Solving the Game of 5000

The Game of 5000 is a game of dice, with an element of choice. You start by rolling six dice, and scoring as follows, taking the largest score possible:

Six-of-a-kind gives you 5000 points.

A straight (1,2,3,4,5,6) is 2000 points.

Two triples or three pairs are 1500 points.

Three ones are 1000 points.

Triple 2s are 200, triple 3s are 300, etc.

Individual 1s are 100 points.

Individual 5s are 50 points.

You then remove the scoring dice, and consider the remaining dice. If you have unused dice, you must roll again. Otherwise, you have a choice: keep your current score, or roll again with the remaining dice. If those dice score, you add them to your total, and repeat; but it you ever roll a zero, you lose everything, and end your turn. The game ends when someone has reached a score of 5000, and the round has finished so all players have an equal number of turns.

The highest-scoring patterns are only available if you are rolling six dice, so lower-dice-count rolls become riskier. With 3, 4, or 5 dice, you can get single triples and collections of individual ones and fives. With one or two dice, you can only get individual ones and fives, and the probability of rolling a zero and losing everything is high.

So, when should you roll again, and when should you stay? To figure this out, we need to know the expected value of any given roll. That is, if you roll a certain number of dice, what is the average score you would get over a large number of rolls? That turns out to depend on what you have to lose, and to incorporate that we will need to know all the ways that you could score, and how many ways you could roll zero and lose everything.

For 1 die, you have 1/6th probability of rolling 100, 1/6th probability of rolling 50, and 2/3rds probability of rolling zero. The sum of all possible scores is 150. Multiplying scores by probabilities and adding them together, you get an expected value of 25, with a 2/3rds chance of scoring nothing and 1/3rd chance of scoring something. Note that the expected value, being an average, is not actually a possible score--the fewest points you can score at once is 50! But that fractional value represents the probability that you could get nothing.

With 2 dice, things get a little more complicated. If we labelled the dice, there would be 36 different possible rolls; but in fact, the order or identity of the dice doesn't matter, so rolling 1,2 is the same, as far as scoring goes, as rolling 2,1. To make scoring easier, we can sort the dice results to put every possible roll in a standard form, and then account for the multiple ways to get each of those collections of numbers. There is one out of 36 ways to get two 1s (200 points) or two 5s (100 points). There are 8 ways to get exactly one 1 (100 points) or exactly one 5 (50 points). There are two ways to get one 1 and one 5 (150 points). And that leave 16/36 ways to roll a zero. That works out to an expectation value of 50 points, and 4/9ths chance of rolling a zero, and 1800 as the sum of all possible scores.

To figure out the numbers for more dice, we can get a computer to enumerate all of the possible combinations of die values and automatically score them.

At three dice, we can start getting higher scores. There 6^3, or 216, labeled rolls, which collapses down to 56 distinct collections of die values. 60 of those rolls produce no score, giving a 5/18th (or approximately 27.8%) chance of rolling a zero, and expectation value of ~86.81, and 18750 as the sum of all possible scores.

At 4 dice, there are 1296 labelled rolls but only 126 distinct rolls. 204/1296 rolls are zeros (17/108ths or ~15.7%), the expectation value is ~141.3, and the sum of scores is 183150.

At 5 dice, there are 7776 labelled rolls, 252 distinct rolls, and 600/7776 zeros (25/324ths or ~7.7%). The expectation value is ~215.5, and the sum of scores is 1675800.

With a full 6 dice, there are 46,656 labelled rolls, 462 distinct rolls, and 1080/46,656 (5/216ths or ~2.3%) zeros. The expectation value is 402.2, and the sum of score is 18763500.

To be really accurate, we should account for the fact that scoring on six dice requires re-rolling, and the results of the next roll should be accounted for in the final value of that choice... and recursively, as you might score on all sixes again, with increasingly small probabilities but potentially forever. However, for practical purposes, that turns out not to matter!

Now, these initial values are only the expectation for a single roll with an initial score of zero. If you have already accumulated some previous score (which must be the case for rolling any number of dice other than six), then all of the zero-scoring rolls actually cost you negative points. Which means the expectation value of the next roll can become negative, depending on a combination of the probability of rolling a zero and the value of your existing score.

We can figure out the expectation value of a roll-in-context by replacing the zero scores with the negative of the current score, and recalculating the average. So, if 'a' is the current accumulated score, 'p' is the total number of labeled rolls, 'z' is the number of zeros, and 's' is the sum of positive scores, then the contextual expectation value can be calculated as

e = (s-az)/p

If the expectation value is positive, that means it's a good idea to roll. In such a situation, even if you have a less than 50% chance of scoring, the amount that you could score is high enough that it is worth the risk. But, we can make the calculation simpler by doing a little algebra to see what the accumulated score must be to make the expectation value zero:

0 = (s-az)/p

0 = s/p-az/p

az/p = s/p

az = s

a = s/z 

We can work that out once for each number of dice, and get the following table of cutoff scores:

1 - 37.5

2 - 112.5

3 - 312.5

4 - ~897.79

5 - 2793.0

6 - ~17373.6

And now you can see why the rule about always re-rolling when you have used all six dice doesn't matter--since the game ends at 5000, you will never accumulate a score higher than 17,373, so the expectation value of rolling six new dice is always positive, and you should always do it!

At the other end, it is impossible to get down to 2 dice left without accumulating a score of at least 200; to get in that situation, you must have scored 50, losing one die each time, four times in a row, or scored 2 50s, losing 2 dice each time, twice. Thus, in any game scenario where you could roll 1 or 2 dice, the expectation value is always negative, and we can conclude that you should never roll fewer than 3 dice unless someone else has already reached 5000 and you are on your last turn, in which case you just keep rolling as long as you can until you either get a higher score or don't.

Since scores only come in multiples of 50, we can thus simplify the preceding table as follows:

3 - 300 (~27.8% chance of loss)

4 - 850 (~15.7% chance of loss)

5 - 2750 (~7.7% chance of loss)

If you have 1 or 2 dice, always stay. For 2 dice, the chance of loss is less than 50%, but the potential gains are so small as to be not worth it. If you have 6 dice, always roll (you have to anyway). If you have 3, 4, or 5 dice, stay if your accumulated score is above the cutoff in the table; otherwise, roll. It turns out, executing the optimal strategy just requires memorizing 3 numbers!