Using combinations to calculate odds

[Rynofasho]

Another newb question, sorry, but can someone explain using combinations to calculate basic odds, or lead me to a site so I dont have to ask more annoying questions and can figure it out on my own? Thanks

[uuDevil]

A combination is an arrangement of objects where the order doesn't matter. To calculate probabilities we count combinations and determine appropriate proportions of combinations:

P(X)= (# combinations of X)/(total # of combinations)

An example:

There are 4 aces in a deck of cards. A pocket pair of aces is a combination of 2 aces. AcAd is the same combination as AdAc since the order doesn't matter. There are 6 possible pocket pairs of aces (combinations of 2 aces) we can generate from the 4 available: AcAd, AcAh, AcAs, AdAh, AdAs, and AsAh. Shorthand for choosing a combination of 2 objects from 4 objects is C(4,2).

C(4,2)=4!/((4-2)!*2!)= 4*3*2*1/((2*1)*(2*1))=3*2=6

Where ! indicates "factorial" and n!=n*(n-1)*(n-2)*...1.

So what is the probability of being dealt a pair of aces?
P(AA)=(# combinations of AA)/(total # combinations of 2 card hands in a deck of 52 cards)

P(AA)=C(4,2)/C(52,2)=6/1326=.004525

This is the same as odds of 220:1.

[Rynofasho]

Thanks for the reply, and I follow that logic, but now say you get AA. How would you use a combination from there on out to calculate the odds of A)flopping a set, B)improving on the turn, or C)improving on the river?

[uuDevil]

[ QUOTE ]
Thanks for the reply, and I follow that logic, but now say you get AA. How would you use a combination from there on out to calculate the odds of A)flopping a set, B)improving on the turn, or C)improving on the river?

[/ QUOTE ]

These just take a little thought and some practice:

For A) C(2,1)*C(48,2)/C(50,3)=0.1151 or 7.69:1

Where
C(2,1)= # ways to choose 1 ace from the 2 remaining
C(48,2)= # ways to choose 2 other flop cards from the 48 remaining non-aces
C(50,3)= total # ways to choose 3 flop cards

So this is the probability of flopping precisely a set but not quads.

P(flopping quads)=C(2,2)*C(48,1)/C(50,3)=.002449

Adding this to the first value gives

P(set or quads)=.11755 or odds of 7.51:1 and this is the number you most often see.

I'll let you try B) and C)

Use the combin() function in Excel to simplify the calculations.

[BruceZ]

[ QUOTE ]
[ QUOTE ]
Thanks for the reply, and I follow that logic, but now say you get AA. How would you use a combination from there on out to calculate the odds of A)flopping a set, B)improving on the turn, or C)improving on the river?

[/ QUOTE ]

These just take a little thought and some practice:

For A) C(2,1)*C(48,2)/C(50,3)=0.1151 or 7.69:1

Where
C(2,1)= # ways to choose 1 ace from the 2 remaining
C(48,2)= # ways to choose 2 other flop cards from the 48 remaining non-aces
C(50,3)= total # ways to choose 3 flop cards

So this is the probability of flopping precisely a set but not quads.

P(flopping quads)=C(2,2)*C(48,1)/C(50,3)=.002449

Adding this to the first value gives

P(set or quads)=.11755 or odds of 7.51:1 and this is the number you most often see.

I'll let you try B) and C)

[/ QUOTE ]


You have included full houses in this as well (but not the ones where the flop is all the same rank). It may be ultimately more useful to include them, but if you really wanted the odds of "precisely a set", you would not use combinations to count the other 2 flop cards, but instead do

C(2,1)*48*44*3 / (50*49*48) = 10.78% = 8.28:1.

The 3 in the numerator counts the 3 positions that the matching card can occur. 48*44 is the number of ways to choose the remaining 2 cards so that the board does not pair. Note that this 48*44 counts permutations instead of combinations, that is, it counts the 2 possible orderings of these cards. The denominator is consistent with this, since it counts all permutations of the flop, that is 50*49*48 = P(50,3) instead of C(50,3). Another way to do this is

[ C(2,1)*(48*44)/2 ] / C(50,3).

In this case we divide 48*44 by 2 so that each combination is counted only once, and then we can use combinations instead of permutations in the denominator. Note that this is equivalent to the first method as we have effectively divided numerator and denominator by 3! = 6.

You could also compute the probability of a full house, and then subtract this from what you have. Again, this ignores the case of the flop all one rank.

P(full house) = C(2,1)*12*6 / C(50,3) = 0.73469%.

12 is the number of ranks that can pair, and 6 is the number of possible pair combinations for each rank.

Subtracting this from your 11.51% gives again 10.78% as above.

[uuDevil]

[ QUOTE ]
You have included full houses in this as well.

[/ QUOTE ]

Well I've always been a sloppy counter. Not enough raps on the knuckles, I suppose.

I'll let someone else mind the store for a while. /images/graemlins/blush.gif

[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]

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

[/ QUOTE ]
That factor is the number of combinations of 2 non-aces to complete the flop, assuming the 1st card was an ace.
[ QUOTE ]

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


[/ QUOTE ]

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. /images/graemlins/tongue.gif