5 cards are randomly selected (without replacement) from a standard deck of 52 playing cards (13 ranks: 2, 3, 4, ..., 10, J, Q, K, A, and 4 suits: S, H, D, C). Find the probability that the hand contain at least two cards of the same rank (e.g. {2, 2, 6, A, Q}, {J, J, K, 4, J}, {8, 8, A, A, 6}, ...).

I know that I can use $$1 - \frac{\binom{13}{5}\binom{4}{1}^5}{\binom{52}{5}}$$

Is there any other way to do this if I don't want to use $1 -$ (something).

  • 3
    $\begingroup$ There is but it is much more difficult computationally. $\endgroup$ – Jimmy R. Oct 5 '16 at 15:58
  • 2
    $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Oct 5 '16 at 16:45
  • $\begingroup$ not at least 2 means one of each rank for the 13 cards. $\endgroup$ – user354674 Oct 5 '16 at 17:20
  • 2
    $\begingroup$ In a direct counting approach, count individually the probabilities of exactly one two of a kind, exactly two two of a kinds, exactly one three of a kind, a two of a kind and a three of a kind, and a four of a kind. Add each respective probability. Assuming I didn't forget any cases, the answer will match yours. This is of course very tedious to do comparatively speaking. This is made even worse if asking about a ten card hand instead of just a five card hand. $\endgroup$ – JMoravitz Oct 6 '16 at 2:57

The following simulation in R stataistical software performs your experiment a million times: Use a deck that does not distinguish suits, choose 5 cards without replacement, count the unique denominations; if the answer is less than 5, then you have at least one repeated denomination. My answer is 0.493, which ought to be correct to two, maybe three, places. (My m-vector rpt is a logical vector, with elements taking values TRUE and FALSE after the loop; the mean of a logical vector is its proportion of TRUEs.)

deck = rep(1:13, 4)
m = 10^6;  rpt = logical(m)
for (i in 1:m) {
  hand = sample(deck, 5)
  rpt[i] = length(unique(hand)) < 5 }
## 0.493236

It seems to me that a correct computation of the probability of avoiding a repeat is $(52/52)(48/51)(44/50)(40/49)(36/48).$ Subtracting that from 1, I get 0.4929172, which agrees with the simulation (within the margin of simulation error, about $\pm 0.001$).

| cite | improve this answer | |
  • $\begingroup$ Compute his answer again and you will see it matches yours. Check that you did 4^5, not 4^4. $\endgroup$ – JMoravitz Oct 6 '16 at 2:51
  • $\begingroup$ Another gripe i have with this answer is that it does not answer the question of finding an analytical solution which avoids the "1 minus something" approach which was explicitly asked for by the op. (A brute force search or a simulation are not analytical solutions and not desired either in any case in which an analytic solution is easily accessible) $\endgroup$ – JMoravitz Oct 6 '16 at 3:03
  • $\begingroup$ @JMoravitz: Sorry about the mistake, now removed. Never repentant about checking analytical solutions with simulations, which I often find helpful. But I respect opinions about 'desirability', even when I disagree. BTW: Do you have a solution that is not unbearably tedious and avoids the '1 - something' approach? I agree that would be interesting to see. $\endgroup$ – BruceET Oct 6 '16 at 3:31
  • $\begingroup$ See my comment above on the main post. If I had access to a computer, I might type it up as a full answer including calculations, but I only have access to a mobile phone at the moment. I would still call my solution unbearably tedious (especially in the generalized problem) but still doable. $\endgroup$ – JMoravitz Oct 6 '16 at 3:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.