A few probabilities, simulations and other mathematics.

[RayGarlington]

I always feel a little frisky when I see big overcards like A,K,Q, vs. a measly low pair and no kicker, BUT heads up:

pokenum -mc 500000 -7s ac kd qh - 2c 2d 3h
7-card Stud Hi: 500000 sampled outcomes
cards win %win lose %lose tie %tie EV
Ac Kd Qh 227312 45.46 272668 54.53 20 0.00 0.455
2c 2d 3h 272668 54.53 227312 45.46 20 0.00 0.545

[Andy B]

Beer's numbers are correct for this calculation (3:2 against is the number given in Morehead's book). You can only go by unseen cards at the point in the hand that you want to make the calculation for. If you're going to take into account the number of cards dealt on fifth street, you must also take into account the ones that would make your ahnd and the ones that don't. Also, on fifth, there is the chance that you will make your hand anyway, so you won't much care whether your cards are live or not.

[Andy B]

You had it right the first time. The numerator goes down by 3 because there are only three cards that can pair your fifth-street card. Your calculation still isn't exact, because some of the cards you can catch on fifth or sixth will be somewhat dead. You'd have to create branches for the dead cards and the live, and even branches for when you catch a card that has had two of its rank show, and that frankly strikes me as being a lot of work. Knowing your odds of filling up when you have trips isn't going to help your game very much anyway, because you should usually play trips very strongly, and almost never fold them. Now if I could just figure out how to play two small pair....

[7stud]

[ QUOTE ]
You had it right the first time. The numerator goes down by 3 because there are only three cards that can pair your fifth-street card.

[/ QUOTE ]

The problem is you don't subtract the cards that make your hand from a previous street's numerator--you subtract them from the total cards left, which is the current street's denominator. In other words, on 6th street there are 38 cards left in the deck, and now instead of 4 cards making your hand, there are 3 more that will make your hand for a total of 7 cards that make your hand. Since 7 of the 38 cards left make your hand, then 38-7 = 31 of the cards don't help, from which you get the 31/38 probability of not making your hand. You can't get 31 by subtracting 3 from the previous street's numerator.

The reason subtracting 3 from the previous street's numerator of 35 doesn't work is because that 35 number was gotten from the total number of cards left: 39 - 4 = 35. On 6th street, if you merely subtract the additional 3 cards that make your hand from 35, what you are really doing is: 39 - (4 + 3), but there aren't 39 cards left on 6th street. On 6th street, there are only 38 cards left, so you have to do this calculation instead: 38 - (4 + 3).

[BeerMoney]

Andy, I think 7 stud is right..

Work backwards..
On sixth street, let's say you have:

3 3 3 8 9 q

Your outs are single 3, 3 8's, 3 9's and 3 q's.
That is ten outs..

On fifth street..

3 3 3 8 9

You would have 7 outs.. (same stuff)

And on fourth street you would have four outs..

So, assuming 37 cards unseen by the river..

It would go

1-(27/37)*(31/38)*(35/39)

I think that's right..

[Andy B]

OK, he's right. Why does this stuff have to be so complicated?