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  #1  
Old 12-06-2004, 07:52 PM
wyattjames4 wyattjames4 is offline
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Default the odds of flopping a set

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.
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  #2  
Old 12-06-2004, 07:58 PM
gaming_mouse gaming_mouse is offline
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Default Re: the odds of flopping a set

[ 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
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  #3  
Old 12-06-2004, 08:15 PM
wyattjames4 wyattjames4 is offline
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Default Re: the odds of flopping a set

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.
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  #4  
Old 12-06-2004, 08:55 PM
gaming_mouse gaming_mouse is offline
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Default Re: the odds of flopping a set

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
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  #5  
Old 12-07-2004, 04:41 PM
wyattjames4 wyattjames4 is offline
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Default Re: the odds of flopping a set

thanks a lot for that answer. your one smart mother f-er. i really appreciate it. thanks.
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  #6  
Old 12-07-2004, 05:31 PM
Dave H. Dave H. is offline
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Default Re: the odds of flopping a set

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
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  #7  
Old 12-07-2004, 06:12 PM
gaming_mouse gaming_mouse is offline
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Default Re: the odds of flopping a set

[ 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
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  #8  
Old 12-07-2004, 06:21 PM
Dave H. Dave H. is offline
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Default Re: the odds of flopping a set

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.
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  #9  
Old 12-07-2004, 07:27 PM
gaming_mouse gaming_mouse is offline
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Default Re: the odds of flopping a set

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
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  #10  
Old 12-08-2004, 01:08 AM
MortalWombatDotCom MortalWombatDotCom is offline
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Default Re: the odds of flopping a set

[ 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.
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