We draw three cards from a deck. We reveal the one with the smallest value. What is the expected value of a card that we draw next?
My approach:
We know that E[x] for a standard 52 sided deck with cards numbered 1 - 13 is simply 7.
I think that the Expected Value of the card we draw next will also be 7 but then not sure about the conditional fact that the previous card is the smallest of the 3 and how do we incorporate that fact.
Define three random variables $X$, $Y$ and $Z$:
Then it is easy to see that $\mathbb{E}[Y] = 7$ and $\mathbb{E}[Z] = 7$, but the problem is asking about $\mathbb{E}[Z \mid X]$, the expected value of the next card with the added information provided by the revealed value, and that's not going to be 7 exactly, although it's going to be close.
Consider an extreme example: imagine that the revealed card is a king. Since it is the smallest of the three drawn cards, then all three drawn cards must be kings. So you know that the next card is going to be drawn from a deck of 49 cards that contain only one king, and four of every other card; so the expected value of the next card in this deck is obviously lower than the usual 7; and in fact: $$\mathbb{E}[Z \mid X = 13] = 6.5 \times \frac{48}{49} + 13 \times \frac{1}{49} \approx 6.63.$$
In general you should expect that if the revealed card is high, then the expected value of the next card is going to be lower than 7, because you'll be drawing from a deck with fewer high cards; and if the revealed card is extremely low, then the expected value of the next card is going to be higher than 7, because you'll be drawing from a deck with fewer lower cards.
I suggest you first calculate $\mathbb{E}[Y \mid X]$, and then $\mathbb{E}[Z \mid Y]$, and finally combine the two formulas to get $\mathbb{E}[Z \mid X]$.
