Probability about cards

Question

Eight hands of five cards each are dealt from a shuffled deck of cards without replacement so that there are twelve undealt cards. Find the probability that at least one of these hands has no diamonds.

The solution uses the inclusion-exclusion theorem. Is there any other way to solve this problem without using that theorem?

Thanks.

Answer 1

Comment. A simulation of a million iterations of this game approximates the answer as $0.9248\pm 0.0005.$

In the R code below, in deck Diamonds are 1s, cards of other suites are 0; dlt is a 40-vector of cards dealt; h1 is $0$ if there are no Diamonds in Hand 1, and so on for the other seven hands. Thus. nd has a million elements: $0$ if one or more hands has no Diamond. Finally, mean(nd) is the proportion of iterations in which one or more hands had no Diamond. With a million iterations we can expect simulation error of $\pm 0.001.$

deck = c(rep(1,13),rep(0,39))
set.seed(2020)
m = 10^6;  nd = numeric(m)
for(i in 1:m) {
 dlt = sample(deck, 40)
 h1 = sum(dlt[1:5]);    h2 = sum(dlt[6:10])
 h3 = sum(dlt[11:15]);  h4 = sum(dlt[16:20])
 h5 = sum(dlt[21:25]);  h6 = sum(dlt[26:30])
 h7 = sum(dlt[31:35]);  h8 = sum(dlt[36:40])
 nd[i] = h1*h2*h3*h4*h5*h6*h7*h8 }
mean(nd==0)
[1] 0.924804
2*sd(nd==0)/1000
0.0005274149    # aprx 95% margin of sim error

Note: Consider the following very rough combinatorial approximation. The probability that one 5-card hand avoids Diamonds is ${{13\choose 0}{39\choose 5}}/{{52\choose 5}} \approx 0.22.$ Then, if the eight hands were independent, the probability of at least one hand without Diamonds would be about $1 - .78^8 \approx 0.86.$

Answer 2

You can do it by computing the probability that each hand has at least one diamond, and subtracting from $1$. While this doesn't use inclusion-exclusion, I don't think it's easier, and I do think it's more error- prone.

You'd have to start by figuring out the ways that the $13$ Diamonds can be distributed. We have $9$ numbers that add up to $13$, that is is we add up the number of Diamonds in each hand, and the number remaining in the undealt portion of the deck. The numbers in the hands must be positive, but the number in the deck can be zero. That is, we must consider the $5$ partitions of $13$ into $9$ parts, and the $7$ partitions of $13$ into $5$ parts. Also, we must take cognizance that a partition like $5,1,1,1,1,1,1,1,1$ corresponds to $2$ cases: one where each hand is dealt $1$ Diamond, and one where $7$ hands are dealt $1$ Diamond and the fifth is dealt $5$. The partition $4,2,1,1,1,1,1,1,1$ corresponds to $3$ different cases.

I think you'll agree that inclusion-exclusion is a superior method.

This assumes that you want the theoretical exact answer. Simulation is a practical alternative. Personally, if I did it by inclusion-exclusion, I would check my answer by simulation.