Question Regarding Three-Of-A-Kind Odds...

First off, I am currently reading David Sklansky's "Theory Of Poker" and I have to say that it has really opened my eyes. I am an above-average player and I love tournament play, and this book has proved to be an extremely valuable investment. One small thing in the book left me confused, however, and I wanted to see if anyone could answer what is probably a simple question.

On the first page of the chapter about implied odds, toward the bottom, Sklansky says that beginning with a low-pair in the hole, the odds of hitting trips on the flop are 8-to-1. Is this really the odds of hitting trips on the flop in this circumstance? I'm thinking of this situation the following way: if you have two sevens in the hole, say, there are only two other sevens left in the deck. If you are playing a 10-handed game, there are 20 cards out of the deck at the moment, thus there are 32 cards left in the deck. It seems to me that the odds then of hitting that three-of-a-kind would be 32-to-2 (the 2 being since there are two sevens left in the deck), which reduces to 16-to-1. In a four-handed game, the odds seem even worse (44-to-2, which reduces to 22-to-1). Please tell me, where is my thinking flawed here? Sklansky didn't say how many people are at the table in that particular example, but 8-to-1 odds to hit trips seemed a bit better than what I would think it is. Thanks in advance for your response(s)!

[stigmata]

2 begginers flaws in your thinking:

1) It doesn't matter how many people are in the game. You do not know what cards other people hold. Therefore there are 50 unseen cards, period. 2 out of 50 cards can make your hand. You do not know whether these cards are in someone elses hand or are the next card in the deck. Therefore YOU HAVE to treat all unseen cards as equal.

2) There are three cards on the flop. The way you calulated it (e.g. 2 out of 50 cards help our hand) would be correct if the flop was only 1 card. However, when there are 3 cards on the flop, your odds of hitting a matching card with your pocket pair are actually about 7.5 to 1.

Hmmm...can you possibly detail how you arrived at that mathematically, though? This is what I'm having trouble understanding...[img]/images/graemlins/confused.gif[/img]...

there are 3 cards on the flop, so it should be 2/50+2/49+2/48 minus the probability you flop quads (since we want a set exactly) and since you only have information about 2 cards so far. this comes out to be .12248 - (2/50*1/49 + 2/49*1/48 + 2/50*1/48), which ends up as .119, and 8 to 1 is .111 repeating.

and 7.5 to 1 is .1176, so yeah this should be right.

OK, so that's exactly what I wanted to see. Is there any Web site you know of or any part of this Web site that shows the odds of hitting certain hands with certain hole cards, all things equal, just for reference? Thanks again.

[stigmata]

ladeeda

most of those are for outs, not what he's looking for.

Awesome, thank you very much. I knew there was some kind of problem with the way I was looking at the example. It's not really a huge thing either way, since you will probably play a pair of hole cards to see the flop, but I just wanted to make sure I understood how Sklansky arrived at 8-to-1 odds to hit those trips.

Well, that's true, but that Web site still serves as a good reference tool. You explained how that 8-to-1 thing worked out though, so I thank you for that.

[stigmata]

The first hit explains quite well how to calculate most of the common things that you need. You should be able to go from there...