I was studying for probability class that I'm taking, and ran into a problem that I came up with a solution for, but was wrong. I feel like my intuition is correct, however, but would like to ask for some advice. This particular exercise problem is from Introduction to Probability (2e) - Blitzstein & Hwang.

Question (paraphrased):

Alice and Bob are wondering if they have any mutual friends. Out of a total of $1000$ people, each have $50$ friends. Assume that Alice's Bob's $50$ friends are randomly sampled, and knowing who Alice's friends are gives no information about who Bob's friends are.

Let $X$ be the number of mutual friends they have. Find the PMF of $X$.

My Approach:

The probability that each person will be a mutual friend is $0.05 \times 0.05 = 0.0025$, and we are going through a total of $1000$ people and determining whether they are a mutual friend or not. We could say that $X$ follows a Binomial distribution with $n=1000$ and $p=0.0025$. The PMF of $X$ would therefore be:

$$P(X=k)\ =\ \binom{1000}{k}(0.0025)^k(0.9975)^{1000-k}$$

Textbook Approach:

Condition on who Alice's friends are among the $1000$ people. Then, count the number of ways that Bob can be friends with exactly $k$ of them.

$$P(X=k)\ =\ \frac{\binom{50}{k} \binom{950}{50-k}}{\binom{1000}{50}}$$

Although I also understand the solution and it seems more intuitive, I am wondering if my approach would be completely flawed or not. Both the Hypergeomtric and Binomial distributions give an expected value of $2.5$ mutual friends.

Any feedback is appreciated. Thank you!

  • 1
    $\begingroup$ You can see a difference looking at the probability of having no mutual friends ($k=0$). After Alice has chosen her 50 friends, Bob's first choice must avoid any of them: probability $p_1=950/1000$, for the second friend he has $p_2 = 949/999$, and so on, $p_{50} = 901/951$. The total probability is the product which agrees with the textbook formula. Your formula is just $(0.9975)^{50}$. It is justified if there were "replacement". Hence the difference in the approach is drawing with vs. without replacement. But there is no "replacement" of friends already taken by Bob. $\endgroup$ – Dr. Wolfgang Hintze Oct 20 '18 at 9:55
  • $\begingroup$ Ah thank you very much! I should have thought more carefully about how I was sampling the friends. :) $\endgroup$ – Seankala Oct 20 '18 at 10:16

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