Optimal strategy to a card game There are $3$ face down cards in front of you. On the side facing the table (you cannot see this), there are labels on each of the three cards. They are $n$, $n + 1$, and $n + 2$. You are not told what $n$ is.
You get to flip a card at random. After flipping the first card, you can either stay or choose another card at random. Then, you again have the option to stay or flip the last card. 
Once you flip one card, you cannot go back to the previous card's value. 
How can you maximize the value of the card you choose to stop flipping on? 

My thoughts:
You can achieve an expected value of $n + 1$ by just stopping on the first card every time.
If you flip another card and it is one less than the first one you flipped over, the last card is $n + 1$ with probability $\frac{1}{2}$ (this occurs when the first card you flipped was $n$), and the last card is $n$ with probability $\frac{1}{2}$. So, I think that the expectation of the last card given the second card you flipped is one less than the first one is equal to $n + \frac{1}{2}$. 
If you flip another card and it is one more than the first one you flipped over, again, I think that the expected value of the last card is $n + \frac{1}{2}$
Anyone have an optimal strategy?
 A: Here is a basic strategy that should do better than just flipping one card and staying with it:
Always flip a second card after the first. If the second card has a higher value than the first, stick with the second card, otherwise flip the third card.
Why does this better than staying with just the one first card?
Well, as you say, just flipping one card has an expected value of $n+1$
But for the other strategy:
There are 6 cases to consider, each of which is equally likely:
First card is $n$, second is $n+1$. The strategy says to stick with the second card, so outcome is $n+1$
First card is $n$, second is $n+2$. Stick with card. Outcome is $n+2$
$n+1$ followed by $n$. Strategy says to flip third card. Outcome $n+2$
$n+1$ followed by $n+2$. Stick with card: $n+2$
$n+2$ then $n$. Flip third card: outcome is $n+1$
$n+2$ then $n+1$. Flip third card: $n$
Since each of these events is equally likely, the expected outcome of this stratey is just the average of these outcomes, which is $n+\frac{4}{3}$
OK, so that's indeed a better strategy. Is it optimal? I don't know.
