Picking cards sequentially vs consecutively We have a pack of 6 cards over the table. Cards are: {A, A}, {B, B}, {C}, {D}.
There are 3 players (Papa, Pepe, Popo) sat around the table. Cards are all upside down, so the players cannot see which card they are picking.
Now we are presented with two scenarios:


*

*Case 1: each player picks one card at a time. That is, first Papa picks a card, then Pepe, then Popo, until there are no cards over the table (i.e. cards are being picked sequentially).

*Case 2: each player picks their 2 cards directly, that is: Papa picks 2 cards, then Pepe, then Popo. (i.e. cards are being picked consecutively).


I am being asked if the probability that Papa has the two A's is the same (or not) for both cases.
My knowledge on statistics and probability is not very advanced, but I'd say that the chance for Papa picking the two A's will not be the same, but I don't have a convincing argument in favour. Maybe someone could point me towards the right direction in solving this problem.
 A: The probabilities that Papa picks two A's are actually the same.


*

*Case 1


Probability of Papa picks A: $\frac{2}{6}$
Probability of Pepe picks cards other than A: $\frac{4}{5}$
Probability of Popo picks cards other than A: $\frac{3}{4}$
Probability of Papa picks A again: $\frac{1}{3}$
So, probability of Papa picks two A's : $\frac{2}{6}\times \frac{4}{5} \times \frac{3}{4} \times \frac{1}{3}=\frac{1}{15}$


*

*Case 2


Probability of Papa picks A: $\frac{2}{6}$
Probability of Papa picks A again: $\frac{1}{5}$
So, probability of Papa picks two A's : $\frac{2}{6}\times \frac{1}{5}=\frac{1}{15}$
A: Case 1
For Papa to get two As, Papa needs to pick an A first out of the $6$ cards, then Pepe needs to pick not an A out of the remaining $5$ cards, then Popo needs to pick not an A out of the remaining $4$ cards, and then Papa needs to pick the other A out of the remaining $3$ cards.
Case 2
For Papa to get two As, he needs to pick an A first out of the $6$ cards and then he needs to pick the other A out of the remaining $5$ cards.
In both cases, we simply need to take the probabilities of each card pick happening and multiply them together, which shouldn't be too hard. Then, once you've done that for both cases, you can compare them to see if they are the same or not.

Now, as you found, both cases have the same probability of $\frac{1}{15}$. To understand why, let's consider the general case where the picking order could anything, as long as Papa picks two cards.
Instead of thinking of it like this, however, let's think of this as a string of letters including $AABBCD$ in the order that the cards are picked. For example, the string $ABCABD$ means that someone picks $A$, then $B$, then $C$, then $A$, then $B$, then $D$. However, these "someone"s could be anyone as long as two of these people are Papa.
Now, in the first case, Papa picks the first and fourth card. The probability that the first card is an A is $\frac 2 6$. Now, given that, there are now five other cards in the picking order and one A left, so the probability that the fourth card is an A is $\frac 1 5$. Thus, the probability for this case is $\frac 2 6*\frac 1 5=\frac 1 {15}$.
In the second case, Papa picks the first and second card. The probability that the first card is an A is $\frac 2 6$. Now, given that, there are now five other cards in the picking order and one A left, so the probability that the second card is an A is $\frac 1 5$. Thus, the probability for this case is $\frac 2 6*\frac 1 5=\frac 1 {15}$.
Notice that for both cases, we have exactly the same logic, just different spots Papa is picking the cards from. This is because we are thinking about this in terms of letters in a string instead of cards picked in an order, so we don't need to worry about all of the letters before or in between. Thus, as long as each person picks two cards, the probability that Papa gets two As is $\frac 1 {15}$.
