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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.

user157902
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  • 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 02:44
  • @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 02:57
  • @RossMillikan you're correct, not the strategy...the winnings do though – afedder Jun 19 '14 at 03:55
  • @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 04:16

2 Answers2

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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.

Ross Millikan
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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.

Ben Grossmann
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