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Homework Help: The probability of a Snap! with two decks plus a single joker

  1. Mar 14, 2012 #1
    1. The problem statement, all variables and given/known data
    So here's a little background: This isn't exactly a homework question. I work for a small software company, and everyday I distribute a status update to disclose where we stand with our current software patch development. To keep it interesting, I ask a daily riddle/problem and those who answer correctly receive a prize. Yesterdays question caused a company wide argument as to what the correct probability of the following question is.

    Two players each have 1 standard deck of cards (52 cards) PLUS 1 Joker per deck. If each player flips a card over per turn, what are the odds that they will flip the same card value (joker included) before both players have gone through all 53 cards. After nobody answered correctly, I posted the following home-grown, and highly debated solution.

    2. Relevant equations
    If we were trying to calculate the probability of each player flipping the same card (suit and value), then the riddle book presents this solution: 1 - ((52/53)^53)

    3. The attempt at a solution
    1 - ( (13/14)^53) equaling roughly 98%.

    Using a Ruby program simulation, I found that 9814 out of 10,000 trials, the players WOULD flip the same card (value only, jokers included) at least once. The simulation accounts for non-replacement, and uniquely generates two unique numbers to produce an index for the two 53 element arrays (containing 52 standard cards, and 1 joker).

    My coworkers argue that roughly 98% is completely incorrect, but cannot produce a solution of their own that they are confident in. Can anyone shine some light on this? If you would like me to produce the code for the Ruby program then let me know.

    Last edited: Mar 14, 2012
  2. jcsd
  3. Mar 15, 2012 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    The solution you quote from the riddle book, seems incorrect. If the two flipped cards must agree in both suit and value then you have the classical "matching" problem, discussed on pages 107-108 of Feller, Introduction to Probability Theory, Vol. I, Wiley 1968. We say a match occurs if the two cards drawn at some stage are the same in the two decks. Feller derives results for the probability of k matches in two decks of N cards, for k = 0, 1, 2, ..., N; in the current case, N = 53. The probabilities are: P{no matches} = 1 - 1 + 1/2! - 1/3! + 1/4! - ... +- 1/N! = first (N+1) terms in the series expansion of e^(-1). Furthermore, P{exactly k matches} = (1/k!)[1 - 1 + 1/2! - 1/3! + ... +- 1/(N-k)!] =~= e^(-1)/k! for large N. So, the number of matches is almost Poisson with mean = 1; in fact, one can show that for any N the exact expected number of matches = 1.

    Here is a table of the number of matches, k, and their probabilities (exact and Poisson approximation) for two decks of size N = 53:

    k exact Poisson
    0 0.367879 0.367879
    1 0.367879 0.367879
    2 0.183940 0.183940
    3 0.061313 0.061313
    4 0.015328 0.015328
    5 0.003066 0.003066
    6 0.000511 0.000511
    7 0.000073 0.000073
    8 0.000009 0.000009
    9 0.000001 0.000001
    10 0.000000 0.000000

    So, the probability of no matches at all is about 37%, while the probability of exactly one match is about 37%, etc.

    The problem of matching by value only (not suit) seems to be much harder, and I have not seen the solution presented anywhere. Off hand, I cannot see how to deal with it other than by a simulation.

    Last edited: Mar 15, 2012
  4. Mar 15, 2012 #3
    Yes, that seems to be the consensus in the office as well. I appreciate the insight though. I will certainly pass the information along.
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