Link to Problem: HERE

Two cars are drawn successively from a deck of 52 cards. Find the probability that the second card is higher in rank than the first card.

There are three possible outcomes to the described experiment.

  1. same: the cards have the same rank.
  2. higher: the second card is higher than the first
  3. lower: the second card is lower than the first

Thus the probability distribution becomes \(1 = P(s) + P(h) + P(l)\)

There are \(52 \times 51 = 2652\) possible outcomes from this experiment. \(156\) of them result in a tie. Thus, \(P(s) = \frac{156}{2652}\). Given any card for the first selection there are always 3 out of the remaining 51 cards that will result in a tie. \(\frac{3}{51}\times2652=156\)

Next \(P(l) = P(h)\). Of the remaining \(2652-156=2496\) combinations half are P(l) and the other half are P(h). Given any first card, second card combination that satisfies \(P(l)\) the card’s orders can be reversed to find a combination that satisfies \(P(h)\). The same is true in reverse. Combine this with the fact that each two card combination is equaly likely and clearly \(P(l)=P(h)\).

Substituting into the original equation:

\(1 = P(s) + P(h) + P(l)\)

\(1 = \frac{156}{2652} + 2P(h)\)

\(\frac{1 - \frac{156}{2652}}{2} = P(h)\)

\(P(h) = \frac{2496}{5304} \approx 0.4705882\)

Below is a simluation written in R to verify the solution:

set.seed(1)
same <- 0
lower <- 0 
higher <- 0

for(i in 1:100000){
  result <- sample(rep(1:13, 4), 2)
  if(result[1] == result[2]){
    same <- same + 1
  }else if(result[1] < result[2]){
    lower <- lower + 1
  }else{
    higher <- higher + 1
  }
}
print(paste0("Same: ", same, ' Prob: ', same/100000))
## [1] "Same: 5939 Prob: 0.05939"
print(paste0("Lower: ", lower, ' Prob: ', lower/100000))
## [1] "Lower: 47010 Prob: 0.4701"
print(paste0("Higher: ", higher, ' Prob: ', higher/100000))
## [1] "Higher: 47051 Prob: 0.47051"