Now if you were to do this MythBuster's style:

1. Calculate the average force by applied by yourself and then your child
2. Set up a robot that could apply that force in alternating spins
3. Set up a web cam to snap pictures at regular intervals
4. Write software to intepret the picture to both determine which number is landed on and when to trigger the robot (dial is not spinning)
5. Let it run over a prolonged period of time
6. Provide us all with the results and tell us if we can expect the same with our games at home

This is way cool by the way!

You are not nearly geeky enough.

My wife would disagree. But, my son is young enough that he doesn't realize his father is one yet. I wonder how many years I have left before it dawns on him.

This all assumes that you have a "fair" dial

Don't get me started on the dial! The width of the line separating the numbers is comparable to the width of the arrow-head. This ambiguity forces far too many re-spins (that's harder to simulate with Perl :). Additionally, this ambiguity is a function of the person spinning the dial. My son claims he must re-spin when it is close and the number does not suit his needs!

My son claims he must re-spin when it is close and the number does not suit his needs!

Which you clearly agree with... meaning you are first Dad and only second geek. You may be in luck: your son may never discover you are a geek. You'll merely be his hero (at least until some long off Saturday night when you absolutely positively refuse to hand over the car keys - but that will be only temporary).

Pretty cool to have a Dad serve up an entire post of obscure stats in honor of his son - I imagine in the next installment, you'll start hand-crafting computer games for him :-) There are many ways to love a child - all of them count.

Best, beth

The width of the line separating the numbers is comparable to the width of the arrow-head. This ambiguity forces far too many re-spins (that's harder to simulate with Perl :).

Oh come on! (Width of line * number of lines) / ( 2 * pi * radius of arrow ) gives you a probability that a spin will land on a line. What's so hard about that? (I believe the width of the arrow point is irrelevant, or at least ignorable.)

Additionally, this ambiguity is a function of the person spinning the dial. My son claims he must re-spin when it is close and the number does not suit his needs!

So a per-player weighting factor needs to be applied to the above-mentioned probability. I will grant that assigning a "direction" to the weighting factor, based on the relative merits of two adjacent values on the dial, makes for a more challenging decision process.

I know... there's a certain degree of general fatigue induced by parenting young children, and this tends to limit the amount of time and effort that can be expended on purely geek pursuits. Been there, done that.

Maybe that's why we use a die. The chances of not hitting one of the 6 faces is too remote to worry about. You would have to account for die being thrown across room and under the sofa.

you forgot the part where they rig the game with explosives just to blow something up :-)

duff

The maximum number of moves is infinite, provided sufficiently poor luck.

Proof: Starting on square 6, it is possible to spin an infinite sequence of fives, causing you to slide down the 16 -> 6 chute an infinite number of times and resulting in an infinite number of moves.

There are also several other infinite loops possible, due to the various chutes and the multiple sequences of spins that can take you from the bottom of any given chute to its top, but the 6-11-16 loop is the simplest to lay out.

(Disclaimer: I do not have a copy of the board readily available, so I have inferred the board topology from the posted source code. If there is no 16 -> 6 chute, or if there is a chute/ladder on 11, the specific example used will not work, but the general principle remains.)

Thanks for the clear explanation. I updated the OP with a link to your reply.

I do not have a copy of the board readily available

The wiki link in the OP includes a photograph of the board.

Update: The wiki photo differs slightly from my game board at home (and the code posted in the OP): the wiki board has a chute that starts on square 47 and ends on 26; the chute in the code is 48-26.

If a player were to keep rolling the same number, where would they wind up?

As the board is a finite size, there's only two possible kinds of answer, they'd wind up going round in circles, or they'd get to the end and (as this is a game for one) win.

For example, if a player keeps rolling 1, then they'd get into a loop involving ladder 28=>84 and snake/chute 87=>24. (Caution: calculated using head, not Perl.)

So what are the results for rolling only 2s .. 6s? Is it possible to win the game? Which number results in the most convoluted loop? Which number carries on longest before it gets into a loop.

Grr - damn this paid employment lark! I have to stop before I can actually try any of this stuff out!

Thanks for the ideas.

You reminded me that I wanted to do some more analysis. While you are interested in infinity, I am still obsessed with ending the game in the minimum number of spins (7).

In another sample of 1 million games, 1509 ended in 7 spins. Of those games, there were 421 unique sequences of 7 spins. Some sequences only occurred once (38 out of 421). Other sequences occurred multiple times, up to a maximum of 10 times.

These 2 sequences occurred 10 times each:
    4->3->6->5->6->5->5
    1->4->4->5->5->4->4
[download]

It is not possible to climb ladder1 or ladder80 more than once in a game.

What fascinating statistics can you dial up?

But the most important statistics is missing. The entire game is almost without decisions (except that you are allowed to skip a roll when rolling 6, and aren't forced to climb a ladder). Which leaves one big decision: do you go first, or second? Obviously, going first increases the chance of winning, but by how much? Said it differently, if you play chutes and ladders (2 player game), the loser paying the winner 1 USD, how much would you pay to go first?

The wikipedia article indeed does mention those facts. But, it also mentions that there are other variants of the game. The version of the game I have at home has no such restrictions. This Perl code is based on the game manufactured by Hasbro (formerly Milton Bradley) in the United States in 1999. Your post has prompted me to do a little more web searching to find a site more specific to this version, including the rules. I apologize for the confusion.

That being said, the variant you pointed out is slightly more complex, and hence, would make for more interesting analysis.

In case of the simple rules, you can't get to the square 100 before the 7^th turn:

#!/usr/bin/perl
use warnings;
use strict;
use feature qw{ say };

my %special = ( ... ); # copy from the OP

my %reachable = ( 0 => undef );

my $turn = 0;
while (not exists $reachable{100}) {
    my %next;
    for my $square (keys %reachable) {
        for my $spin (1 .. 6) {
            my $target = $square + $spin;
            next if $target > 100;
            undef $next{$target};
            undef $next{ $special{$target}{end} } if $special{$target}
+;
        }
    }
    undef @reachable{ keys %next };
    say ++$turn, ': ', join ' ', sort { $a <=> $b } keys %reachable;
}
say "$turn";
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لսႽ† ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ

What is the probability of running a game that uses no ladders or no chutes?

Winning games which use 0 ladders and 0 chutes: 1178 out of 1 million, or about 0.12%, which is rarer than winning the game in the minimum number of moves.

Nothing. It is an unused variable. I was probably using it at one time, but forgot to remove it. As has been discussed many times here at the Monastery, there is no simple way to identify unused variables which have been declared with my.

It is harmless, but I will update the code in the OP to remove the variable. Thanks for the good catch.