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:

Comments

  • 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
  • wait wait wait wait wait....

    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?

    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......

    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
    :confused:
  • why divided by (3*2*1)?

    This represents the 6 different ways of ordering 3 distinct objects (i.e. the 3 flop cards).
    huh@#$? why (3/1)?

    Sorry, I botched that one. It should have been

    2 C 1 = 2/1
    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
  • Hey scotty! thanks a million!
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