What is the fair price of the game below? You are playing a fair die game with 'n' die rolls. You have decided on the "fair" price of the game. So, to take part in this game you need to pay upfront this value. Then you get to roll a die, n times. If you ever role a die more than or equal to this value you instantly cash out, and the game ends. If not, you keep rolling and take whatever is the nth die roll value. For ex:
3 Die Roll Game. You decided the fair value is 3 (this is not the actual fair price, just an illustrative example.) A person rolls a 2 in his first attempt. It is less than 3, he continues. Again rolls a 2, again continues. In his 3rd and last attempt he rolls a 1. He has to cash out now with a value of 1, and hence a net loss of 2 because his 3 rolls are done. (Note, at equality when the die's roll is 3 in this case, he cashes out)
What is the fair value of a game with n rolls, such that the expected profit should be zero.
P.S. This is not the simple choose your maximum out of n die rolls game. I have tried working it out, but I am missing it. Can anyone give me the basic logic that I need to write a code for this. I am looking for a cleaner proof than enumerate all the possibilities. I was thinking on the lines of Recursion.
 A: For $n=1$ the fair price is $3.5$, because that is the expectation of the single roll.  As you increase the number of rolls, you should stop whenever the current roll exceeds the expected value of the following series. For $n=2$, you will stop if you get $4,5,6$ on the first roll, and roll again otherwise.  The value is then $\frac 12 \cdot 5 + \frac 12\cdot 3.5=4.25$  Then if $n=3$ you should only stop on the first roll with a $5,6$, so the value is $\frac 13 \cdot 5.5 + \frac 23 \cdot 4.25=\frac {14}3$
A: For reasonably large $n$ I think it is clear that the answer you want, call it $E$, is between $5$ and $6$.  
Assuming this, we see that the effective cutoff is $6$.  That is, you stop if you ever throw a $6$, otherwise you get the last value.  
The probability of reaching the last value is $(\frac 56)^{n-1}$.  The probability that you get a $6$ earlier than the last value is $1-(\frac 56)^{n-1}$.  Thus the expected payout of this game is $$E_n=(1+2+3+4+5+6)*\frac 16*(\frac 56)^{n-1}+6*(1-(\frac 56)^{n-1})$$
This is the answer you seek, provided $5≤E_n≤6$ .  For $n=11$, say, this comes to about $5.125$, which is satisfactory. For $n=10$ the value is less than $5$ (barely) so you'll need to try a lower cutoff.
