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Card fun...

Wilmer

In Memoriam
Mar 19, 2012
376
OK you probabilitizers:
21 cards are labelled: 1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,7,8,9,10,11
You pick 5 cards at random, no replacement.
To "win", you need at least 3 DIFFERENT cards < 6.
Example: two 3's counts as one 3; three 4's counts as one 4.
So, as example, 5,2,5,9,5 is not a winning combo...got that?
But 3,1,3,3,5 is a winner...ok?
What's the probability of picking a winning combo?​
 

Wilmer

In Memoriam
Mar 19, 2012
376
No takers? Solution:
Relabel the cards: 1a 1b 1c 2a 2b 2c 3a 3b 3c 4a 4b 4c 5a 5b 5c 6 7 8 9 10 11
so there are C(21, 5) equally likely hands.

Sort the hand. Then a winning combination must be one of the following.

A)

Two high cards and three different low cards
C(6,2) * C(5,3) * 3 * 3 * 3 = 150 * 27 = 4050
(The last factors of 3 are the selection of a, b, or c.)

B)

One high card, three different low cards, and a matching low card
C(6,1) * C(5,3) * 3 * 3 * 3 * 3 * 2 / 2 = 60 * 81 = 4860
(The division by 2 is to allow for the order of the match.)
One high card and four different low cards
C(6,1) * C(5,4) * 3 * 3 * 3 * 3 = 2430

C)

Two pairs of matching low cards, and a different low card
C(5,2) * 3 * 3 * C(3,1) * 3 = 810
(The middle factors of 3 are the omission of a, b, or c.)
A triple low card, and two different low cards
C(5,1) * C(4,2) * 3 * 3 = 270
Four different low cards, and a matching low card
C(5,4) * 3 * 3 * 3 * 3 * 4 * 2 / 2 = 1620
Five different low cards
3 * 3 * 3 * 3 * 3 = 243

So the probability of a winning combo is
(4050 + 4860 + 2430 + 810 + 270 + 1620 + 243) / C(21,5)
= 14283 / 20349
= 1587 / 2261
= 70.2 %
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
If you want to post challenge problems to which you already know the solution then we have a special forum for that. I'll move this there now.