MarkD
MarkD
  • Threads: 2
  • Posts: 5
Joined: Jan 6, 2014
January 6th, 2014 at 8:45:06 PM permalink
On the Wizard of Odds website I found this explanation:

The probability of not making a set is (48+combin(48,3))/combin(50,3) = 17,344/19600 = 88.49%. So the probability of making a set is 11.51%

I don't understand where this formula comes from, what is the meaning of these numbers and why we cannot use this one which seems to me more making sense
combin(2,1)*combin(48,4)/combin(50,5)=0.18 or 18%
If we are talking about getting a set at flop I would put
combin(2,1)*combin(48,2)/combin(50/3)=0.115 or 11.5% It is actually giving the same answer, so it is also correct?

I was also trying to calculate probability of getting a flush at preflop in Texas Holdem if I'm holding a suited hand. So i use this
combin(11,3)*combin(39,2)/combin(50,5) which gives 0.0577 or 5.8% It seems wrong because it is well known that suited hand adds you about 4% equity.
Where I'm wrong?
Thank you!
sodawater
sodawater
  • Threads: 64
  • Posts: 3321
Joined: May 14, 2012
January 6th, 2014 at 8:57:19 PM permalink
I disagree with the wizard's probability of flopping a set.

I think the correct calculation is ((2/50)*(48/49)*(44/48))*3 = 10.776%

Perhaps Wizard is including full houses and quads in his odds?



For making a flush by the river starting with 2 suited cards:

((COMBIN(11,3)*COMBIN(39,2)) + (COMBIN(11,4)*COMBIN(39,1)) + (COMBIN(11,5)))/((50*49*48*47*46)/(5*4*3*2*1)) = 6.4%

according to http://www.homepokergames.com/odds.php
MarkD
MarkD
  • Threads: 2
  • Posts: 5
Joined: Jan 6, 2014
January 6th, 2014 at 10:31:35 PM permalink
Yes, I'm sure that at least this formula
combin(2,1)*combin(48,4)/combin(50,5)=0.18
includes full houses (as technically, full house is part of set) but not quads.
For quads we need to add this one
combin(2,2)*combin(48,1)/combin(50,3)=0.002 or 0.2%

Then yes, the formula fromgives us the same result if we add full houses here.

As for flush, yes this is the same formula I just didn't add the situation when we get 4 and 5 suited cards from flop to river.

But this is what surprises me a lot, because if you enter to Pokerstove (or any other simulator) any suited hand against non suited it will give you 55.5% vs 47.5% thus 5% equity. Not 6.4%

So, now it is pretty clear with the sets even though I still don't understand where this formula comes from (48+combin(48,3))/combin(50,3) = 17,344/19600 = 88.49%
Can anybody explain it to me?

But for the flush I'm still puzzled. Why it gives 6.4% not 5%?
charliepatrick
charliepatrick
  • Threads: 39
  • Posts: 2951
Joined: Jun 17, 2011
January 7th, 2014 at 2:56:27 PM permalink
Quote: MarkD

...The probability of not making a set is (48+combin(48,3))/combin(50,3) = 17,344/19600 = 88.49%. So the probability of making a set is 11.51%...

The method used is as follows.

Say you have A A then to get trips (or better) then one way of working it out is (a) Two Aces and another (b) One Ace and another two compared with (c) Zero Aces and others. You'd do three calculations.

However consider it a different way that is how likely you are NOT to get another Ace on the flop. That's easier as there are 48 non-Aces, so the total of these is combin(48,3). The total of all possible hands is combin(50,3).

A similar method of (1-X) is used to answer the question what is the chance of throwing one or more sixes in four rolls of a fair die. Using reverse logic you throw no Sixes with probability of 5/6 x 5/6 x 5/6 x 5/6. So one or more sixes is 1-(625/1296)=671/1296=51.77%.
MarkD
MarkD
  • Threads: 2
  • Posts: 5
Joined: Jan 6, 2014
January 7th, 2014 at 9:11:09 PM permalink
Thank you for your explanation! This impelled me to do some math... I had spent a couple of hours.... I was in university some time 30 years ago, so I'm slow.
My understanding is that probability of NOT getting any Ace at flop would be simple
combin(48,3)/combin(50,3)=17296/19600=0.882449
Then probability of getting TWO Aces at flop would be combin(2,2)*combin(48,1)/combin(50,3)=48/19600=0.002449
And then if we add probability of NOT getting any Ace and probability of getting TWO Aces we will get the above formula and numbers
(48+combin(48,3))/combin(50,3) = 17,344/19600 = 0.884898
So the above formula (48+combin(48,3))/combin(50,3) shows probability of NOT getting exact the set, so 1-(48+combin(48,3))/combin(50,3) would give us probability of getting the set, but not quads.
I'm OK with that, thanks again for you explanation.

But could anybody clarify the situation with the flush which I described above? I understand that for practical use the difference between 5% and 6.4% is miserable but I'm still intrigued with that.
  • Jump to: