the odds of flopping a set

[wyattjames4]

the way ive always conceived the odds of flopping a set is the traditional 2 outs outta 50 cards, three times, equals 6 outs outta 50 equals a 12 percent chance or 7 and 1/3 to 1. but im wondering if thats the right way to figure it out. shouldnt the question be, of the remaining 50 cards in the deck, how many 3 card combinations contain one or both of the same ranked card in your hand? i dont know, you tell me. and if that is the question anyone know an easy way to work the math on it.

[gaming_mouse]

[ QUOTE ]
shouldnt the question be, of the remaining 50 cards in the deck, how many 3 card combinations contain one or both of the same ranked card in your hand? i dont know, you tell me. and if that is the question anyone know an easy way to work the math on it.

[/ QUOTE ]

Yes, that is the correct question. Your solution give a good approximation, though. The correct calculation is:

1 - (48 choose 3)/(50 choose 3) = .1175, still close to 12%.

gm

[wyattjames4]

Im not even gonna try to figure the math on that, but i do have another question. And this is what brought me to my original question. Lets say your playing some weird game where you got dealt 13 cards. and lets say you got dealt all 13 diamonds so that the remaining 39 cards consist of clubs, hearts, and spades. now lets say you ask the question what are the odds that the flop will contain a club. the math would be 13 clubs outta of 39 cards is 13 outs 3 times. that equals 39. so 39/39 is 1 which means their is a hundred percent chance the flop will have a club. but obviously there are many flops that wouldnt contain a club, so where did the math break down there. i have no clue so it wont bother me if no one else has an answer to this, but id appreciate an answer.

[gaming_mouse]

When you have 13 outs 3 times, you can't just add your chances together. In general, if you are trying to find the probability of event A of event B, you can add the chances together ONLY if the events are MUTUALLY EXCLUSIVE -- that means it's impossible for them to happen at the same time.

In your case, you are looking at the three flop cards as three separate events. Each event has a chance of 13/39 of occurring. However, it possible that the events happen at the same time: ie, the first card AND the second card could both be clubs. Because of this, you are not allowed to add the probabilities together.

The correct way to solve this kind of problem is to calculate the chance that NO clubs appear on the board, and then subtract your answer from 1. Starting with the first flop card, there are 26 non-club cards out of 39 cards total. Thus:

1 - (26/39)*(25/38)*(24/37)= .71

is the answer.

gm

[wyattjames4]

thanks a lot for that answer. your one smart mother f-er. i really appreciate it. thanks.

[Dave H.]

gm,

To carry this a bit further, does this mean that to determine the probability of making a set by the turn (assuming a pocket pair), one would simply compute

1 - (probability of NOT making the set on the flop AND NOT making the set on the turn)= 1 - ((C(48,3)/C(50,3)) * 45/47) = 1 - (.8825 * .9574) = 1 - .8450 = .1550

and then to compute the probability of making a set by the river, one would simply compute

1 - (probability of NOT making the set on the flop AND NOT making the set on the turn AND NOT making the set on the river)= 1 - ((C(48,3)/C(50,3)) * 45/47 * 44/46) = 1 - (.8825 * .9574 * .9565) = 1 - .8082 = .1918

[gaming_mouse]

[ QUOTE ]
gm,

To carry this a bit further, does this mean that to determine the probability of making a set by the turn (assuming a pocket pair), one would simply compute

1 - (probability of NOT making the set on the flop AND NOT making the set on the turn)= 1 - ((C(48,3)/C(50,3)) * 45/47) = 1 - (.8825 * .9574) = 1 - .8450 = .1550

[/ QUOTE ]

Yes. Note that you also just do:

1 - (48 choose 4)/(50 choose 4) = .155, same answer you got.

gm

[Dave H.]

Ahhh...I see!...

OK, can we now make it a bit more difficult please? I still have that pocket pair, but now I want to know about quads, i.e. the probability that I will make quads on the flop, then by the turn, then by the river.

[gaming_mouse]

This should be easy.

First, how many flops contain quads? Well, two of the cards are already determined. That only leaves the final card, which can be any of the remaining 48 cards in the deck.

So 48/(50 choose 3) is the chance of quads on the flop.

On the turn we can make our quads 48*47 possible ways, since we have 2 free cards which may vary.

So 48*47/(50 choose 4) is the chance of quads by the turn.

River is similar.

gm

[MortalWombatDotCom]

[ QUOTE ]
This should be easy.

First, how many flops contain quads? Well, two of the cards are already determined. That only leaves the final card, which can be any of the remaining 48 cards in the deck.

So 48/(50 choose 3) is the chance of quads on the flop.

On the turn we can make our quads 48*47 possible ways, since we have 2 free cards which may vary.

So 48*47/(50 choose 4) is the chance of quads by the turn.

River is similar.

gm

[/ QUOTE ]

i think it should be (48 choose 2)/(50 choose 4) for the turn, which is half of your number.

[gaming_mouse]

[ QUOTE ]


i think it should be (48 choose 2)/(50 choose 4) for the turn, which is half of your number.

[/ QUOTE ]

Yep. You're right.

gm

[Dave H.]

Hmmm...I'm always trying to make these more difficult than they are, I guess. The flop was easy. And the turn turned out to be OK too.

I can see now, based on your explanation, that the river is just C(48,3)/C(50,5)...CORRECT?

But when I was thinking about it (quads by the river) prior to seeing your explanation, I was saying to myself:

I could get one of my paired cards on the flop, miss on the turn, and get it on the river. Or I could get none of my paired cards on the flop, but get one on the turn and the river, etc. Then I started the "Are they mutually exclusive" stuff, and then I got twisted up in my shorts...This probably just comes with practice, but if you have any "mental process" that you can come up with, it would be appreciated. I realize that's a pretty nebulous request, but I guess it can't hurt to ask.

I suppose that, after seeing your formulas, what I should have said to myself (about the river) was:

a. How many 5 card combinations are there in the unseen 50 cards?

b. Two of the 5 cards must be my pocket denomination. So that leaves 48 cards to be chosen 3 ways.

[gaming_mouse]

Dave,

Yes, your river answer is correct.

As far as mental process, I think you're getting it. There's no magic formula except for seeing lots of problems solved -- after that, you'll start to see recurring patterns.

With poker calculations, I would say there are easily less than 10 "types" of problems, all requiring only basic combinatorics. Just keep messing around with problems, and they'll get easier and easier. I can already tell you know alot more than you did after your first post.

gm