From the March 11, 2022 “Riddler”:
We’re playing a game where you have to pick four whole numbers. Then I will roll four fair dice. If any two of the dice add up to any one of the numbers you picked, then you win! Otherwise, you lose.
For example, suppose you picked the numbers 2, 3, 4 and 12, and the four dice came up 1, 2, 4 and 5. Then you’d win, because two of the dice (1 and 2) add up to at least one of the numbers you picked (3).
To maximize your chances of winning, which four numbers should you pick? And what are your chances of winning?
Some first thoughts:
- You want numbers that are common as the sums of two dice – middling numbers, numbers near seven.
- The problem has a reflection symmetry. The dice values
win with the target sums
if and only if the dice values
win with the target sums
.
Putting these together, a symmetric set of middling numbers seems likely to be the best target set – something like 5, 6, 8, 9. This is a nasty case analysis, but it’s easy to do by brute force in R.
library(tidyverse)
dice_and_targets = expand.grid(d1 = 1:6, d2 = 1:6, d3 = 1:6, d4 = 1:6,
t1 = 2:12, t2 = 2:12, t3 = 2:12, t4 = 2:12) %>% filter(t1 < t2 & t2 < t3 & t3 < t4) %>%
mutate(s12 = d1 + d2, s13 = d1 + d3, s14 = d1 + d4,
s23 = d2 + d3, s24 = d2 + d4, s34 = d3 + d4)The data frame `dice_and_targets` has a row for every possible combination of dice results (d1 … d4) and targets (t1 … t4), and the sums of the dice (s12 … s34). It’s a big data frame, with 
Let’s take a look at a sample of this data frame, consisting of 10 randomly selected rows:
set.seed(1)
dice_and_targets$idx = sample(nrow(dice_and_targets))
dice_and_targets %>% filter(idx <= 10) %>% select(-idx)
d1 d2 d3 d4 t1 t2 t3 t4 s12 s13 s14 s23 s24 s34
1 1 2 3 5 2 3 5 8 3 4 6 5 7 8
2 4 1 6 5 2 3 4 9 5 10 9 7 6 11
3 2 3 4 5 3 6 7 9 5 6 7 7 8 9
4 2 1 5 6 3 5 8 9 3 7 8 6 7 11
5 1 6 4 6 2 6 9 10 7 5 7 10 12 10
6 3 4 4 6 3 6 7 11 7 7 9 8 10 10
7 2 2 4 2 4 8 10 11 4 6 4 6 4 6
8 1 2 3 5 2 3 4 12 3 4 6 5 7 8
9 2 6 6 1 2 6 8 12 8 8 3 12 7 7
10 1 6 3 6 6 8 11 12 7 4 7 9 12 9Consider, for example, the first row. In this case we roll 1, 2, 3, and 5; the targets are 2, 3, 5, and 8; the pairwise sums are 1+2 = 3, 1+3 = 4, 1+5 = 6, 2+3 = 5, 2+5 = 7, and 2+6 = 8; and we win the game, in fact three times over, since the pairwise sums include three of the targets, namely 3, 5, and 8.
Next we can work out which rows win, leveraging some bitwise operations because how often do I get a chance to use these?
dice_and_targets = dice_and_targets %>% mutate(target_bits = bitwOr(bitwOr(2^t1, 2^t2), bitwOr(2^t3, 2^t4)))
dice_and_targets = dice_and_targets %>%
mutate(sum_bits = bitwOr(bitwOr(bitwOr(2^s12, 2^s13), bitwOr(2^s14, 2^s23)), bitwOr(2^s24, 2^s34)))
dice_and_targets = dice_and_targets %>%
mutate(win = bitwAnd(target_bits, sum_bits) > 0)In this case target_bits has the bit corresponding to 
sum_bits has the bit corresponding to 
bitwAnd(target_bits, sum_bits) has a nonzero bit if and only if we have a winning combination.
