Tricky probability problem (2 possible solutions ?!)

Question

The problem is stated as follow:

You are trying to complete the World Cup sticker album, for that you need distinct 600 normal stickers and 80 distinct holographic stickers. It is known that the probability of getting a holographic in a sticker pack is $\frac{1}{5}$ of the probability of getting the normal sticker. Suppose that the sticker packs comes with 2 stickers.

(a) Given that you need only 1 normal sticker to complete the album, what is the probability of getting this sticker in the next pack?

(b) The same problem as (a) but now it is a fact that no sticker pack comes with 2 equal stickers.

PS: (a) is pretty straightforward but item (b) seems to me to have two apparently correct reasonings that gives different answers:

• 1. Define the events:

A: You get the normal sticker you need

B: The sticker pack comes with different stickers

$P = P(A|B) = \frac{P(A \cap B )}{P(B)} = \frac{\frac{5}{3080}\frac{3075}{3080} + \frac{80}{3080}\frac{5}{3080} + \frac{2995}{3080}\frac{5}{3080}}{\frac{3000}{3080}\frac{2995}{3080} + \frac{3000}{3080}\frac{80}{3080} + \frac{80}{3080}\frac{3000}{3080} + \frac{80}{3080}\frac{79}{3080}} = 0.003246643551268461$

• 2. If you think about choosing with no replacement:

$P = \frac{5}{3080} + \frac{80}{3080}\frac{5}{3079} + \frac{2995}{3080}\frac{5}{3075} = 0.0032466983793436$

PPS: I arrived at this problem when I was trying to solve this puzzle: https://gscap.com.br/puzzle/. It asks the expected value of the money you need to complete the album. Some of you may be interested in solving it!

Answer 1

Consider a much smaller sample so we can brute force the calculations as necessary. We have $\{A_1, A_2, B, C\}$.

• Scenario 1: We draw two letters from the set independently and uniformly at random. If they are both the same letter ($A_1$ and $A_2$ both count as $A$'s) we try again from the start.

You will find that each of $\{A_1,B\}, \{A_1, C\}, \{A_2, B\}, \{A_2, C\},$ and $\{B,C\}$ are equally likely to have occurred. In particular, $\{B,C\}$ occurs with probability then $\dfrac{1}{5}$

• Scenario 2: We draw one letter from the set uniformly at random. We remove that letter (both copies in the case of being letter $A$) from the set and then draw again from whatever remains.

Here, we will have the same outcomes however we can no longer claim they are equally likely to have occurred. You will see in particular that $\{B,C\}$ will occur with probability $\dfrac{2}{4}\times \dfrac{1}{3} = \dfrac{1}{6}$

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It is a direct result of the difference in selection processes which caused the difference in probabilities. In the same way, your selection processes implied by your calculations above were using these two respective scenarios in their attempted calculations. Both appear to be calculated correctly however they are correct answers to their own respective different problems.