Shaun Higgins
United Kingdom London

I've been reading Knizia's Dice Games Explained and have a couple of questions around a tactic he proposes for the dice game of Par. Below is a short overview of the game and tactic (in Knizia's words) followed by a couple of questions I have. I will be interested to know your response...
The Game
Roll five dice with the aim of getting a minimum of 24 or more. You can reroll the dice as long as you set aside one die each time. Any dice set aside cannot be used for any future throw. Your turn ends when you have set aside five dice.
Scoring
 Below 24: lose points equal to amount below 24  24: nil points gained or lost  above 24: total points above 24 equal your "hit number". Now reroll all five dice and any rolling your hit number are added together to produce your final score
Tactics
The key challenge is to decide upon which dice to keep and which to reroll. Most importantly, the decision does not simply rest upon the number on any given die, but also the numbers upon any other given dice. If some dice need improvement, you may need to reroll some of your dice to increase the overall number of your possible rerolls. If you have three dice left and set aside two of them, you have only one die to roll again thereafter. However, if you instead only set aside one die, you have two dice to roll next and then one again, thereafter. Consequently, it is often advantageous to reroll 4s or even 5s.
Tactical Rules, based on Probability
i) If you have more than two dice available that do not show 6s, only keep the 6s and reroll the rest ii) With only two available dice reroll both if none offers a 5 or 6, otherwise only reroll any 1, 2, or 3 iii) Rethrow a single die if offers a 1, 2, or 3
There are two exceptions to these rules:
iv) Do not reroll at all if your result contains at most one 4 and all dice are 5s and 6s v) If you have three dice left and they show 553, 552 or 551, only reroll the lowest die and keep the 5s
Example: Accept 55554 and don't reroll. In contrast, although 65553 has the same total, the 3 is below average, so reroll all dice including the 5s, but not the 6.
Questions
1) In the above example, I understand why the 3 is below average and therefore should be rerolled, but why would the 5s also be rerolled if they are well above average? Does it also not contradict Tactical Rule (v)?
2) In Tactical Rule (ii), would you not reroll any die offering a 4?


1) The chance to roll only 4s or lower with 4d6 is smaller (19.75%) than the chance to roll a 1 or 2 with 1d6 (33%). It doesn't contradict rule v because you have more than three dice left.
2) With two dice available, the chance to increase the value of a die that is already a five is slim (30.56%), so you keep 5s and 6s. The chance to roll at least one five is much better (55.56%). Only on a single die a four is already above average.

'Bernard Wingrave'
United States Wyoming Ohio

I am not a probability expert, but I'll weigh in on your first question.
"Accept 55554 and don't reroll. In contrast, although 65553 has the same total..." 65553 is presumably the first roll (assuming that 55554 is going to be accepted asis with no reroll, it looks like 65553 is also a first roll).
Tactical rule v is about when you have 3 dice left, which is not the situation here (you could argue that you have 5 dice left or that you could choose to set one aside and have 4 dice left  either way, you aren't in the rule v situation).
Tactical rule i is about when you have 2 or more dice available that do not show 6. That is the case for 65553 (since there are 4 dice available that do not show 6).
You also asked why the 5's in 65553 would be rerolled since they are above average. I do not have a complete answer to this aspect of tactics, but I suspect is has something to do with possibly rolling 6's on those dice and if they don't come up 6's you could still reroll them to a point (see rule i).

Shaun Higgins
United Kingdom London

wrote: Tactical rule 1 is about when you have 2 or more dice available that do not show 6. That is the case for 65553 (since there are 4 dice available that do not show 6).
Thanks both for your replies here. My only thought on the 65553 example is that we cannot know how many of these dice have already been set aside, which changes the number/type of dice available for rerolling.

Graham Muller
South Africa Cape Town

Shaun Higgins wrote: Example: Accept 55554 and don't reroll. In contrast, although 65553 has the same total, the 3 is below average, so reroll all dice including the 5s, but not the 6.
Questions
1) In the above example, I understand why the 3 is below average and therefore should be rerolled, but why would the 5s also be rerolled if they are well above average? Does it also not contradict Tactical Rule (v)?
2) In Tactical Rule (ii), would you not reroll any die offering a 4?
Thought would just play around with it a bit:
For: 1) a: 55554 removing 1 dice gives you 5 total with 4 dice potential b: 55553 removing 1 dice gives you 6 total with 4 dice potential
From rule i: If you have more than two dice available that do not show 6s, only keep the 6s and reroll the rest Chance of rolling atleast 1: 6 (1: 5 or 6) on 3 dice is 42% (70%) Chance of rolling atleast 1: 6 (1: 5 or 6) on 4 dice is 52% (80%)
So the chance to stay the same or improve for atleast one of the 5 dice is 70% on 3 dice and 80% on 4 dice. Assuming a 3.5 average on the last 2 dice it means off 4 dice we have an expected result of about 18 which is exceeded by the 19 of 5,5,5,4
EDIT: From the example you should assume that this is the first roll EDIT2: What I am seeing is that Rule 1 is saying if you have 6's take them, and as you are ahead of the game, attempt to go for more 6's. For the exception note that it stipulates 1 4, the reason for this is more than 1 4 will take you to or below the expected result using rerolls.

Jeremy Lennert
United States California

This may be a helpful way of looking at it:
One possible strategy would be to always lock your highest single die and reroll the rest. If you do this, the dice you get are going to be:
#1: The highest of 5 dice #2: The highest of 4 dice #3: The highest of 3 dice #4: The highest of 2 dice #5: a random die
Let's take that as our "default" strategy, and ask when we should deviate from it.
You've just made your first roll and locked your #1 die. For your #2 die, you now have the following choice:
option A: reroll 4 dice, and lock the highest result option B: lock the highest result that's already showing
When you look at it like that, it's pretty obvious why you wouldn't lock a 4 without rerolling: your alternative isn't "a random die", it's "the highest of 4 random dice", which has an expected value of 5.24.
In fact, these are the expected values:
highest of 5 dice: 5.43 highest of 4 dice: 5.24 highest of 3 dice: 4.96 highest of 2 dice: 4.47 1 die: 3.5 (total expected sum for original strategy: 23.6)
So a simple improvement on our "default" strategy would be to say that, if we have the opportunity to reroll 4 or more dice, we should only lock 6s, and if we have the opportunity to reroll at least 2 dice, we should lock 6s and 5s, but if we see 44 we should reroll both.
This is still not a complete analysis; if we wish to devise a perfect strategy, we need to consider how a reroll will affect all dice, not just the next one we're about to lock. But this is already nearly the same as Knizia's proposed strategy.
(On a side note, the above analysis assumes you are simply going for the highest sum, and ignores the complicated scoring rules from the OP. This is probably fine, because those complicated scoring rules produce very nearly the same average result as "your score is the sum of all your dice".)

Shaun Higgins
United Kingdom London

Jeremy
Thanks very much for this. This is a very straightforward & accessible approach to the problem and also one that parallels Knizia's general analytic approach in Dice Games Properly Explained.
I will dwell upon this more ....


