet when I win.

Not perfect but close I think.

Regards,

Rick

[mikelow]

You came close. Where I tripped up (and others also) is where where you lose two bets 1/4 of the time you improve and win two bets 3/4 of the time you improve.

That should be an expectation of one bet everytime you improve.