As part of a course, I came across this question
"A player is offered to play against a dealer who rolls 6 dice simultaneously. If there's a 'street' (each of the numbers from $1$ to $6$ appears on each of the dice), then the dealer wins $60\$$, else, the player wins $1\$$. Is this game profitable in the long run?"
For this, I tried writing the expression for the expected value after $n$ plays and tried to examine the expression in the limit, and whether it is always positive. What I got is the following
$$E(n,p) = \sum_{i=0}^n (n-61i){n \choose i}p^i(1-p)^{n-i}$$
Where $p$ is the probability of a street occuring (which would be $\frac{6!}{6^6}$)
I am stuck on this expression, don't know how to make any conclusions from this. Another interesting side note is to quantify the effect of changing the winning/losing amounts on $E$, to design a game where it is break even in the long run
Edit - corrected the probability of obtaining a street, forgot to account for permutations
I derived the expression by considering the cases of $i$ losses and $(n-i)$ wins in $n$ trials and summing over $i$ weighted with the return from each possible trial