Flopping a straight flush

[3rdEye]

In a home game s&g I played earlier, I called a min-raise with JT of hearts from the BB. The flop came KQ9 of hearts. The turn and river came AK, which was nice, since my opponent had AA (which he foolishly slowplayed on the flop and turn, especially considering he did not have Ah).

After the hand, we were trying to determine how often one could flop a straight flush with my hand, and I calculated that this would occur about 1 in 20,000 times (3/50*2/49*1/48).

Am I on the right track?

[MickeyHoldem]

Your on the right track, but remember you can flop the straight flush 4 different ways when you hold JTs...

AKQ
KQ9
Q98
987

so there are 4 / c(50,3) = 0.000204082 = 1 in 4900 which is 4 times as likely as your calculation.

[joytoxin68]

Hi. This is my first post. Thanks to all the knowledgeable posters who have helped me out up till now when I was lurking.

I think there may be a mistake in the first calculation. I agree there are four ways of flopping, given your hand. These are as noted,
AKQ
KQ9
Q98
987

However, if you hold 10J Hearts, the prob. flopping any of these is (1/50)*(1/49)*(1/48), isn't it? For example if you have 10J hearts, the only 987 that will help is 9H, 8H, 7H. Since there is only ONE 9H, chance it falls is 1/50. So P(9H-8H-7H given JH-10H) is 8.5 * 10^-6. Cumulative P = 4 times this.

[elitegimp]

[ QUOTE ]
Hi. This is my first post. Thanks to all the knowledgeable posters who have helped me out up till now when I was lurking.

I think there may be a mistake in the first calculation. I agree there are four ways of flopping, given your hand. These are as noted,
AKQ
KQ9
Q98
987

However, if you hold 10J Hearts, the prob. flopping any of these is (1/50)*(1/49)*(1/48), isn't it? For example if you have 10J hearts, the only 987 that will help is 9H, 8H, 7H. Since there is only ONE 9H, chance it falls is 1/50. So P(9H-8H-7H given JH-10H) is 8.5 * 10^-6. Cumulative P = 4 times this.

[/ QUOTE ]

not quite: consider the "AKQ" flop --

there are 50 cards left in the deck, and 3 of them are either the Ah, Kh, or Qh. Hence, 3/50 chance of flopping one of those.

Once one card comes, there are 49 cards left in the deck, and only two cards to help you make this particular straight flush (The Ah and Kh if the Qh was the first card seen on the flop, etc).

Welcome to the forum!

[joytoxin68]

I understand now. Thanks!

[3rdEye]

[ QUOTE ]
Your on the right track, but remember you can flop the straight flush 4 different ways when you hold JTs...

AKQ
KQ9
Q98
987

so there are 4 / c(50,3) = 0.000204082 = 1 in 4900 which is 4 times as likely as your calculation.

[/ QUOTE ]

I hadn't thought of that! Thanks.

[Ringo]

Long time lurker, first time poster - hi guys!

Yesterday I flopped a royal flush. Holding AhQh, the flop came 10h-Jh-Kh. When the turn came another heart, I had the deck crippled and won a measly $7.. [img]/images/graemlins/blush.gif[/img]

..but am I right in saying then, that with that particular starting hand, since there is only one flop that can give me a straight flush, it's as simple as c(50,3) then?

[elitegimp]

[ QUOTE ]

Long time lurker, first time poster - hi guys!

Yesterday I flopped a royal flush. Holding AhQh, the flop came 10h-Jh-Kh. When the turn came another heart, I had the deck crippled and won a measly $7.. [img]/images/graemlins/blush.gif[/img]

..but am I right in saying then, that with that particular starting hand, since there is only one flop that can give me a straight flush, it's as simple as c(50,3) then?

[/ QUOTE ]

assuming you mean 1 in c(50,3), then yes -- c(50,3) is just the number of possible flops assuming order of cards doesn't matter (so Ah Kh Jh is the same flop as Ah Jh Kh), and since there's only one resulting in the desired outcome (flopped straight flush), 1/c(50,3) is the probability that it happens.

edit: and welcome to the forum [img]/images/graemlins/smile.gif[/img]