The expected payoff of a dice game with n above 3 or in general any number For 3 cases, i particularly referred here and probably the best answer but when I take n=4 i.e. no of times dice can rolled, for I get expected off as 4.94 and for n=5, i get 5.12 that means for n=6, logically it means payoff i expect should be 6, and for any case above 6, for example here n=1000, the fair value which i should pay is 6 as that will be expected payoff. Am i right here, not missing something?
 A: The expected payoff will never be exactly $6$, but will approach it in the limit of infinite rolls.
For $4$ and $5$ dice the exact expected payoffs are $\frac{89}{18}$ and $\frac{277}{54}$ respectively. Since $\frac{277}{54}>5$, the expected payoff for $n$ dice from $n\ge6$, $E_n$, is computed from $E_{n-1}$ as
$$E_n=\frac66+\frac16E_{n-1}=1+\frac56E_{n-1}$$
So $E_6=\frac{1709}{324}$, for example. It is easy to show by induction that $E_n<6$ for all $n\ge1$, and that the sequence $\{E_n\}$ is strictly increasing, so $\lim_{n\to\infty}E_n=6$.
An intuitive reason for the expected payoff never being $6$ is that there is a nonzero probability of not getting a $6$ at all in any finite number of rolls, so sometimes you will have to settle for less than that.
A: Adding the Parcly’s answer 
In the given strategy you might not even toss the coin n times and choose to settle with something less than 6. That is, say you’re in your $(n-2)^{th}$ toss according to the strategy you would quit if $(n-2)^{th}$ toss is a 5. Since you have only 2 more tosses left and your expected pay off given this situation is less than 5, so even though you had $n$ tosses to begin with you’re happy to let go off with a 5 now given there are only 2 more tosses left.
Hence your expected payoff would always  be less than 6 for a given finite number of tosses.
