Three cards from a standard pack

[GwydionM]

With a standard pack, if you randomly pick three cards, there would be 132600 possible combinations, counting each card as unique. (52 * 51 * 50). But if the order doesn't matter, there would be rather less. Only 22100, I think, because there are six ways to line up three items.

Now supposing we are ignoring most of the difference and looking for just four patterns from a random selection of three cards.
A) No court cards (court cards being the King, Queen and Jack).
B) One court card
C) Two court cards
D) Three court cards.

Obviously (A) would be the most common and (D) the rarest. But can anyone figure the exact odds?

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An old man who lives in the UK

 

[ck1999]

For option D

52 / 12 * 51/11 * 50/10

or

13 / 3 * 51/11 * 5


If this is correct it is a 1 in 100 chance of getting 3 court cards if you select 3 cards from a deck of 52

ck1999

 

[AlexCuse]

[COLOR=white]3 Non-Face	0.447058824
1 Face	        0.423529412
2 Face	        0.119457014
3 Face	        0.009954751[/color]

I was pretty well along with this yesterday but checked with the statistical analyst I work with this morning. He gave me a valuable hint to use Excel's COMBIN function to make life easier

Basically you will want to find the number of possible combinations (example, for 2 face cards you will want the number of ways you can pick 1 of the 12 face cards times the number of ways you can pick 2 of the 40 non-face cards, divided by the number of ways you can take 3 of the 52 cards)

Or in excel:

[COLOR=white]=(COMBIN(12, 2) *COMBIN(40,1))/COMBIN(52,3)[/color]

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[AlexCuse]

My '2 face cards' example above (the long drawn out one) is actually the formula for 1 face card. I switched the 1 and 2 around.

At least the excel formula was correct

[small]----signature below----[/small]
Who are you, and why should I care?
Side A: We're a community of outdated robots who refused to upgrade and came here
*changes sides*
Side B: to live a simpler existence. Free of technology.

 

[GwydionM]

Seems about right. I tried a different method, (40/52) * (39/51) * (38/50) for the chance of dealing three non-court cards. That gave just the same answer.

Thanks.

------------------------------
An old man who lives in the UK

 

[ChrisHunt]

No excel required, though I did use it to work out the (approximate) percentages:

A) No court cards (court cards being the King, Queen and Jack).

(40 * 39 * 38) / (52 * 51 * 50) = 45%


B) One court card

((12 * 40 * 39) + (40 * 12 * 39) + (40 * 39 * 12)) / (52 * 51 * 50) = 42%


C) Two court cards

((12 * 11 * 40) + (12 * 40 * 11) + (40 * 12 * 11)) / (52 * 51 * 50) = 12%


D) Three court cards.

(12 * 11 * 10) / (52 * 51 * 50) = 1%




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