probability – Calculating the Expected Value of Sum of Cards in Probability Games

Question

If each card on a regular 52 deck card has points that corresponds to their number (like 2 of hearts is 2 points, 7 of clubs is 7 points), the Jack, Queen, King each being 10 points and you keep drawing without replacement until sum of all points is 10 or more….what's the mean of sum of points?

Answer 1

By your definition, you have $16$ cards ($10$, $\text{J}$, $\text{K}$, $\text{Q}$) that are worth $10$ points, so with probability $16/52$ you get $10$ points in a single draw. Since $9+\text{anything}=10$, then if we take into consideration that there is $4$ nines, than we instantly know that with probability greater than $20/52$ you finish in two draws. However, return of two draws is simple to obtain by enumerating all $52 \choose 2$ combinations of card pairs and summing their scores.

unique_cards <- c(1:10, 10, 10, 10)  # A, 1, 2, ..., 10, J, Q, K
unique_cards <- rep(unique_cards, 4) # each appears 4 times

comb <- combn(unique_cards, 2) # take all possible combinations of card pairs

what gives $79\%$ probability of obtaining score of at least $10$ in two draws

> sum(colSums(comb) >= 10)/choose(52, 2) # accepted / all combinations
[1] 0.7888386

Lazy solution for more than two draws can be obtained by a simple simulation, where whole deck is shuffled and then cards are drawn until their total score is at least $10$.

set.seed(123)

sim <- function(target = 10) {
  res <- cumsum(sample(unique_cards)) # shuffle, draw and sum
  n <- which.max(res >= target)       # take first score >= 10
  c(sum = res[n], n = n)
}

R <- 1e4
res <- replicate(R, sim())

and the result is that on average you have to draw two cards and the average total score is $12.77$

> apply(res, 1, summary)
          sum     n
Min.    10.00 1.000
1st Qu. 10.00 1.000
Median  12.00 2.000
Mean    12.77 1.946
3rd Qu. 15.00 2.000
Max.    19.00 6.000

Moreover, as expected, with approximately $30\%$ probability you finish with one draw, but with $79\%$ probability you finish with two draws and you rarely get over three draws:

 > cumsum(table(res[2,])/R)
     1      2      3      4      5      6 
0.3006 0.7910 0.9653 0.9968 0.9999 1.0000