Posted Fri Aug 13, 2004 6:56 am GMT by skel
What are the odds at a 9 player game that
1) Flop will have atleast 2 suited cards
2)someone will have 2 of the same suited cards that flopped
3)they will hit their flush if 1 and 2 are true
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Posted Fri Aug 13, 2004 11:48 am GMT by Sean_in_NJ
Odds of any flush are about 6%, I think.
If you flop a 4 flush, then you've got 9 outs and two cards to come, and you're about 38% to make it on the turn or river.
Posted Sun Aug 15, 2004 3:24 am GMT by StarlightCoast
I thought getting the flush in this case was a 4 to 1 shot. If the percentage is 38% would that not make the odds less than 3 to 1?
Posted Sun Aug 15, 2004 2:15 pm GMT by Sean_in_NJ
| StarlightCoast wrote: | | I thought getting the flush in this case was a 4 to 1 shot. If the percentage is 38% would that not make the odds less than 3 to 1? |
It's about 4-1 for each card. The probability of hitting it on the turn OR river is about 38%.
Posted Mon Aug 16, 2004 9:39 am GMT by skel
| Sean_in_NJ wrote: | | StarlightCoast wrote: | | I thought getting the flush in this case was a 4 to 1 shot. If the percentage is 38% would that not make the odds less than 3 to 1? |
It's about 4-1 for each card. The probability of hitting it on the turn OR river is about 38%. |
edit
Posted Mon Aug 16, 2004 11:53 am GMT by Sean_in_NJ
| Sean_in_NJ wrote: | | StarlightCoast wrote: | | I thought getting the flush in this case was a 4 to 1 shot. If the percentage is 38% would that not make the odds less than 3 to 1? |
It's about 4-1 for each card. The probability of hitting it on the turn OR river is about 38%. |
My calculation was a little simplistic. I took the odds for making it on the turn and added it to the odds for making it on the river. In actuality, the percentage for making it on either card is 35%, and you have to calculate that by figuring the odds of NOT making it on either card, and subtracting that number from 1.
So, it's 1 - ((38/47) * (37/46)) = .349
Posted Mon Aug 16, 2004 1:00 pm GMT by skel
| Sean_in_NJ wrote: | | Sean_in_NJ wrote: | | StarlightCoast wrote: | | I thought getting the flush in this case was a 4 to 1 shot. If the percentage is 38% would that not make the odds less than 3 to 1? |
It's about 4-1 for each card. The probability of hitting it on the turn OR river is about 38%. |
My calculation was a little simplistic. I took the odds for making it on the turn and added it to the odds for making it on the river. In actuality, the percentage for making it on either card is 35%, and you have to calculate that by figuring the odds of NOT making it on either card, and subtracting that number from 1.
So, it's 1 - ((38/47) * (37/46)) = .349 |
Ok I suck at math. Ill take your word for it. What would be the odds of hitting it on either the turn or river? Im assuming you mean its a 35% chance of hitting on the turn and another 35% of hitting it on the river.
Posted Mon Aug 16, 2004 2:09 pm GMT by Sean_in_NJ
| skel wrote: | | Ok I suck at math. Ill take your word for it. What would be the odds of hitting it on either the turn or river? Im assuming you mean its a 35% chance of hitting on the turn and another 35% of hitting it on the river. |
No, I mean it's 35% to hit it on the turn or river. It's 19% (or 4-1) to hit it on the turn, and then another 19% to hit it on the river.
Posted Thu Aug 19, 2004 8:37 am GMT by infarom
1) For a specific symbol, if m is the number of your own pocket cards with that symbol (m=0, 1 or 2), then the probability for the flop to contain minimum 2 of that cards is C(13-m,2)*48/C(50,2).
If you want the probability for any two suited cards to be contained in the flop, you have to sum four of such above terms (you will have there 4 parameters as m).
2) If you want the probability for an opponent (a fixed one) to hold two suited cards (with specific symbol) (after the flop), that is
C(13-s,2)/C(47,2), where s is the total number of seen cards with that symbol (from own hand and board).
If you want the probability for at least one opponent to hold such cards, the formula is a little bit more complicated:
P= n*C(13-s,2)/C(47,2) - C(n,2)*P(two opponents hold such cards) + C(n,3)*P(three opponents hold such cards) - ... and so on, etc.
For all Hold'em odds, see Do not post links to commercial areas of sites, that was clearly a plug for that book
Posted Sat Aug 21, 2004 3:54 am GMT by vegasholdem
[What are the odds at a 9 player game that
1) Flop will have atleast 2 suited cards-8 to 1 against
2)someone will have 2 of the same suited cards that flopped-2.4 - 1 against
3)they will hit their flush if 1 and 2 are true-4 to 1 against(with 1 card to come)
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