This was a problem that arose from a game that me and my friends play following a game of golf to determine who picks up the post game refreshments in the clubhouse. Let me first explain the game, then I will give you the question that we have come up with. For simplicity, assume that there are only two players, A and B. The first thing that they do is rollone die, highest roll going first (Let's say player A). Player A rolls 5 die and makes the best "poker hand" that he can (disregard straights and full houses, the only thing that we care about is pairs, three of a kind, four of a kind and the five of a kind. Assume that he rolled a 2,3,5,5,6. He now has several options. The first is to stay put, meaning that player B would have to beat a pair of fives (nothing but the high pair or three of a kind etc. matters--no two pair, no high card) in one roll. Or, player A can pick up all the die and roll again, after which he is confronted with the same options. A player can only roll 3 times, after which his roll is what he is stuck with. Or, player A can leave the two fives (or any other combination for that matter, and re-roll as many as he would like). In this case, he would obviously only be rolling three die, but already have at least a pair banked, hoping to build on them WITH ONLY ONES OR FIVES, as there is nothing else that can help a pair of fives. So that is the game. Let me know if there are any questions or if I did not make myself clear enough.(adsbygoogle = window.adsbygoogle || []).push({});

Here is the question: If, after the first roll, you have nothing but a one and a six (assume 1,6, 4,5,3), obviously you keep the one, but what about the six? Here is the way that I started to solve it and was wondering if there would be a way that would not require to do it all by hand as I did the first part.

Assume you are playing the same game, only with 2 die and only two rolls. So there are 9 possible outcomes. The worst that you can get is a three high, and the best that you can get is a pair of sixes. Assign a value to each of the outcomes in ascending order of what they beat:

high 3- 1

high 4- 2

high 5 -3

etc. etc. on to Pair of sixes - 9.

Then you do the old math thing and figure every possible roll out and the value that they provide, and since ever roll is as likley as the next, you average all the rolls out and you find that the average on one roll of two dice is 4.83, or right in between a high six and a pair of twos. So, if you are going first and you roll anything higher than a pair of twos, you stay, and have that amount in one roll and your opponent must beat it in one roll. But if you have anything less than a pair of twos, you must roll again. But How many Die to re-roll? Assume that you have rolled a 3-high, and you are going to keep the three; your average will be a 3.6 compared the the 4.83 that we figured was the average when rolling two die on one roll. So you runn all the numbers for everything less than a pair of twos and find that if you have a three or four high you pick up both, five high is a push and 6 high means you keep the 6 and roll one.

So (I think) that is the problem solved for 2 die, but can anyone figure out a way to do it with five die? There are some interesting patters that develop, but I could not find a pattern that would carry over. I thought about Monte Carlo Simulation and setting n to 100,000 and letting it go by itself so I would at least know the answer, but I am having trouble setting it up. I realize that this is long winded, but this has been bothering me for two weeks and I am in desperate need of help. Any guidance would be greatly appreciated. I have another more genreal question that I will post later on, but I have to at least look like I am trying to work some of the time.

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# Dice Game Problem (Hard)

Can you offer guidance or do you also need help?

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