February 2018

[axn]

Originally Posted by VBCurtis

Consider a 6-sided die for you, and an 8-sided die for me. There are quite a few plays of the game where I'll finish in fewer tries than you will, as sometimes I'll get lucky with 10 or 12 throws and you'll still be waiting for your last number. The closer the smaller dice are to the big die, the more often this happens, making for quite a gross conditional-probability calculation.

I'm in LaurV's boat here...

If the expectation of dice with n1 sides is e1, and expectation of dice with n2 sides is e2, Why is the combined expectation not simply max(e1, e2) ?

I am not saying it is so. I am asking for an explanation, why it isn't so. And giving examples of individual cases doesn't help. We know that event though the expectation of n1 is e1, there are many trials where the actual value will be > e1. Yet that doesn't change the expectation. So simple handwaving wouldn't cut it.

[uau]

Originally Posted by axn

If the expectation of dice with n1 sides is e1, and expectation of dice with n2 sides is e2, Why is the combined expectation not simply max(e1, e2) ?

Consider two dice with six sides. They won't always finish on the same turn right? The second die will sometimes last longer. Those times, the game will last longer than it would have with the first die alone. Since you expect that the game will sometimes last longer with two dice than just the first die, and never shorter, the combined expectation for two equal dice must be longer than the expectation for just the first die alone.

[axn]

Originally Posted by uau

Consider two dice with six sides. They won't always finish on the same turn right? The second die will sometimes last longer. Those times, the game will last longer than it would have with the first die alone. Since you expect that the game will sometimes last longer with two dice than just the first die, and never shorter, the combined expectation for two equal dice must be longer than the expectation for just the first die alone.

Thank you. Simple explanation that even I can grok. I guess I owe an apology to VBCurtis - I got lost in the words, I failed to understand the concept.

[a1call]

Originally Posted by uau

Consider two dice with six sides. They won't always finish on the same turn right? The second die will sometimes last longer. Those times, the game will last longer than it would have with the first die alone. Since you expect that the game will sometimes last longer with two dice than just the first die, and never shorter, the combined expectation for two equal dice must be longer than the expectation for just the first die alone.

I don't think that makes any sense.
Mathematically speaking:
* The expected time it takes to roll all 6 faces at least once is an absolute numerical constant value.
* This is independent of any other die toss.
* Tossing 100 different dice will not change the expected game completion time of any single one of them.
* Plus, any die with less faces than the one with the largest number of faces will have a game completion time which is less than it and will not contribute to the total expected simultaneous game duration.

[a1call]

is the mean or expectation of the distribution (and also its median and mode),

https://en.m.wikipedia.org/wiki/Normal_distribution

Without given values for Standard-Deviation or Variance, the expected duration can only be a constant number and not a range. Without these values the range would be infinitely wide and range overlaps do not make sense.

[VBCurtis]

A1call-
We are provided a specific number as the expected duration; you cite a wiki definition to inform us that the expected duration is a single number.

Huh?

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?

[a1call]

The bell curve of probability is clipped on the lower side by the minimum number of trials which is the number of the faces and unbounded to infinity on the other side. Without somehow clipping the upper bound the range overlaps of each player will be infinitely large. Bell curves of each player will have a mean expectancy value which is different than the other and independent from it. A simultaneous game expectancy duration based on these 3 independent mean values, can have a statistical value based on infinite number of trials of the simultaneous game. But that is certainly beyond my statistical math skills.

[Dieter]

Originally Posted by uau

Anyone have an idea what the '*' marks in answer list mean? There doesn't seem to be any visible bonus objective...

I have a solution n_1 < n_2 < n_3. Maybe there is at least one better solution (the expected time is closer to 2569) with a bigger n_3 and smaller n_1, n_2. But I cannot compute it because the computation exceeds the binary64-range - it requires too many throws.
Such a solution could get a "*". But this is only a guess!

[LaurV]

Originally Posted by uau

Consider two dice with six sides. They won't always finish on the same turn right? The second die will sometimes last longer. Those times, the game will last longer than it would have with the first die alone. Since you expect that the game will sometimes last longer with two dice than just the first die, and never shorter, the combined expectation for two equal dice must be longer than the expectation for just the first die alone.

Actually, this makes a lot of sense. But I still don't know how to solve it, beside simulation, like playing a million cases in your favorite tool, like excel, pari, etc, vary the dices +/-, blah blah, for this I already got the best solution, and I still don't know how to write down a conditioned probabilities formula to get these numbers, even if I have the numbers already. Probabilities theory is a bitch... I will wait for a solution from the site.

[LaurV]

Hey Watson? ...
....(bzzzzz)....

...Watson?...
....(bzzzzz)....

...Reception...
....(bzzzzz)....

...Haifa?...
....(bzzzzz)....

Hm... it is not just me... we tried repeatedly in the last ~18 hours and it looks the same. We even tried the old haifa site, which shows unregistered now... should we start selling our IBM shares?



And we only wanted to see how many of you solved the problem... (well, not exactly, we were more interested in possible updates, but do not tell to the other guys)

[LaurV]

Site is back, we think they were scared by our previous post.
We think we solved the mystery of the star: actually this problem is extremely simple to solve by simulation, our code is just about 10 lines in excel VBA (including the nice interface in the sheet), so we assume that you got your name on the wall if you send a solution, but for the star you have to prove that the solution is the best (i.e. formulas, which we don't know yet how to write, but we didn't try very hard). We may post our code here after the ETA.