Goal-Driven Dice-Rolling Difficulty Question

Suppose I have ten dice. Suppose I am required to set aside one or more of them into a set of “locked dice” and reroll the rest. Suppose that out of the rerolls, I am again required to set aside one or more into the set of “locked dice,” and reroll the remainder (if there is a remainder). Suppose I am required to do this until all the dice have been locked.

Now, suppose my goal is to get as many dice in the final locked set to have the same number on them as possible. (For example, I might be happy to get six threes, but then I’d be even happier on another try if I got seven ones.)

My question may not be answerable, and it’s a little vague. But basically, I’m trying to figure out* “how hard” this task is for various numbers of same-kind dice in the final locked set. For example, suppose someone came up with seven of a kind. Should I think they got lucky? Or should I think that’s an easy result?

*Ha, that’s a lie. I’m asking others to figure it out for me. And thanks in advance!

An answer is highly dependant on the rules.
It may be impossible, through to guaranteed.

The rules would be simple … well commonly they are … But determining the result can be problematic…

Assume a simple strategy: Find the most populous face-number after the first roll and lock only that number thereafter. In case of tie, pick one of the tied values at random. If your number doesn’t appear on a given roll, lock one arbitrary die.

In the (unlikely? :stuck_out_tongue: ) event my calculations are correct, your result chances are then:
[ul]
[li] 3 alike - 0.3%[/li][li] 4 alike - 2.8%[/li][li] 5 alike - 10.1%[/li][li] 6 alike - 22.1%[/li][li] 7 alike - 29.7%[/li][li] 8 alike - 23.5%[/li][li] 9 alike - 9.8%[/li][li] 10 alike - 1.6%[/li][/ul]

(The strategy described is not optimal. For example, if you roll two Deuces, then eight Sixes, the stated strategy would have you throw seven of the Sixes away! :smack: But I think an optimal strategy will do only slightly better than the stated strategy, on average.)

(The rules you describe are very very vaguely reminiscent of a dice game called 5000. Googling now, I see it was marketed as Farkle, starting a decade after I played the game.)

The list is not complete. There is a greater than 0 chance that you end up with 2 alike.

I’m too lazy to try to do the combinatorics all the way, but just doing it for the first roll where you are not allowed to have more than pairs (5 pairs, or 4 pair and 2 singles) you end up with the chance being 21/3003 = 0.006993. You still have at least 4 more roles (or 8 more if we stick to your strategy), so the end percentage gets pretty low. I would guess somewhere below 1*10^(-9).

So I suppose with such a negligible chance we will never see it actually happen, but it is possible.

Yes, to the precision I was showing it was zero, so I omitted it. Here’s the list with a bit more precision:

[ul]
[li] 2 alike - 0.0016%[/li][li] 3 alike - 0.33%[/li][li] 4 alike - 2.81%[/li][li] 5 alike - 10.11%[/li][li] 6 alike - 22.07%[/li][li] 7 alike - 29.69%[/li][li] 8 alike - 23.57%[/li][li] 9 alike - 9.83%[/li][li] 10 alike - 1.60%[/li][/ul]

I tried to give a complete set of rules in the OP. What did you think was missing?

Okay, thanks, so it looks like seven is the most common result! (Using that simple strategy, which is probably what gets used usually anyway. BTW I should have specified the game I’m talking about is To Court a King, but I am only talking about a very specific subset of the rules.)

And seven-or-more in a row will occur more than half the time apparently.

The game offers you opportunities both to gain extra dice and to gain powers to manipulate the dice in various ways, but it really looks like if you get behind by more than one die you’re practically doomed, no matter what dice-manipulation powers you may have gained. (I had a game where I had seven dice and powers sufficient to do practically anything with them, my opponent had ten dice and no powers at all, and in the end he got seven sixes while I could only manage to build six of them so I lost, and I was surprised by this. I know you guys don’t know all the rules here but I think what I’ve gestured at here gives you the flavor of the situation.)

Can you keep the ones and reroll the threes?

also, how many rerolls do you get?

