| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1722.1 | | CSC32::D_DERAMO | Dan D'Eramo, Customer Support Center | Sun Feb 28 1993 22:55 | 14 |
| n!
With the notation C(n,k) = ---------
k! (n-k)!
there are C(52,6) possible hands. If m of them have
three distinct pairs then the probability is m / C(52,6).
So what is m? You can choose 3 of the 13 ranks in C(13,3)
ways. A pair can be selected from the four suits in C(4,2)
ways. So m seems to be C(13,3) * C(4,2)^3.
Does anyone else count them differently? :-)
Dan
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| 1722.2 | Counting can be hazardous | VMSDEV::HALLYB | Fish have no concept of fire. | Mon Mar 01 1993 09:05 | 19 |
| Well, we agree on the numerator. That's half the problem :-)
Here's another denominator. It's different. It's wrong.
How many possible 6-card, 3-distinct-pair hands are there?
I count thusly:
Card #1 : 52 choices
Card #2 : 3 choices, the possible pairs for Card #1
Card #3 : 48 choices, anything not of the rank of card #1
Card #4 : 3 choices matching the rank of Card #3
Card #5 : 44 choices, anything not the rank of cards #1 or #3
Card #6 : 3 choices matching the rank of card #5
_______
2965248 possible hands, roughly 1/7th of C(52,6).
Gee, I didn't know the chances were so good. What's wrong here?
(Hint: 6! does not divide 2965248, so that isn't it).
John
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| 1722.3 | | CSC32::D_DERAMO | Dan D'Eramo, Customer Support Center | Mon Mar 01 1993 14:03 | 8 |
| The enumeration in .2 would count each 3-distinct-pair
hand in each of the 3! orders of choosing cards 1, 3,
and 5. Can you check if it is six times the numerator
from .1?
Dan
p.s. We agreed on the denominator. :-)
|
| 1722.4 | | RUSURE::EDP | Always mount a scratch monkey. | Mon Mar 01 1993 14:11 | 7 |
| .2 counts each set of three ranks six times (once for each ordering of
the set) and counts each of the three pairs twice (once for each
ordering of cards in the pair). If it is divided by 6*2^3, the result
is the same as .1.
-- edp
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| 1722.5 | Thank you | NECSC::S_BUSH | | Mon Mar 01 1993 20:47 | 5 |
| My wife thanks all of you for your responses. She understands your
answers even if I do not. She would like to know if she could call on
your expertise later if needed?
Steven Bush (Lost in Math)
|
| 1722.6 | | CSC32::D_DERAMO | Dan D'Eramo, Customer Support Center | Mon Mar 01 1993 21:48 | 3 |
| Yes. Feel free to post notes here any time.
Dan
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| 1722.7 | | VMSDEV::HALLYB | Fish have no concept of fire. | Tue Mar 02 1993 08:46 | 5 |
| > Yes. Feel free to post notes here any time.
Changes the situation from "No answer" to "Many answers".
John (& thanks to EDP for his observation)
|