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Old Oct 07,2004, 08:38 PM   #1
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small pairs and mathematics

I understand there is a 7.33:1 (or 8:1) chance of flopping trips with pairs in the hole. How is this calculated?? What's the formula? :banghead:
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Old Oct 07,2004, 08:58 PM   #2
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small pairs and mathematics

After taking your 2 hole cards out of the deck the total number of flops from the 50 unseen cards is 50 C 3 (or "50 choose 3") which is equal to

(50*49*48)/(3*2*1)

There are 2 cards of the remaining 50 in the deck which will make you a set. The number of flops containing exactly one of those 2 cards is

(2 C 1) * (48 C 2)

= (2/1) * ( (48*47)/(2*1) )

The probability of flopping a set is the ratio of these, which can be simplified to:

(47*3*2)/(50*49)

= 0.115 = 11.5%

Converting percentage to odds against gives about

88.5 to 11.5

which reduces to

7.7 to 1

ScottyZ
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Old Oct 07,2004, 09:25 PM   #3
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small pairs and mathematics

wait wait wait wait wait....


Quote:
After taking your 2 hole cards out of the deck the total number of flops from the 50 unseen cards is 50 C 3 (or "50 choose 3") which is equal to

(50*49*48)/(3*2*1)
why divided by (3*2*1)? I figured the 50*49*2 <-- cause there are two of your missing cards in the deck, right?

Quote:

There are 2 cards of the remaining 50 in the deck which will make you a set. The number of flops containing exactly one of those 2 cards is

(2 C 1) * (48 C 2)

= (3/1) * ( (48*47)/(2*1) )
:banghead: huh@#$? why (3/1)? I'm lost on the dividing parts. I took this s#it in high school......

Quote:

The probability of flopping a set is the ratio of these, which can be simplified to:

(47*3*2)/(50*49)

= 0.115 = 11.5%

Converting percentage to odds against gives about

88.5 to 11.5

which reduces to

7.7 to 1
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Old Oct 07,2004, 10:38 PM   #4
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small pairs and mathematics

Quote:
why divided by (3*2*1)?
This represents the 6 different ways of ordering 3 distinct objects (i.e. the 3 flop cards).

Quote:
huh@#$? why (3/1)?
Sorry, I botched that one. It should have been

2 C 1 = 2/1

Quote:
I'm lost on the dividing parts.
This is just using the formula for combinations:

n C r = n!/( (n-r)!*r! )

Proving that this really is the correct formula for "r objects taken from n without replacement disregarding order" takes a lot of work.

Check out

http://mathforum.org/dr.math/faq/faq.comb.perm.html

or

http://mathforum.org/dr.math/faq/faq.mcdonalds.html

for some more information on combinations (i.e. the "50 choose 3" stuff).

ScottyZ
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Old Oct 07,2004, 10:43 PM   #5
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small pairs and mathematics

Hey scotty! thanks a million!
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