Using combinations to calculate odds

[Rynofasho]

Well I hate to disappoint, but I cannot find my answers Here is what I tried for on the turn:
Following your logic, my first Combo was still C(2,1) since there are 2 ways to pick one ace. From there im a bit fuzzy since in your example you multiplied that value times the combo of ways to not flop an A, and then divided by number of ways to choose one card out of the remaining 47. So. . .
C(2,1)*C(45,0)/C(47,1)=0.042553 where
C(2,1)=Number of ways to choose one A of the two remaining
C(45,0)=Not sure
C(47,1)=Number of ways to choose one turn card

I noticed that the sum of the numerator values equal the combo values on the bottom, so I followed that logic for part of that, although now I cant really interpret my answer since I dont know what the odds really are of making it on turn. Please correct me, and thanks

[uuDevil]

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....in your example you multiplied that value times the combo of ways to not flop an A

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That factor is the number of combinations of 2 non-aces to complete the flop, assuming the 1st card was an ace.
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C(2,1)*C(45,0)/C(47,1)=0.042553 where
C(2,1)=Number of ways to choose one A of the two remaining
C(45,0)=Not sure
C(47,1)=Number of ways to choose one turn card


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This is right. It is superfluous to include the factor C(45,0) here but it represents the fact that after we have picked the turn card, we are picking 0 cards of the remaining 45 (i.e. we only deal 1 turn card).

So the answer is .0425 which is odds of (1-.0425)/.0425 to 1, or 22.5:1. This is what an odds chart will tell you your odds of improving are on the next card when you have 2 outs.

Some general comments:

There is usually more than 1 way to do these problems, so you can check your own answers. For example, in this case you can just say: "On the turn there are 2 cards that make a set out of the 47 remaining, and 2 out of 47= 2/47=.0425."

Or "2 cards make a set and 45 do not, so the odds are 45:2 or 22.5:1."

Combinations are just one tool out of many. You should try to use the best (most efficient) tool for the problem at hand. Sometimes you have to use more than one tool. Try to solve the problem different ways. This will give you confidence in your answer.

Cheat. Look the answer up somewhere and if your number is different, try to figure out why. On the other hand you should realize that the answer at the back of the book is sometimes wrong. This goes for numbers you find on these forums too. [img]/images/graemlins/tongue.gif[/img]