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Number Cards Before Face Cards (Posted on 2019-05-02) Difficulty: 3 of 5
If you turn over a well shuffled deck of cards one card at a time, what is the probability that you'll see at least one of each of the denominations of the numeric cards, ace through ten (1 - 10) before seeing any face card (J, Q, K)?

  Submitted by Charlie    
Rating: 3.0000 (1 votes)
Solution: (Hide)
Use the inclusion/exclusion technique: add the individual probabilities of seeing ace, plus of seeing deuce, etc. before seeing a face card. There are of course ten probabilities to add, all of them the same: 4/16 = 1/4; since there are ten of them the total is 10/4 = 5/2.

Then subtract out all the pairwise probabilities of seeing an ace or a deuce, an ace or a trey, etc. before the first face card. Each pairwise probability is 8/20 = 2/5, and there are C(10,2) = 45 of them so the total is 90/5 = 18 to be subtracted out.

Continue to triples, quadruples, ... , any of all ten, alternately adding and subtracting the totals.


-let individual   how many     sum

1       1/4          10         5/2
2       2/5          45         18
3      12/24        120         60
4      16/28        210        120
5      20/32        252       315/2
6      24/36        210        140
7      28/40        120         84
8      32/44         45       360/11
9      36/48         10        15/2
10     40/52          1        10/13

The probability of getting at least one of each numeric denomination before any face card is therefore

5/2 - 18 + 60 - 120 + 315/2 - 140 + 84 - 360/11 + 15/2 - 10/13 = 1/286

~= 0.00349650349653097

Comments: ( You must be logged in to post comments.)
  Subject Author Date
ThanksSteven Lord2019-05-19 16:55:51
Preliminary solutionSteven Lord2019-05-13 11:22:48
Hints/TipsHintCharlie2019-05-06 07:51:34
Full solution: part 1 formatting correctionFrankM2019-05-05 11:35:30
Full solution: part 1FrankM2019-05-05 11:29:29
A start....Steven Lord2019-05-02 17:44:25
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