Pot odds question

Hi all


Have I misunderstood something about pot odds, or is there something wrong in the book. I`d read the book "seven card stud poker" by kontantin othmer on page 127 the book (table) says, that you can call trip tens vs trip aces 6.street, if there are 7 units in the pot, or the same as 70$. Assumed the hand is checked on the river. One unit is equal to 10$ in a 10/20$ game.


I have furthermore simulated the hand in pokerwiz, and it says that trip aces vs trip tens is 85% vs.15%. If we assume that this is true:

I call 20$ and win 15 times 70$ + my opponent 20$, the same as:

15 * 90$ = 1350$


I lose 20$ 85 times, the same as:

20 * 85 = 1700$


1350 - 1700 = -350$


For me that must be a call with negative expectation, and therefore I cant call. Have I misunderstood something ??


I appreciate any help.

I'm not sure if I can figure it out either, but here it goes. Trip tens verses trip Aces means your

beat at this point. Assuming each player's hands are equally live, each opponent has the same

chance of fulling up on the end. Therefore, He either fulls up and you don't, You fill up and he

doesn't, you both full up, or neither full up. Out of those four scenerios, you lose three times out

of four. Therefore, on this simplistic analysis, which I hope some of you math majors can correct

me on, you should call if the pot is giving you four to one -- But wait!


Your call on sixth street almost always commits you to call on the river as well, unless we

assume you are confident enough in your read to trash trip tens or if the trip aces are exposed. In

that case you will only call if you improve. Therefore, you call if you full up and he fulls up, or if

you full up and he doesn't full up. This means you will actually be putting in two bets rather than

just one at least half the time.


Ok, the math is too much for me. But assuming you have a one in four shot at beating him, but

you have to put in two bets at least half the time to find out, having at least six bets in the pot is

necessary, and seven in probably right. One less bet than you really need, i.e. eight (two times

four) because you can fold when you don't improve, but must call when you do.


Notice, there are no real implied odds here.

I haven't read this book. As Rich points out, the only way for you to win is for you to hit and for your opponent to miss. Considering each hand in a vacuum, there are ten helping cards and 36 non-helping cards, so the odds against filling are 36:10 or 3.6:1. You would, of course, adjust this based on the cards that have shown. The rough probability that you will hit while your opponent misses is (10/46)*(36/46) which is .17, or about 5:1 against. You also stand to lose an additional bet or more on those occasions that you both improve, but I guess that's being ignored.


I think that in a $10/20 game, a "unit" would be $20, not $10. $20 is the bet you're going to have to call, so I'd want $100 and change in there if I really thought the other guy had trip Aces. In practice, I almost never fold trips, so the point is moot. /images/smile.gif

I haven't read the book, but pot odds aren't the issue anyway. It's implied odds.


If your sim is right, then you're an 85:15 dog which comes out to 5.7:1. You know your opponent will bet/call the river no matter what (nobody's going to lay down trip aces) whereas you are going to fold if you don't fill up, so you can include your opponent's river bet in the current calculations. If the bet size is $20, then you need only $100 in the pot now, plus the $20 your opponent will put in on the river, to show a small theoretical profit.


It's not that simple, however, because the cards that are out could profoundly affect both your and your opponent's chances of filling up. So the sim figure is fairly useless except as an extremely general rule of thumb.


TRLS

Hi


I appreaciate your answers, but I still dont understand. Because the book says on page 127 "the table assumes the hand is checked on seventh street and that an average number of your side-cards have been seen in other hands".


Furthermore the book says´"the following table shows the size of pot necessary to call a two unit bet in situations where you are an underdog" Remember we are talking about 6. street, thats why I know that 2 unit must be the same as 20$ in a 10/20$ game, and therefore 7 units must be the same as 70$.


Morten

It's not going to be checked on the river, but assuming it is, the calculations do seem a little off, but only by one bet. I think you need at least $80.00 in the pot, and perferably $100.00, because you are a 5 to 1 shot.


Your 85/15 stat is probably based on earlier streets. With only one card to come, either you improve or you don't.

My simulation is on 6. street and i have used pokerwiz. I have simulated exactly this hand:


As-Ah-Ad-2d-5k-12s vs. 10s-10h-10d-2k-5h-12k


pokerwizz says: 85% - 15% on 6. street, after 5M hands simulated.

OK im sorry, now i understand.


I have simulated a new hand with a little difference, the first simulation was too dead for the tens, because its the hand that mostly need to improve. The new simulation goes like this:


Ak-As-Ah-2h-8d-Ks vs. Ts-Tk-Th-2d-9h-Jd


pokerwizz says now: 81,5% vs. 18,5%


wins: 18,5 * 9 = 166,50$

losses: 81,5 * 2 = 163$


Now the needed cards are more live for the tens, and that makes the difference. And now it would be right to call in the example decribed in the book. im sorry I bother you guys.