sorry for this beginner question but:
I have been trying to find the prob of floping 2 pair or better (2pair -full house) with two unpaired cards and It just doesnt seem right to me.
please if anybody can point out my mistake It would be awsome .

why is this wrong?

6x5x48/3!= 240 240/19600= 0.0122 ansx100 = 1.22%

using both my 2 cards

You have to do the full houses (and quads) separately since multiplying by 48 will count them multiple times. For 2-pair or trips you multiply by 44 unpaired cards instead of 48, and then divide by 2 not 3! since there are 6*5/2 combinations of 2 cards that match your hand, times 44 ways to choose the unpaired card. Assuming you want 2-pair made with both hole cards, trips, full house, or quads, it is

(6*5*44/2 + 6*5*4/3!)/19600

=~ 3.47%

where the 6*5*44/2 is for 2-pair or trips, and the 6*5*4/3! is for a full house or quads. It's not clear whether you wanted to include trips or quads. It looks like you were trying to include trips, but then you say that you are using both hole cards. If you mean that you have to pair both hole cards, then you can't make trips or quads. You can also calculate each hand type separately:

2-pair: 3*3*44/19600 =~ 2.02%

trips: 2*(3*2/2)*44/19600 =~ 1.35%

full house: 2*(3*2/2)*3/19600 =~ 0.0918%

quads: 2/19600 =~ 0.0102%

================================
total =~ 3.47%


as before. Note that the factor of 2 for the trips and full houses is for the 2 ranks that can make the trips.

Hi Bruce,
Thanks alot for your answer!

so from what I gather because there are only two selections of a specific amount of cards from the pool it is devided by 2!=2 right?
ya I tried to do 6x5x48/2 ans/19600 and what i came up with was 3.67
so is it fair to say that the full houses and quads were counted 3 times??? cause
3.67~ -0.2~= 3.47~ .
hey again thank you for your detailed answer I wasnt sure that I can add them up like that but now that I know this I can do what I wanted to do.

The divide by 2 is for the 6*5/2 combinations of 2 cards that pair your hand since you don't want to count each combination twice. Then each of those 2 card combinations can be combined with 44 other cards, so multiply by 44 to get total 3-card combinations. The 19600 in the denominator doesn't count different orderings (permutations) of the cards, so the numerator must also be combinations without order. Note that the full houses and quads divide the 6*5*4 by 3! since in that case there are 3! ways to order all 3 cards.


Yes, that will count each full house 3 times, and each quad 3 times. When the 3rd card makes a full house or quads, that combination will also be counted 2 other times when that same card was one of the first 2 cards.

6*5*48/2/19600 =~ 0.036734694

which exactly matches what you get if you take

2-pair + trips + 3*(full house) + 3*quads

from my previous post.


You can add the probabilities because the hands are mutually exclusive. That is, a given hand can only be in one of the categories.

Can't you add straights and flushes?

If your 2 unpaired cards are 3-gapped at most, you could flop a straight.
If your 2 unpaired cards are suited, you could flop a flush.

P(<=3 gapped|2 unpaired) * P(straight on flop)
+ P(suited|2 unpaired) * P(flush on flop)
- P(<=3 gapped && suited|2 unpaired) * P(straight flush or royal flush) => double count of straight flushes/royal flushes

Or

1 - (1 - P(<=3 gapped|2 unpaired) * P(straight on flop))
* (1 - P(suited|2 unpaired) * P(flush on flop))

You can simply add those to the 3.47% because they are mutually exclusive (straights/flushes and all paired scenarios).

Keep in mind:
P(<=3 gapped|2 unpaired) * P(straight on flop) =
P(0 gapped|2 unpaired) * P(straight on flop)
+ P(1 gapped|2 unpaired) * P(straight on flop)
+ P(2 gapped|2 unpaired) * P(straight on flop)
+ P(3 gapped|2 unpaired) * P(straight on flop)

I also assume P(straight on flop) includes straight and royal flushes.
And P(... on flop) is the probability given a certain starting hand.

This would give the probability for some unspecified unpaired hand. I think he wants to know the probability for any particular hand, and that depends on the exact hand in question. It didn't look like he was asking that question, but yes the flushes and straights are each mutually exclusive of the other hands, so you can add the probability of straights and flushes to the other hands, but be careful not to double count straight flushes. Note that flush draws and straight draws are also mutually exclusive of the other hands. That's a much more complicated calculation due to the flush draws that are straight draws which are not just straight flush draws, and due to paired flops. I've shown that calculation here before for connectors and offsuit gappers.

cyber, my idea was to do str8 and flushes separately so that I can them to the spacific hand: suited so +0.84 to flop a flush, mabe I should also include to flop a f draw. If connected +1.224 to flop a str8 without flushes on bord.
I figured that knowing these numbers will make me more aware of my implied odds and the amount that I should be putting in to see the flop against certain players.
I know what you read in books is prob based on the same calculations that have been done before but mabe understanding this will will help me make more acurete decisions based on the specific situations. I really hope that I am not doing all of this for nothing
BRUCE
after reading your last post I realize that 0.84 is prob including str8 flushes, how significant is this??if at all?

The probability of flopping a straight flush when it's possible is just 1/19600 =~ 0.005%.

If you want flush and straight draw numbers, see this post. You can find calculations for gappers, connectors, connectors that make fewer than the maximum numbers of draws, and even gutshot draws, but you have to follow the links, sometimes several levels deep starting with that first post. Many people and at least one book that I know of screw up the calculation of the straight draws and the combined straight and flush draw probabilities by over counting.

?

you have to have certain cards in your hand to make it possible

That's for a particular straight flush. You multiply that by the number of straight flushes that are possible, but that number depends on your cards. The probability of flopping any straight flush can be 0, 1/19600, 2/19600, 3/19600, or 4/19600.

Connectors 54s-JTs can flop 4 different straight flushes.
43s and QJs flop 3.
32s and KQs flop 2.
A2s and AKs flop 1.

1-gappers 53s-QTs flop 3.
42s and KJs flop 2.
A3s and AQs flop 1.

2-gappers 52s-KTs flop 2.
A4s and ATs flop 1.

3-gappers A5s-ATs flop 1.

>3-gappers flop 0.

You can use these for the straights too, just multiply by 4³ ways to choose the suits, and subtract the straight flushes.