Yes.

No limit is specified, but the number of dice (ten) imposes a necessary limit.

I misunderstood what was ‘vague’. I was thinking it referred to the ‘suppose’
clause.

here’s the rules.

  1. take a set of 10 dice, to form the “in play set”.

  2. roll the "in play "set.

  3. Lock at least one away into the result set. (at the first roll , there’d have to be two to lock away. )

  4. If the in play set is not empty, goto 2.

  5. the goal is to get n of a kind - the larger n is better.
    The % chances given by others look reasonable to me, its going to be that skewed bell curve.

So I can just reroll until I get 10 of a kind, Double Yahtzee?

Pretty much exactly, though there’s not any reason your first lock-in would have to be at least two dice.

Indeed, I admit I was surprised at the high chance of getting seven of a kind.

No, you lock in at least one die each round.

Cool! The chance is significantly higher than I would have thought. One or two out of every 100,000…still better than the chance that the Cubbies will take the Series (next year). :slight_smile:

For the record, I’d better clarify. Always being a shirker happy to let the “other guy” do the heavy lifting, I let my laptop do 50 million simulated plays and just reported its totals. Although the strategy was as described, at the end of each complete play, the software did, correctly, check all face-numbers for highest total, not just the target set after the first roll.

If instead we insist that only that selected target can be used for the final total, the chance of getting only 2 alike increases to 0.009525%, or, if greater accuracy is needed, precisely (7⋅5[sup]38[/sup]) ÷ (2⋅6[sup]40[/sup]).

On the first roll, the probable maximum number of any particular digit is just under 4. (about 3.8. I determined that by a random number generator, 30 trials of ten dice). Those, of course will be the 4 that you save (for simplicity, let’s assume whole numbers) . So assume that on your first roll, you will lock in four dice that matched. From that point on, the remaining rolls of 6 5 4 3 2 1 (21 dice) will yield 3.5 that match your original ~3.8, so the median expected number is ~7.3.

Note that, if you get a good pair on the first remaining roll, your rolls would be 6 4 3 2 1 (or fewer) instead of 6 5 4 3 2 1. The yield averages 3.2456, not 3.5. (Sorry if the smallish difference seems like nitpicking.) It may seem slightly paradoxical that the average yield goes down due to cases where the early yield is high.

An easy way to see that the 21 → 3.5 argument cannot be correct, is that the statistic on which it is based would allow 21 matches! (6 matches is obviously the maximum.)

Subquestion:

On you first roll, your best number is a pair (we’ll use sixes as an example) so you lock the two sixes and roll the other eight dice. You don’t get any sixes but you roll three fives. Should you lock the three fives and switch to fives (being stuck with the already locked pair of sixes) or should you lock a useless number and keep trying to build upon the sixes you’ve already started with?

Suppose you got four fives on your second roll instead of three?

I considered this when I ran my simulation and even planned, masochistically, on deriving a complete strategy table! :smiley: I abandoned that plan when I saw how little effect such deviations would have on the final result.

But to answer your question in the (still unlikely :rolleyes: ) case where my software is bug-free:

The default strategy I described yields 6.9600 matches on average.
Switching to triplet (or better) after roll 2 yields 6.9538 on average.
Switching to quadruplet (or better) after roll 2 yields 6.9591 on average.
Switching to quintuplet (or better) after roll 2 yields 6.9602 on average.
IOW, and perhaps surprisingly, you need quintuplets before locking them is best!

(The effects as presented are small, because the averages are over all runs, not just the rarish cases where you actually get the triplet or better.)

21 is the maximum number of rolls, if there are four matches in the first roll, because you have to lock at least one after each roll. So if there are less than 6 additional rolls, it is because a) there were two matches in one of them, or b) there were a lucky five matches in the opening roll, which leaves you with at least six, no matter now many rolls remain. So fewer matches need to be made in later rolls, offsetting the disadvantage if having fewer later rolls… Once you have that six, you only need one more match in all your remaining rolls to make the seven. That can be done in 3 2 1.