As a couple of solvers noted, this is an easy problem that’s really just

about

counting. In fact, you can solve the entire problem without ever

rolling a die.

The trick is really in how you think about the problem. Basically we

want to

enumerate all of the possible rolls from:

[1, 1, 1, 1, 1, 1]

to:

[6, 6, 6, 6, 6, 6]

Let me show you that one more time though, minus some Ruby syntax

characters:

111111 to 666666

Now it starts to look like counting, right? It almost is. If we

counted

straight between those two numbers though, we would pick up some digits

that

aren’t on our dice. But if we drop the digits on both sides by one:

000000 to 555555

Now that is counting, in base six, and it’s just a quick transliteration

to

adjust the digits up. We’re going to look at a solution that does just

that.

While I put the code together that I’m going to show you, it’s heavily

inspired

from many of the submissions I liked. Let’s just say that this week’s

showcased

solution belongs to everyone.

Here’s the argument processing code:

#!/usr/bin/env ruby -wKU

MODE = if ARGV.first =~ /-([sv])/

ARGV.shift

$1

end

unless ARGV.size == 2

abort “Usage: #{File.basename($PROGRAM_NAME)} [-s|-v] NUM_DICE

MIN_FIVES”

end

NUM_DICE, MIN_FIVES = ARGV.map { |n| Integer(n) }

# …

This code should be pretty easy to break down. We check to see if the

first

argument looks like a switch and store it in the MODE constant when it

does.

Note that the code doesn’t provide an else branch for this check, but

Ruby will

default the constant to nil for us when the if condition fails to match.

The next bit of code just ensures that we have the two required

arguments,

converts them or throws an error if they are not what we expected, and

stores

them in constants for future use.

We’re now ready for the counting code:

# …

desirable_outcomes = 0

POSSIBLE_OUTCOMES = 6 ** NUM_DICE

POSSIBLE_OUTCOMES.times do |i|

outcome = i.to_s(6).rjust(NUM_DICE, “0”)

if desirable = outcome.count(“4”) >= MIN_FIVES

desirable_outcomes += 1

end

if MODE == “v” or (MODE == “s” and (i % 50_000).zero?)

puts “%#{NUM_DICE}d [%s]#{’ <==’ if desirable}” %

[i + 1, outcome.tr(“0-5”, “1-6”).gsub(/(\d)(\d)/, ‘\1,\2’)]

end

end

# …

This is the heart of the code and, as you can see, still quite simple.

We begin

by defining the range we will iterate over, from zero to the maximum

count on

the requested number of dice.

The times() iterator does the actual count. Remember that we really

want these

numbers in base six. The first line of the iterator does the

conversion,

padding with zeros to ensure we have the right number of “dice.”

The first if branch is our test for the minimum number of fives. When

found, we

bump a global counter that can later be used to generate results.

The final if statement handles printing for either of the inspection

modes. If

we’re in verbose mode, or sample mode and on a 50,000th iteration, the

code

prints the result index, the dice values, and an indicator arrow for

rolls that

were flagged as desirable. This is where you see the code to bump all

of the

digits so they will match pips on a die.

The final bit of code, just prints the results:

# …

puts if MODE

puts “Number of desirable outcomes is #{desirable_outcomes}”

puts “Number of possible outcomes is #{POSSIBLE_OUTCOMES}”

puts

puts “Probability is #{desirable_outcomes / POSSIBLE_OUTCOMES.to_f}”

This is as easy as it gets. We show the desirable, the possible, and

calculate

the odds from the two of them. The first line just sneaks in a line of

padding

between any sampling and the end results.

Again, this code combined things I liked from many of the solutions, so

they are

definitely worth a look. Especially don’t miss Alex LeDonne’s second

solution

which uses some math to skip counting at all, when it can get away with

it.

As always, my thanks to all who took the time to solve this quiz. I

feel like

we always get creative results for the simple problems.

Tomorrow we will tackle one of the all time famous tough problems, and

see if we

just can’t solve it reasonably well with minimal knowledge of the

domain…