Find the probability that the hand contain at least two cards of the same rank.

Question

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).

Answer 1

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 }
mean(rpt)
## 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 $\pm 0.001$).