MHB Probability of Winning with 21 Cards: OK You Probabilitizers!

  • Thread starter Thread starter Wilmer
  • Start date Start date
  • Tags Tags
    Cards Probability
AI Thread Summary
The discussion focuses on the probability of winning with a specific set of 21 cards labeled from 1 to 11, where a winning hand consists of at least three different cards valued less than 6. Participants are challenged to calculate the probability of drawing five cards at random without replacement that meet this winning criterion. An example illustrates that combinations with repeated values do not count towards the winning condition. The thread concludes with a note that challenges with known solutions should be posted in a designated forum. The conversation emphasizes the importance of understanding the rules for determining winning combinations in this card game scenario.
Wilmer
Messages
303
Reaction score
0
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?​
 
Mathematics news on Phys.org
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 %
 
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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top