what are the odds that your turn/river a boat after a flopped set?

[theonlycheese]

As the subject line says .. what are the odds that you turn/river a boat after a flopped set?

[jay_shark]

Given that you have flopped a set , the probability using the 4X rule is 1*4=4% . You have one out and you expect to see two more cards .

If you only peel off one card then you use the 2x rule which is about 2% .

[BruceZ]

[ QUOTE ]
Given that you have flopped a set , the probability using the 4X rule is 1*4=4% . You have one out and you expect to see two more cards .

If you only peel off one card then you use the 2x rule which is about 2% .

[/ QUOTE ]

You don't just have 1 out to make a boat, you have 6 outs on the turn, and 9 outs on the river if you miss on the turn. The probability that you make it on the turn is 6/47 =~ 12.8% or 6.8-to-1. The probability that you make it on the river after you miss on the turn is 9/46 =~ 19.6% or 4.1-to-1. The probability that you make it on the turn OR miss on the turn AND then make it on the river is

6/47 + (1-7/47)*9/46 =~ 29.4% or 2.4-to-1.

We multiply the 9/46 by the probability that we miss the full house and quads on the turn (1-7/47). Note that you can also make quads, and the probability that you make a full house or quads by the river is

7/47 + (1-7/47)*10/46 =~33.4% or 2-to-1.

EDIT: Changed the factor (1-6/47) to (1-7/47) in the first equation for full house alone, since we don't have outs to make a full house if we make quads on the turn. The second equation for full house or quads is really the important one.

[jay_shark]

oops , I meant quads Bruce and I confused quads with boats .

More coffee for me ...

[Voltaire]

Let's see if I can do this right. I'll do the probability for making a full house AND getting four of kind, which is really what you want to know.

After the flop you have seven cards which will improve your hand, the fourth of your set and three of each of the other two cards on the flop. There are 47 cards left in the deck. So the probability that you will fill up or make four of a kind on the turn is 7/47 or 14.89%.

Now if you don't improve your hand on the turn, there will now be 10 cards that can help your hand, the fourth of your set and three each of the other three cards on the board. Since there are now 46 cards left in the deck, the odds of helping your hand with the river card are 10/46 or 21.74%.

The tricky part is to figure the probability of helping your hand on EITHER the turn or river. I will do this by figuring the probability that you won't help your hand and then subtracting that from 1.

Since you have 7/47 ways to help your hand on the turn, you therefore have 40/47 ways of not helping your hand. For the river the chances you won't help are 36/46. Multiplying these together we get the probability that you will face a showdown with only a set: 66.60%. Subtracting from 1 we get 33.40%. That is the probability on the flop that you will make four of a kind or fill up before the hand is over.

[Voltaire]

Actually my answer in the previous post is not QUITE correct. The probabilities I gave for the turn alone (14.89%) and for the river alone (21.74%) are correct. But in figuring the probability of helping your hand on EITHER the turn or the river, I forgot to figure in the probability of a full house coming from the infamous RUNNER-RUNNER! In other words, the turn and the river cards could be a pair different from any cards on the flop and you would also have a full house.

Consequently the way to figure this is a little different from the above. Here's how I did it:

First figure the probability of NOT getting the case card of your set on either the turn or the river. That is 46/47 times 45/46 equals 95.74% (rounded). Subtract that from 1 = 4.26% (rounded).

Next figure the probability of NOT getting another of one of the other cards on the flop: 44/47 times 43/46 and subtract that from 1 = 12.49%.

Do the same with the other card. Also equals 12.49% (rounded).

Now figure the probability of getting a runner-runner from the ten ranks that are not represented on the flop. Since there are 1081 ways to get 47 cards taken two at a time [47C2 = (47!/45!) = (47*46/2) = 1081], and since there are 6 ways to make a pair 4C2 = 6 and since there are 10 of these pairs possible, the probability of a runner-runner pair is 6*10 over 1081 which equals 5.55%.

Add 4.26% to 12.49% to 12.49% to 5.55% (all rounded figures) and the total probability of getting four of kind or a full house by the river is 34.78% which is a little higher than the 33.40% that I gave in the previous post.

There probably is a more elegant want to do this. If somebody knows it, please advise.

By the way, how to figure combinations, which is the essences of figuring poker probabilities, is given in Sklansky's "Getting the Best of It."

[Voltaire]

I think you made the same error I did in my first post: you forgot to figure for a runner-runner full house!

Note my second post where I give the problem the full treatment--well, at 99% treatment. I THINK i got it right!

[BruceZ]

[ QUOTE ]
I think you made the same error I did in my first post: you forgot to figure for a runner-runner full house!

[/ QUOTE ]

I didn't forget anything, and neither did you until your 2nd post which is the only error here. I took the runner-runner case into account when I changed the number of outs from 7 to 10 when we miss on the turn, and you changed the probability of a non-out from 40/47 on the turn to 36/46 on the river which is the same thing. The extra 3 outs are for the river card pairing the turn card.

[ QUOTE ]
Note my second post where I give the problem the full treatment--well, at 99% treatment. I THINK i got it right!

[/ QUOTE ]

Your second post is completely unnecessary, and it is also incorrect because you can't add those probabilities together due to overlaps. That is why you got an answer that is too big.

The original calculation and the 33.4% result are correct and well-known.

[Voltaire]

Thanks Bruce. I see that you are right.