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I learned the following dice game from another forum. It was not solved there. The dice game is as follows.

Toss 6 dice. Each toss you can keep one or more up to the total number of dice tossed. You continue to toss until you have no dice remaining. You may not reintroduce a die once it has been kept.

In order to get a score in this game you must have retained both a 2 and a 4. You get no score for them but without them you get no score at all . Your score is the sum of the remaining 4 dice.

The question is how to act in order to maximize your expected score.

user123981
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    I think this is the same question as http://math.stackexchange.com/questions/661166/what-is-a-good-strategy-for-this-dice-game – miracle173 Feb 02 '14 at 20:57
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    It would surprise me if there is a nice, insightful answer to this question. You tagged it dynamic-programming and I'd suggest you use that: write a computer program that solves this for you - using dynamic-programming. The search space is small enough that this is quite feasible. – Magdiragdag Feb 03 '14 at 08:58

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With out a restriction on how many times you can throw the dice you don't keep(or a cost for rethrowing), it is simple. Throw all six dice until you get a 2 and a 4(You could keep the 2 or 4 from previous throws and continue with the remaining dice, it doesn't matter). Then rethrow the remaining 4 dice until you get all sixes(Again you could keep sixes from previous throws as well).

h4nusGT
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I don't think it is easy, but here are some thoughts. Let's first ignore the 2 and 4 requirement and think about throwing four dice for maximum sum. On the next to last throw, you throw two dice. You must keep at least one, which will obviously be the higher. Should you keep the lower? On average, it will be a $3.5$, so you should keep it if it shows four or above. Now imagine you are throwing three dice. You will keep the highest. The question is whether to keep the second. You can do a more complicated calculation in this vein to get the answer.

My guess would be to start by keeping a 2, a 4, and whatever sixes you get. When you get to three dice, keep fives if you already have the 2 and 4. This is clearly not the final answer-if you throw all sixes on the first throw you can't keep them all.

Ross Millikan
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Perhaps some useful insight for a good strategy in the original problem can be obtained from the solution of the following simpler problem in which you start again with rolling six dice. You are allowed no more than six rounds to roll the dice and after each round you have to put aside at least one of the dice, but you never put aside more dice than necessary in order to reach your goal. Your goal is to have put aside a 2 and 4 within the six rounds . Let $p(0,r)$ be the probability of reaching your goal from the state in which you have neither a 2 nor a 4 and still $r$ rounds are allowed . Also, let $p(1,r)$ be the probability of reaching your goal from the state in which you have put aside either a 2 or a 4 and still $r$ rounds are allowed. These probabilities can be numerically obtained by calculculating absorption probabilities in a Markov chain with 13 states including two absorbing states. I found that the probability $p(0,r)$ has the values 0.9567, 0.8725, 0.6943, 0.4180, and 0.1389 for $r$=6, 5, 4, 3, and 2. The probability $p(1,r)$ has the values 0.9350, 0.8384, 0.6651, 0.4212, and 0.1667 for $r$=5, 4, 3, 2, and 1.

user123981
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