combination or permutation??

[BBD]

Can sombody show me the math on the chances of a JTo making a straight by the river? Any help is much appreciated.

[BBD]

Ok, instead say we're talking about AKo making a straight. Would the math look like 4 * 4 * 4 * 38C2 / 50C5 that is 4 tens, 4 jacks and 4 queens in the deck of which we want one of each on the board? this doesn't seem right but I really don't know. thanks again for any help anyone might be able to give.

[KJL]

Yea it seems right but, I think it should b 47C2 instead of 38C2.

[AaronBrown]

You're both almost right, but it's a bit trickier than this. If you use 38 then you're computing the probability of getting a straight without pairing any of the needed cards. But if you use 47, you'll double count those occurances (and more than double counts one with two pair or threes).

To do this correctly you start with 38, then you compute the probability of getting a straight and pairing one of the cards; getting a straight and pairing two of the cards; and getting a straight with three of a kind.

That's:

4*4*4*C(38,2) (no pairs)
3*4*4*6*38 (one pair)
3*4*6*6 (two pairs)
3*4*4*4 (three of a kind)

[BBD]

Thanks for the clarification, appreciate it. can you tell me how we arrive at the math for the pair/two pair/trips and making a straight? not sure I understand the equation.
Thanks! also for JTo would it be 4^6 * 26C2 / 50C5 and the same pair/two pair/trip math?? thanks again then!

[BBD]

can anyone explain these factors:
3*4*4*6*38 (one pair)
3*4*6*6 (two pairs)
3*4*4*4 (three of a kind)

thanking you in advance

[BruceZ]

[ QUOTE ]
You're both almost right, but it's a bit trickier than this. If you use 38 then you're computing the probability of getting a straight without pairing any of the needed cards. But if you use 47, you'll double count those occurances (and more than double counts one with two pair or threes).

To do this correctly you start with 38, then you compute the probability of getting a straight and pairing one of the cards; getting a straight and pairing two of the cards; and getting a straight with three of a kind.

That's:

4*4*4*C(38,2) (no pairs)
3*4*4*6*38 (one pair)
3*4*6*6 (two pairs)
3*4*4*4 (three of a kind)

[/ QUOTE ]


And we should really subtract off the flushes too. Here is what I had in my "play area". It's the same as yours except for the last line of the numerator which subtracts off the flushes:

[
4^3*C(38,2) +
3*C(4,2)*4^2*38 +
C(3,2)*C(4,2)*C(4,2)*4 +
3*C(4,3)*4^2
- 2*[9*38 + C(9,2)] - 18*C(9,2)
] / C(50,5)

=~ 2.6%


To answer the OPs questions,

[ QUOTE ]
3*4*4*6*38 (one pair)

[/ QUOTE ]

That's the same as my

3*C(4,2)*4^2*38

There are 3 ranks that can pair (T,J,Q) and C(4,2) = 6 ways to make a pair with each rank. Then there are 4^2 ways to choose the remaining 2 straight cards, and 38 ways to choose the 5th card so that it doesn't pair the board.


[ QUOTE ]
3*4*6*6 (two pairs)

[/ QUOTE ]

This is the same as my

C(3,2)*C(4,2)*C(4,2)*4

There are C(3,2) = 3 ways to pick the 2 ranks out the 3 ranks T,J,Q which will pair. There are C(4,2) = 6 ways to make each pair. Then there are 4 ranks which make the remaining card of the straight.


[ QUOTE ]
3*4*4*4 (three of a kind)

[/ QUOTE ]

This is the same as my

3*C(4,3)*4^2

There are 3 ranks that can make trips on the board (T,J,K), and there are C(4,3) = 4 ways to choose 3 of one rank. Then there are 4^2 ways to choose the remaining 2 cards of the straight.

JTo is trickier because there are 4 possible straights, but you can't just multiply by 4 because it is possible to make 2 or 3 of these straights at the same time by making a 6 or 7 card straight, and you don't want to count these more than once. I suggest separating out the 5,6, and 7 card straights.

[BruceZ]

[ QUOTE ]
can anyone explain these factors:
3*4*4*6*38 (one pair)
3*4*6*6 (two pairs)
3*4*4*4 (three of a kind)

thanking you in advance

[/ QUOTE ]

See my post above.