What is the Optimal Strategy for a Dice-Based Poker Game?

In summary: This method can be applied to any number of dice and rolls, making it a versatile solution for this game. In summary, the conversation discussed a game played with dice to determine who picks up post-game refreshments. The game involves rolling dice and trying to get the best "poker hand" with pairs, three of a kind, four of a kind, and five of a kind. After the first roll, players have the option to keep their current hand or re-roll some or all of the dice. The conversation also included a question about what to do with a one and a six after the first roll, and a potential solution using a Monte Carlo simulation to determine the best strategy for different numbers of dice and rolls. This method can be
  • #1
Mudrow
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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.

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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  • #2
One potential solution to this problem is to use a Monte Carlo simulation. You would set up a program that generates random rolls of five dice and then assigns a value to each roll. You can then run the simulation multiple times and calculate the average value for each roll. This will give you an idea of what the expected value for each roll is, and you can use this information to decide whether or not to keep certain dice in your hand or re-roll them. Additionally, you could also run the simulation multiple times with different starting parameters (e.g. the number of dice rolled each time) to see how this affects the expected value of each roll.
 
  • #3


First of all, thank you for providing a thorough explanation of the game and the question at hand. It is clear that you have put a lot of thought and effort into trying to solve this problem.

To answer your question, there is a mathematical concept that can help in solving this problem - expected value. Expected value is the average value that we can expect to get from a random event. In this case, the random event is rolling the dice.

To calculate the expected value of a roll, we need to first determine the possible outcomes and their corresponding probabilities. In this game, the possible outcomes are: nothing, a pair, three of a kind, four of a kind, and five of a kind. The probabilities of each of these outcomes can be calculated by considering all possible combinations of the dice.

For example, the probability of getting a pair is (6/6)(5/6)(1/6)(1/6)(1/6) = 5/216. This is because there are 6 different ways to get a pair (such as 2,2,3,4,5 or 1,1,3,4,5) out of a total of 6^5 = 7776 possible combinations.

Once we have calculated the probabilities of each outcome, we can assign a value to each outcome (as you have done in your solution) and then calculate the expected value using the formula: expected value = probability of outcome x value of outcome.

For example, the expected value of getting a pair is (5/216) x 9 = 5/24. This means that on average, if you roll the dice, you can expect to get a pair 5/24 times.

Using this concept, we can solve the problem for any number of dice. The expected value of a roll with five dice will be the sum of the expected values for each possible outcome (nothing, pair, three of a kind, four of a kind, five of a kind).

However, I understand that this can be a tedious process, especially for larger numbers of dice. In that case, using a Monte Carlo simulation can be a helpful tool. This involves running a large number of simulations (such as 100,000) and recording the results. The average of these results will give us an estimate of the expected value.

I hope this helps in solving the problem and finding a more efficient solution.
 

1. What is the Dice Game Problem (Hard)?

The Dice Game Problem (Hard) is a mathematical problem that involves rolling a set of dice and trying to achieve a specific result. It is a popular problem in the field of probability and has been used as a brain teaser in many settings.

2. How does the Dice Game Problem (Hard) work?

In the Dice Game Problem (Hard), the player rolls a set of dice and tries to achieve a specific result by manipulating the numbers on the dice. The difficulty of the problem lies in the fact that the player is only allowed a limited number of moves to achieve the desired result.

3. What is the strategy for solving the Dice Game Problem (Hard)?

The strategy for solving the Dice Game Problem (Hard) involves understanding the probability of rolling certain numbers on the dice and using that information to make strategic moves. It also involves trial and error and being able to think ahead to anticipate the outcome of each move.

4. Is there a solution to the Dice Game Problem (Hard)?

Yes, there is a solution to the Dice Game Problem (Hard). However, it is a difficult problem and may require multiple attempts to solve it. The solution involves making strategic moves and understanding the probability of rolling certain numbers on the dice.

5. Can the Dice Game Problem (Hard) be applied to real-life situations?

While the Dice Game Problem (Hard) is a mathematical problem, the concept of understanding probabilities and making strategic moves can be applied to many real-life situations. It can be useful in decision-making processes and in calculating risks and potential outcomes.

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