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:


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*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.

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 : $\frac{2}{6}\times \frac{1}{5}=\frac{1}{15}$
