There are 42 actors in our movie set. We have 2 rounds of random assignments to teams of 3. In each round, we will randomly divide the actors into 14 total groups of 3 persons each.

(A) Suppose that we have that A and B are teammates in Round 1. What's the probability that they are teammates again in Round 2?

(B) What's probability that no pair of teammates in Round 1 are teammates again in Round 2?

I tried to use Bayes formula for part A but it appears that the two events are independent of one another. So therefore, is it the case that the given information isn't needed?

for B, I tried to do $1-P(\text{at least one pair of teammates})$, does this approach work?

My work for A:

$P(\text{teammates in round 2} \mid \text{teammates in round 1}) = P(\text{teammates in round 2 and teammates in round 1})/P(\text{teammates in round 1}) = P(\text{teammates in round 2})$ (since these are independent of each other).

The probability of being teammates in round 2 is just

$\frac{1}{14} * \frac{1}{14}*14$, where we have 14 ways of arrangements.

So the final probability for part A is $\frac{1}{14}$

  • 2
    $\begingroup$ I don't think your answer for (A) can be correct. The probability that A and B are teammates in round 2 isn't 1/14. There are 42 slots. One is occupied by A. Of the 41 other slots only two of them are in the same group as A. $\endgroup$
    – Glen_b
    Commented Feb 11, 2014 at 6:39

1 Answer 1


Part A: Your answer is not correct. The assignment of teams in the two rounds are independent, so this is just the probability that two specific actors A, B do get assigned to the same team of three. This can be calculated as $$ \DeclareMathOperator{\P}{\mathbb{P}} \begin{align} \P(\text{A,B assigned to same team})&=& \\ \sum_{j=1}^{14}\P(\text{B assigned to team $j$} \mid \text{A assigned to team $j$}) &=&\\ \sum_{j=1}^{14} \frac1{14}\cdot \frac{3-1}{42-1}&=& 14\cdot \frac1{14} \cdot\frac{2}{41} \end{align} $$ Part B: What is the probability that no actors in same round 1 team gets assigned together in same team in round 2? Since, again, assignments in each round are independent, so it is enough to see at round two. Formulate as a problem of colored balls in buckets: We have 14 colors, three balls of each color. Throwing randomly into 14 buckets, what is probability that no bucket has two or three equal-colored balls? I cannot see fast an simple analytical solution (though one could be programmed), so first a simple simulation (in R) to get an idea:

N <- 14L # Number of teams/colors
p <- 3L  # NUmber of team members

# simulate one round,  return number of buckets with no same-colored pair:

one_round <- function() {
    balls <- rep(1:N, rep(p, N))
    new_buckets <- sample(balls, length(balls)) %>% matrix(p) # buckets as cols 
    lu <- new_buckets %>% apply(2, function(x) length(unique(x)))
    sum(lu < p)                           

set.seed(7*11*13) # My public seed
B <- 1e6L
res <- replicate(B, one_round() )

[1] 0.120614

So the probability is around 12%. Some ideas for an analytical solution (or approximation) would be nice! A similar question (without a complete answer).


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