Probability of getting 2 cards

[moracca]

Well, I'm by no means a math expert, but for a second i thought i was getting a hang of this probability stuff. My initial problem was that I was interested in finding out the odds of getting 2 more cards to complete a flush after the flop. for example, I get A [img]/images/graemlins/heart.gif[/img]K [img]/images/graemlins/heart.gif[/img] and the flop is 7 [img]/images/graemlins/spade.gif[/img] 2 [img]/images/graemlins/club.gif[/img] 5 [img]/images/graemlins/heart.gif[/img]. I want to know the probability of seeing a heart on the turn, and also a heart on the river. I have no problem calculating those by themselves, i just do not know how to combine those to find the odds of getting BOTH.

If i'm correct in my thinking, the odds of getting a heart on the turn are 10/47 = ~21%. then IF that happened, the odds of getting another heart on the river are 9/46 = ~19.5%. But i dont know how to figure out the odds of both.

I looked around, and started thinking.. I decided I should use C(n,r). I decided that since there are 47 cards left, and 10 of those are hearts, C(37,2) would give me the # of combos that wouldnt help me. Then I realized I was probably wrong, since that would be giving me the combos that wouldnt contain 1 heart. So, i'm back to square one (totally lost). I would very much appreciate any help on this.

Moracca

PS. I would also eventually like to be able to figure out the odds of getting 2 cards to complete a straight... but I can do that later assuming i get an understanding of this. Thanks again

[elitegimp]

[ QUOTE ]
Well, I'm by no means a math expert, but for a second i thought i was getting a hang of this probability stuff. My initial problem was that I was interested in finding out the odds of getting 2 more cards to complete a flush after the flop. for example, I get A [img]/images/graemlins/heart.gif[/img]K [img]/images/graemlins/heart.gif[/img] and the flop is 7 [img]/images/graemlins/spade.gif[/img] 2 [img]/images/graemlins/club.gif[/img] 5 [img]/images/graemlins/heart.gif[/img]. I want to know the probability of seeing a heart on the turn, and also a heart on the river. I have no problem calculating those by themselves, i just do not know how to combine those to find the odds of getting BOTH.

If i'm correct in my thinking, the odds of getting a heart on the turn are 11/47 = ~23%. then IF that happened, the odds of getting another heart on the river are 10/46 = ~22%. But i dont know how to figure out the odds of both.

I looked around, and started thinking.. I decided I should use C(n,r). I decided that since there are 47 cards left, and 11 of those are hearts, C(36,2) would give me the # of combos that wouldnt help me. Then I realized I was probably wrong, since that would be giving me the combos that wouldnt contain 1 heart. So, i'm back to square one (totally lost). I would very much appreciate any help on this.

Moracca

PS. I would also eventually like to be able to figure out the odds of getting 2 cards to complete a straight... but I can do that later assuming i get an understanding of this. Thanks again

[/ QUOTE ]

you're really close in both methods --

1) There's an 10/47 chance (not 11/47 [img]/images/graemlins/smile.gif[/img]) of seeing a flush on the turn, and _if it hits_ a 9/46 chance of hitting on the river. Multiply those together to get the probability of both of them happening. (10/47 * 9/46 = 90/2162, or 4.1%)

2) There are C(10,2)=45 ways the next two cards are both hearts. There are C(47,2)= 1081 possible next cards. Probability of both hearts = C(10,2) / C(47,2) = 45/1081 = 90/2162 (multiply by 2/2) = 4.1%

[moracca]

hah yeah -- i edited my post to change that -- but you were too quick for me. Thanks though, That helped me out a lot. I started thinking about multiplying them, but wasnt sure if that was right or not. with the C formula, i guess i looked at too many posts of using it to find the number of cards that DONT help, instead of those that do. Thank you very much. My next opponent is the straight dilemma.

Moracca

[moracca]

[ QUOTE ]
2) There are C(10,2)=45 ways the next two cards are both hearts. There are C(47,2)= 1081 possible next cards. Probability of both hearts = C(10,2) / C(47,2) = 45/1081 = 90/2162 (multiply by 2/2) = 4.1%

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Just out of curiosity, where did the 2/2 come from? because 2/2 = 1... so its like multiplying by 1 (45/1081 = 4.1%). Anyway the reason i was confused when I used this method was because i thought that using C(10,2) would give me the probability of getting ONE of those two cards, not both. I can see my the matching percentages, that I'm wrong

[JRegs]

You can count a backdoor flush as 1 or 1.5 outs.

[imported_Robert Andersson]

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You can count a backdoor flush as 1 or 1.5 outs.

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i count "runner runner" flushes like this

An Ace high flush 2 outs
An King high flush 1,5 outs
Any other high flush 1 outs

The true value is ~1.91 or so i want to remember but be conservative with all not "nuts" flushes

/Robert

[elitegimp]

[ QUOTE ]
[ QUOTE ]
2) There are C(10,2)=45 ways the next two cards are both hearts. There are C(47,2)= 1081 possible next cards. Probability of both hearts = C(10,2) / C(47,2) = 45/1081 = 90/2162 (multiply by 2/2) = 4.1%

[/ QUOTE ]

Just out of curiosity, where did the 2/2 come from? because 2/2 = 1... so its like multiplying by 1 (45/1081 = 4.1%). Anyway the reason i was confused when I used this method was because i thought that using C(10,2) would give me the probability of getting ONE of those two cards, not both. I can see my the matching percentages, that I'm wrong

[/ QUOTE ]

I got the "2/2" term (which is 1 [img]/images/graemlins/smile.gif[/img]) to show you that the results from both methods give you the same probability. (I wasn't sure how obvious it was that 45/1081 and 90/2162 are the same.)

[Carmine]

Method #1 was simple enough to understand. Can someone explain what C(n,r) means. I have no idea what the C or the n(10) or the r(2) means.

[Little Lew]

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Method #1 was simple enough to understand. Can someone explain what C(n,r) means. I have no idea what the C or the n(10) or the r(2) means.

[/ QUOTE ]

Here is a link to a pretty clear (even to me) paper on combinations and permutations Link on combinations .

I hope this is of interest and use to you.