Cardgame 36 cards paradox Suppose we split a deck of $36$ cards in four. You take one fourth and so do I... There's a $50\%$ chance that one of us has heart ace. If by discovering my cards I don't have it, does it mean that there's still a $50\%$ chance that it is in your hand, considering that you have $9$ cards and $18$ are still on the side?
On the other side the ace must be in one of the remaining $27$ cards and we still have $9$, so it would be $33+\frac{1}{3}\%$.
thank you for your answers!
 A: (This problem has nothing to do with the Monty Hall problem)
Your second thought is correct. There are still 27 cards on the table and you hold 9. So the probability that you hold the A of hearts is $ \frac{9}{27}=\frac{1}{3}$.
In order to make it more clear, this is how a "monty hall" problem would be in your case:


*

*You take 1 of four piles, and I take one.

*Since there are 2 piles left, one of them certainly does not contain the Ace of Hearts

*Someone (who can see all the cards) reveals this pile. So now there are 3 piles remaining (our 2 selected and one more on the table)


$ \Rightarrow $ Still your initial claim is correct "There's a 50% chance that one of us has heart ace"
The difference between these two examples, is the word "certainly". 


*

*In this second case, we did something that we could always do. So this cannot change the result.

*In your example, when revealing your pile, you might have found the Ace of Hearts. So you cannot always reach the final status of 27 cards (containing the A of H).

