Ad Display Strategies in Newsfeed

Solution & Explanation

Option 1: One ad is shown for every 25 stories

In this strategy, the placement of ads is deterministic. For every 25 stories, exactly one ad is shown. Therefore, in a sequence of 100 stories:

• Expected number of ads:

  $\mathbb{E}\left(\text{No. of ads in 100 stories}\right) = \frac{100}{25} = 4$

  This means that on average, a user will see 4 ads in 100 stories.

Option 2: Each story has a 4% probability of being an ad

In this strategy, each story independently has a 4% chance of being an ad. This follows a binomial distribution with parameters $n = 100$ (number of trials) and $p = 0.04$ (probability of success, i.e., a story being an ad).

• Expected number of ads:

  The expected value for a binomial distribution $\text{Binomial}(n, p)$ is given by:

  $\mathbb{E}[X] = n \cdot p$

  Substituting the given values:

  $\mathbb{E}[\text{No. of ads in 100 stories}] = 100 \times 0.04 = 4$

  Like the first strategy, the expected number of ads in 100 stories is also 4.

• Probability of exactly one ad in 100 stories:

  This requires calculating the probability of exactly one success (ad) in 100 trials (stories), which is given by the probability mass function (PMF) of the binomial distribution:

  $\mathbb{P}(X = 1) = {n \choose x} \cdot p^x \cdot (1-p)^{n-x}$

  Substituting the values:

  $\mathbb{P}(X = 1) = {100 \choose 1} \cdot (0.04)^1 \cdot (0.96)^{99}$

  • Calculation:

    ${100 \choose 1} = 100$

    $(0.04)^1 = 0.04$

    $(0.96)^{99} \approx 0.01757$

  • Result:

    $\mathbb{P}(X = 1) = 100 \times 0.04 \times 0.01757 \approx 0.0703$

  Therefore, if the second strategy is chosen, the probability that a user sees exactly one ad in 100 stories is approximately 7.03%.

Summary

• Both strategies result in an expected number of 4 ads in 100 stories.
• The probability of seeing exactly one ad in the second strategy is approximately 7.03%.