Let’s look at those randomly selected rows, now with the wins figured out:
dice_and_targets %>% filter(idx <= 10) %>% select(-idx)
d1 d2 d3 d4 t1 t2 t3 t4 s12 s13 s14 s23 s24 s34 target_bits sum_bits win
1 1 2 3 5 2 3 5 8 3 4 6 5 7 8 300 504 TRUE
2 4 1 6 5 2 3 4 9 5 10 9 7 6 11 540 3808 TRUE
3 2 3 4 5 3 6 7 9 5 6 7 7 8 9 712 992 TRUE
4 2 1 5 6 3 5 8 9 3 7 8 6 7 11 808 2504 TRUE
5 1 6 4 6 2 6 9 10 7 5 7 10 12 10 1604 5280 TRUE
6 3 4 4 6 3 6 7 11 7 7 9 8 10 10 2248 1920 TRUE
7 2 2 4 2 4 8 10 11 4 6 4 6 4 6 3344 80 TRUE
8 1 2 3 5 2 3 4 12 3 4 6 5 7 8 4124 504 TRUE
9 2 6 6 1 2 6 8 12 8 8 3 12 7 7 4420 4488 TRUE
10 1 6 3 6 6 8 11 12 7 4 7 9 12 9 6464 4752 TRUE
In the first row, here, target_bits is 
sum_bits is 
bitwAnd(target_bits, sum_bits) (not shown) is 
You might get the idea that it’s impossible to lose from this sample. We got a little lucky here: mean(dice_and_targets$win) returns 0.874018. If you pick a random roll of four dice and four random targets out of 2, 3, …, 12, one of the pairwise sums will be in the target set 87% of the time.
But we want to know which target set makes a win most likely.
win_counts_by_target = dice_and_targets %>% group_by(t1, t2, t3, t4) %>% summarize(wins = sum(win)) %>% arrange(desc(wins))
head(win_counts_by_target)
> head(win_counts_by_target)
# A tibble: 6 x 5
# Groups: t1, t2, t3 [5]
t1 t2 t3 t4 wins
<int> <int> <int> <int> <int>
1 4 6 8 10 1264
2 2 6 8 10 1246
3 4 6 8 12 1246
4 4 6 7 9 1238
5 5 7 8 10 1238
6 4 7 8 9 1236
There we go! And not a loop in sight.
Once I have the 4, 6, 8, 10 target set it’s easy to come up with that number 1264. Consider just the dice that show even numbers – you win if there are at least two of these and they’re not all showing 6. Similarly you win if there are at least two odd dice and they’re not all showing 1. So the losing combinations are
and their rearrangements, of which there are 32.
Incidentally, my initial guess (5, 6, 8, 9) wasn’t bad – it wins 1228 times out of 1296, good enough for 14th place out of the 330 possible target sets. And the very worst target set? It’s (2, 3, 11, 12). No surprise there, although even that one wins 776 times out of 1296, nearly 60% of the time.
. So we need to find positive integers x, y, z such that x + y + z = 100 and 
. The three big factors – the two 7s and the 11 – look like they’ll be worrisome. So let’s spread them out: let x and y both be multiples of 7, and let z be a multiple of 11. To get them to add up to 100, we need to find a multiple of 7 and a multiple of 11 that add up to 100; it’s easy to find 44 and 56. So z = 44. 
, so x and y must be 21 and 35. 
will have many factors, and many ways to be written as a product of three integers, making it more likely that one of those will have the right sum.
. 
, and the probabilities of getting n+2, n+3, …, 2n are 
– generalizing the well-known triangular distribution. The probability that you and your friend both roll 2, then, is 
, so we should get a result of order 1/n. We can peservere and do the algebra and we get
. He writes that “the percentage chance of a match score falls off a lot slower than I would have predicted.”
are possible with any appreciable frequency. There are on the order of a few square roots of k of these, so the answer should be 
for some 
. Like Berry, I am too lazy to find the constant 
.
, by approximating both sums as normal and applying the central limit theorem. This is about 
. If we bear in mind that the variance of a single roll of an n-sided die is 
, then we get 
for rolling k n-sided dice.
. This reduces to 1/n, the correct answer, if you let 
and ignore the -1.
, so clearly 
. (All logarithms in this post are to base 10.) Can we approximate the logs of any other primes this way?
. 
. And of course we have 
, so this relation becomes
.
and $\log 4374 \approx \log 4375$, namely
.
.
(a fundamental fact of musical temperament). 
right away, 
, and then 
so 
. This is essentially I. J. Good’s 
, the fact on which our usual musical tuning system is based: 
, and a bit less accurately 
. (Musically these are the facts that the perfect fifth, major third, and minor seventh are the ratios 3/2, 5/4, and 7/4.). 
, which gives 
. This is a less good approximation than those of 
and 
, just as the seventh harmonic isn’t really a note in the equal-tempered scale – 
. Euler proved 
, which we’ll approximate as 
. Thus 
, or rearranging 
; the number of edges (state borders) is about three times the number of states. 
, 
, and 
respectively. Since the second-column product isn’t divisible by 3, the second column must consist of only 1, 2, 4, 5, 7, and 8. The third column isn’t divisible by 4 or 5 so it can’t contain 4, 5, or 8; furthermore it contains only a single even number (2 or 6).
so they must be two 9s and two 8s. That lets us complete the first column, because there is only two possible locations for a 9, two for an 8, and one for a 5. And knowing those first-column values allows us to complete various rows:
God plays dice 
