February 2018

[LaurV]

Originally Posted by VBCurtis

I offer you a game: You play with one 8-sided die and 8 6-sided dice, and I play with one 8-sided die and one 6-sided die. Every time your game takes longer, you pay me $1 per extra throw, while every time my game takes longer I pay you $1.50 per extra throw.
If your logic is right, I'll get crushed. Why wouldn't I?

If you pay me 1.96 every time I win (and I still pay you 1.0 every time I lose), I am game for that... hehe... According with my "simulator", I will have a slightly advantage. If we play a million games, I will make just a few bucks for these premium ratios. This is somehow counter-intuitive, thinking that most of your games will end in 24 throws or so, and I will need just about 26 throws to end the same amount of my games (of course, few games will take longer, on both sides, but anything between 8 and n throws can end any of the games, grrrr..., which looks like an almost fifty-fifty chance... I think that without simulation, I would have taken your bet with $1.5 and lose my shirt...)

With 1.95 and 1.96 to 1, in fact, the game will be "almost" pure luck ("almost" means that the "pure luck" is in my advantage for 1.96, hehe, as I consider myself a lucky guy) - with 1.97 I can make money , and with 1.94 I will lose money.

Another version: I accept your 1.5 to 1, but you have to pay me x every time we draw. How much should x be? I would be quite happy with anything more than 4.15...

As I said, probability theory is a bitch...

[axn]

A question about interpretation. Will the first move be done at t=0 or t=60? With the latter interpretation, I get 4 solutions (i.e. round( expected # of steps * 60) = 2569). With the former, the closest I get is 2571, again 4 of them (i.e. round ( (expected-1)*60 ) = 2571).

This is assuming my fugly, recursive, memoized monstrosity of PARI script was correct.

[tricky]

Originally Posted by axn

A question about interpretation. Will the first move be done at t=0 or t=60? With the latter interpretation, I get 4 solutions (i.e. round( expected # of steps * 60) = 2569). With the former, the closest I get is 2571, again 4 of them (i.e. round ( (expected-1)*60 ) = 2571).

This is assuming my fugly, recursive, memoized monstrosity of PARI script was correct.

This is my first post here, so I hope I don't spoil the puzzle more than is customary.

Assuming your computations are correct, the interpretation has to be t_0 = 60, right? Now, knowing that a star is up for grabs, give it some extra thought.

[axn]

Solution

[LaurV]

Haha, no way to take that star, in a million years... (speaking in my name only)

[Dieter]

My way was a little different, because I didn’t knowthe “coupon collector’s problem”.
I have found the following formula for the probabilityof „a dice with n faces had all outcomes after i throws” in an old schoolbook:
p(n,i) = sum (k=0 to n-1) {(-1)^k*(n choose k)*(n-k)^i}/n^i
This formula holds for i = 0,1,…,n-1, too. Then p=0.
The probability of „three dices with n_1, n_2, n_3faces had all outcomes after i throws” is
P(i) := p(n_1,i)* p(n_2,i)* p(n_3,i)
For the computation of the expected time we need theprobability Q of „three dices with n_1, n_2, n_3 faces had all outcomes at throwi (and not before!)”:
Q(i) = P(i) – P(i-1)
The expected time is
<t> = 60*Q(1) + 120*Q(2) + 180*Q(3) + … (withthe parameters n_1, n_2, n_3)
I have taken into account 200 throws for seeing a sortof convergence…