View Full Version : 97o getting a staight

08-16-2005, 11:10 AM
what are the odds of 97o, all in preflop, getting a straight coming in at the river? Or just getting a straight at all?

I went all in on a guy in a small dfriendly tourny, We were the last two players in. I had 99, he shows 97. He takes home the money with a straight.

08-16-2005, 11:56 AM
You need to be more specific. Does the board coming A-K-Q-J-10 count? What about 3-4-5-6-X? Are you asking about using both cards?

Are you really just looking for the odds that you'll win this hand?

08-16-2005, 12:21 PM
I should have been more specific. I meant making a staright with just the 7.

the chances of winning i think are 80+%. ya.cool.thanks.

08-16-2005, 12:50 PM
Well, with just the 7 you're looking at a board of 3-4-5-6-X where X isn't an 8 or a 7.

=(4 * 4 * 4 * 4 * 39)/COMBIN(50,5) = 0.471219%

Note: This includes flushes/straight flushes.

08-16-2005, 08:53 PM
Where did you get this 39 number?
If we assume that two nines are not taken out, then it has to be 52-4(sevens)-4(eights)-2(9,7) = 42 cards for X.
So, for no 99 case it will be: 4^4*42/C(50,5)= 0.507%.
With 99 exposed: 4^4*40/C(48,5)= 0.598%.

Am I right here?

08-16-2005, 09:18 PM
52-4(99 and 97)-4(3,4,5,6)-4(4 eights)-3(3 remaining sevens)=37 is what I get.

08-16-2005, 10:53 PM

You are wrong here - we don't mind having more than one card from 3-6 rank range (they will not ruin the hand).

08-16-2005, 11:00 PM
Yes, but in order to make the staright we need to use one 3, one 4, one 5, and one 6 for a total of four cards.

08-16-2005, 11:10 PM
oops, a couple of corrections:
1. Three sevens left in the deck, not four (my fault)
2. We are actually not afraid of eights here
3. 4568x(no seven) is winning too, so it is getting a little more complicated.

Ok, let's try again:
52-4(99, 97)-3(777)-4(four cards already picked out)= 41

and the chances to win are: 4*4*4*8*41/C(48,2)= 1.22%

08-16-2005, 11:20 PM
I should have been more specific. I meant making a staright with just the 7.

[/ QUOTE ] .

So we can't have an 8. And with the number 45, you also have to include the 3,4,5,6. So again I think its 37.

08-16-2005, 11:22 PM
yeah, stupid me.

08-16-2005, 11:27 PM
Tell me why we can't have 8? The straight is here even if there is no 9...

08-16-2005, 11:41 PM
The OP wanted the probabilty of a straight using only the seven in the villian's hand.

See the third post in this thread.

08-16-2005, 11:56 PM
I saw the post. /images/graemlins/smile.gif

Does 4568x on the table make "straight using only the seven in the villian's hand"?

I think it does. The presence of 9 does not change probability here.

Edit: after some thinking (English is not my native language) I can see your logic now, but I still think in the author's statement there were no indication that use of 9 is strictly prohibited.
All in all, the difference is subtle. The best opion is to ask the author, but I am too lazy for that.

By the way, will the following cards qualify:

it would add about 0.5%

08-17-2005, 03:50 PM
I eliminated the 3, 4, 5 & 6 as well as all four 7's and all four 8's and the 9 in hand as possible X's. 52 - (4 + 4 + 4 + 1) = 39

08-18-2005, 02:02 AM
So, do you agree that eights have to be excluded?

08-18-2005, 08:47 AM
Based on the way he worded it, yes. If the 8 falls, the 9 must be used.