2
$\begingroup$

Envelopes are on sale for Rs. 30 each. Each envelope contains exactly one coupon, which can be one of four types with equal probability. Suppose you keep on buying envelopes and stop when you collect all the four types of coupons. What will be your expected expenditure?

I tried finding a pattern in the probabilities for each of the mass points taking $X$ as the number of trials required. But as I proceeded, it took an ugly structure with little noticeable pattern. Certainly I am not on the right track. Can someone please help me solving the problem?

$\endgroup$
1
$\begingroup$

One way to approach this would be to break down $X$, the number of trials required, into the sum $X = X_1 + X_2 + X_3 + X_4$, where $X_i$ is the number of trials needed to get the $i^{th}$ unique coupon after having already drawn $i-1$ unique coupons.

It is relatively simple to evaluate $E[X_i]$ for each $i$:

  • $X_1 = 1$ deterministically, since no matter what coupon you draw in the first envelope, it will be the first unique coupon type you have drawn. Therefore, $E[X_1] = 1$.
  • Once you have drawn 1 unique coupon, there is a 75% probability that each new envelope you have will have a new type of coupon and a 25% probability that each new envelope you have will be the type of coupon you already have. Therefore, the number of envelopes you need to draw before you get your second unique coupon is distributed as $X_2\sim \mathrm{Geom}(0.75)$, where $\mathrm{Geom}(p)$ is the geometric distribution that counts the number of trials before the first success, where the probability of success is $p$. Therefore $E[X_2] = 1/0.75 = 4/3$.
  • Similarly to the logic above, we see $X_3\sim \mathrm{Geom}(0.5)$ and $X_4\sim \mathrm{Geom}(0.25)$, meaning $E[X_3] = 2$ and $E[X_4] = 4$.

Pulling this all together, we have $E[X] = E[X_1 + X_2 + X_3 + X_4] = 25/3$. Therefore the expected expenditure is Rs. 250.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ That was brilliant. $\endgroup$ – user666 Apr 24 '16 at 21:37

Not the answer you're looking for? Browse other questions tagged or ask your own question.