What is the best strategy for this dice game?

Given the following game, what is the best strategy to maximize your total score?

Each game play starts with a random point value. You are then given a chance to multiply that value by choosing between 1-5 dice to roll. If you select n dice, all n must come up six. If you don't win the dice toss, you still keep the starting value. The return values for the different number of dice are:

• n = 0: 1x
• n = 1: 2x
• n = 2: 5x
• n = 3: 30x
• n = 4: 200x
• n = 5: 1000x

You are allowed to repeatedly play the game and the total score is cumulative across game plays.

• Call $X$ the random point value. Then, the winnings $W_n$ is a function of $X$. Let $a_n$ be the multiplier for each choice of $n$. Now, see if you can compute the expected value of $W_n$ for fixed $n$. Note that this expectation will depend upon $\mathbb{E}(X)$, but you included no additional information about the random point value. It follows that the strategy is dependent upon the distribution of the random point value. – afedder Jun 19 '14 at 2:44
• @afedder: if the random value is chosen at the start, it just multiplies the score and the strategy doesn't depend on it. – Ross Millikan Jun 19 '14 at 2:57
• @RossMillikan you're correct, not the strategy...the winnings do though – afedder Jun 19 '14 at 3:55
• @user157902 compute the expected value of $W_n$ and this will depend upon $\mathbb{E}(X)$. However, you can treat this as a constant and you can still maximize $W_n$ with respect to $n$ to find the optimal choice for $n$. – afedder Jun 19 '14 at 4:16

Let $P$ be the point value. The expected profit (compared to failure) from setting $n = 1$ is $$E = \frac 16\cdot (2-1)P$$ Similarly, $$n=2: E = \frac 1{6^2}(5-1)P\\ n=3: E = \frac 1{6^3}(30-1)P\\ n=4: E = \frac 1{6^4}(200-1)P\\ n=5: E = \frac 1{6^5}(1000-1)P\\$$ So, in any particular turn, we maximize the expected profit by selecting $n=1$. So, sticking to $n=1$ seems to be the optimal strategy, in the long run.
Hint: think first if you only had the choice of $1$ die or $2$ dice. What is the expectation for $1$ die? Most (how much) of the time you get $x$ and some (how much) of the time you get $2x$. By contrast, if you roll $2$ dice, most (how much) of the time you get $x$ and some (how much) of the time you get $5x$. Calculate the expectation of each and you know what to do. Now do the other numbers of dice similarly.