PDA

View Full Version : flopping a set

03-25-2003, 08:08 AM
It is my understanding that the probability of making a set IF u stay all the way to the river is just under 20%
My question is what is the probability of making a set on the FLOP

03-25-2003, 08:13 AM
I forgot to add while holding pocket pairs.....sorry

SunTzu68
03-25-2003, 09:09 AM

ie.

2/50 (4%) + 2/49 (4.08%) + 2/48 (4.17%) =12.25%

pudley4
03-25-2003, 12:14 PM
approx 11.75% (or 7.5-1)

pudley4
03-25-2003, 01:10 PM
Using your way (adding the probabilities of each card) we need to make a correction:

The chance of hitting the set on the first card = 2/50 = 4%.
The chance of hitting on the second = 2/49 has to be multiplied by the chance of missing on the first (48/50) = 3.918%
The chance of hitting on the third = 2/48 has to be multiplied by the chance of missing the first and second (48/50 * 47/49) = 3.836% Add them up to get 11.754%

Another way: There are 19600 total flops. 17296 (48*47*46/6) do not contain one of your pair. 2304 contain one or both of your cards. 2304/19600 = 11.755%

BruceZ
03-25-2003, 05:27 PM
This is correct for a set or quads. For just a set, you can do this just with fractions as (2/50)(48/49)(47/48)*3 = 11.510% or 7.7-1. The reason we multiply by 3 is because the set can be made on any of the 3 cards.

So this would say the probability of quads must be 11.755% - 11.510% = 0.245%. To check this, the probability of quads is (2/50)(1/49)(48/48)*3 = 0.245% so it checks. Here we multiply by 3 because the extra card can be in 3 places.

BruceZ
03-26-2003, 12:26 PM
This still isn't the probability of just a set, because we are still including flopping a full house. The probability of just a set is (2/50)(48/49)(44/48)*3 = 10.776% or 8.3-1. The probability of a full house is (2/50)(48/49)(3/48)*3 = 0.735%. The probability of quads as we said is (2/50)(1/49)(48/48)*3 = 0.245%. These add up to 11.756% or 7.5-1 for a set or better.

SunTzu68
04-01-2003, 02:25 PM
Pudley or Bruce,

Maybe you can help me with the remedial math. I understand how you got to the percentage, however the odds do not match up for me. Let me walk through what I did, and if you could tell me where I went astray I'd appreciate it.

I took the odds to make the set on each card and added them together (ie. 50 cards left in the deck minus 2 cards for outs gets me to 48-2)....(1/24)+(1/23.5)+(1/23) and I got 7.83-1 to make a set.

SExyBeast
04-01-2003, 08:28 PM
or put another way pp in the hole ur about 8-1 shot to make set /forums/images/icons/grin.gif gl hf

BruceZ
04-03-2003, 07:50 AM
I took the odds to make the set on each card and added them together (ie. 50 cards left in the deck minus 2 cards for outs gets me to 48-2)....(1/24)+(1/23.5)+(1/23) and I got 7.83-1 to make a set.

There are several problems here. First of all, 48-2 odds is not 1/24, it is 2/50 = 1/25. Second of all, this sum adds up to 1 IN 7.83, but that is 6.83-1. Third of all, the second term cannot be 2/47 = 1/23.5 because this is just the probability of making a set on the 2nd card when you miss it on the 1st card. You have to multiply that by the probability that we also didn't make it on the 1st card as pudley did above. This still would include full houses. You can also break it down as I did.

Here's another way to do this. Before the flop, the probability of making a set any given card is 2/50. If we take 3 times this or 2/50 + 2/50 + 2/50 = 6/50, that would be the probability of making it on any of the 3 cards, except that we would be double counting all the times we make quads because when we add the second 2/50, that counts all the times we make quads on cards 2 and 1 which we already counted, and when we add the third 2/50, that counts all the times we make quads on cards 3 and 1 and on cards 3 and 2 which we already counted. Since all quads are counted twice, we can subtract the probability of making quads, which we found above is 0.245%. So 6/50 - 0.245% = 11.76% or 7.5-1 for a set or better.