O8 probability question - AK35 flops a straight?

[franknagaijr]

As an exercise, I'm going through a starting hand and looking at the chances of hitting a favorable flop, component by component. The following question comes up, and I want to make sure I'm not missing anything, because the number seems so small.

Assume you hold A35K rainbow, and 48 cards are unknown. What are the chances of flopping QJT of any suits?

I came up with 0.0037, which seems tiny and I wanted to confirm my logic:

QJT Rainbow - 24 combinations (4 q * 3 J * 2T)
QJT of one suit - 4 combinations
QJT double- suited 36 combinations as follows
QJ-T =4 * 3 (QJ same suit)
QT-J = 4*3 (QT same suit)
TJ-Q = 4*3 (TJ same suit)
That's a total of 64 ways to flop QJT, with possible flop combinations of C(48,3) or 64/17296.

Is that the correct number?

thanks

[fishyak]

Another way: 4 Q's, 4 J's, 4 T's = 4*4*4 = 64. The 4 factorial with 3 slots is your rainbow and then you added back in the non-rainbow combo's to get the same result.

But where did you get your denominator? 48*47*46 = 103776, not 17296. 48*47*46 should be the available combinations of a deck with 48 cards remaining to fill 3 slots. Odds of FLOPPING 3 cards for this straight should come in low, about 6 out of 10,000 hands.

Feel free to tell me if my basic statistics are rusty. They should be...it's been 33 years since my last statistics class.

[fishyak]

Also keep in mind that if you could draw either way to make the straight - say you have 89 instead of AK - you would have 4 times as many ways of flopping a straight because both ends become available. Your odds are very low here because are have limited yourself to precisely one set of 3 cards that complete the straight.

[MattS]

[ QUOTE ]
But where did you get your denominator? 48*47*46 = 103776, not 17296. 48*47*46 should be the available combinations of a deck with 48 cards remaining to fill 3 slots. Odds of FLOPPING 3 cards for this straight should come in low, about 6 out of 10,000 hands.

[/ QUOTE ]

The ordering of the cards on the flop is unimportant. For example 5[img]/images/graemlins/club.gif[/img] 6[img]/images/graemlins/diamond.gif[/img] 7[img]/images/graemlins/heart.gif[/img] is the same Flop as 7[img]/images/graemlins/heart.gif[/img] 6[img]/images/graemlins/diamond.gif[/img] 5[img]/images/graemlins/club.gif[/img]. Therefore you have only C(48,3) = 48*47*46 / (3*2*1) = 17296 possible flops.

[franknagaijr]

Driving in my car, I recognized the 4*4*4=64 combination, and it clicked. I was going to delete the post, but it did get a response.

Fishyak, 48*47*46 is the permutation, (excel=permut(x,y)) and that assumes that we care about the order in which they come. To get to the combination, you've got to divide by the number of ways they can come. For ABC, that's
ABC
ACB
BAC
BCA
CAB
CBA

Or six ways. Took me a while to figure this out, since there are times when you can multiply the three numbers and get the correct combination (4*4*4) springs to mind, but that's only because there are 3 spaces for three distinct cards, and the permutations would be six times that.

The other handy concept when crunching this junk is how combinations combine. If you've got 48 cards, and 11 flush cards open, and you want to figure out how many flops could have 2 or 3 flush cards, then you could break it down like this to confirm the answer

0 flush cards combinations
combin(37,3)
1 flush card combinations
combin(37,2) * combin(11,1)
2 flush card combinations
combin(37,1) * combin(11,2)
3 flush card combinations
combin(11,3)

You add up all those numbers, it should equal combin(48,3) for a check figure.

Wheee!!!!

[Buzz]

[ QUOTE ]
64/17296.

Is that the correct number?

[/ QUOTE ]Yes.

Hi Frank - You'll flop three specific ranks when you have none of those ranks roughly one time in 270.

But that's the flop. The five card board on the river will have three specific ranks (say QJT when you hold AK53) about 3%.

Here's the chart:
<ul type="square">QQQJT 4*4*4
QJJJT 4*4*4
QJTTT 4*4*4
QQJJT 6*6*4
QQJTT 6*4*6
QJJTT 4*6*6
QQJTX 6*4*4*36
QJJTX 4*6*4*36
QJTTX 4*4*6*36
QJTXY 4*4*4*36*35/2[/list]
Whatever that all comes to be (maybe 51312?) is out of C(48,5) = 1712304.

51312/1712304 = almost 3 per cent.

Buzz

[fishyak]

[ QUOTE ]
[ QUOTE ]
But where did you get your denominator? 48*47*46 = 103776, not 17296. 48*47*46 should be the available combinations of a deck with 48 cards remaining to fill 3 slots. Odds of FLOPPING 3 cards for this straight should come in low, about 6 out of 10,000 hands.

[/ QUOTE ]

The ordering of the cards on the flop is unimportant. For example 5[img]/images/graemlins/club.gif[/img] 6[img]/images/graemlins/diamond.gif[/img] 7[img]/images/graemlins/heart.gif[/img] is the same Flop as 7[img]/images/graemlins/heart.gif[/img] 6[img]/images/graemlins/diamond.gif[/img] 5[img]/images/graemlins/club.gif[/img]. Therefore you have only C(48,3) = 48*47*46 / (3*2*1) = 17296 possible flops.

[/ QUOTE ]

Thanks. I get it now.

[BillytheKidd]

And people think poker is just a bunch of degenerates hoping to get lucky! Its amazing how much math CAN be involved in the game for those that choose to use it.