20180202, 14:29  #1 
Aug 2002
2^{3}×5×11×19 Posts 
February 2018

20180202, 16:30  #2 
Aug 2006
3·1,993 Posts 
I can't find a solution  the closest I can find is (25, 15, 15) which has an expected time of 5841.87... seconds which is too large by 0.37.... I must be doing something wrong, because I've exhausted all the possibilities since even (26,1,1) takes > 6000 seconds, so n_i <= 25.

20180202, 16:46  #3 
Jun 2003
12111_{8} Posts 

20180205, 07:47  #4 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2^{4}×613 Posts 
That is the first problem in a long while which looks interesting, and what a pity we don't know how to solve it, haha...
If I have a die with 6 faces and I have to throw it until I exhaust all the possibilities (which are 6), then probabilistically, I will have to cast it at least 6 times. I will have to expect to throw it more than 6 times, to get all faces, unless I am motherfreakinglucky. How many more, well, we may be able to compute that, with confidence intervals and all the stuff, but we do not want to spoil it. But assuming all the other players have dices with less than 6 faces (less than my die), then why should they matter for the game? Wouldn't the game finish when the guy with the largernumberedfaced die ends the game? This part I can not get. If you an you both have dices with 1 face (I assume they have to look like nodding dolls, or rolypoly, to fall always on a side, or stand up, somehow, or work in a different universe...whocares...) and I have a 2faced die, wouldn't the game always end after n>2 minutes (by me), no matter what the rolypoly dices say? I can not understand for the hack of my head, why the guys with the smaller dices count... The game will just end when the guy with the larger die exhausted all possibilities... Or... am I totally wrong? Last fiddled with by LaurV on 20180205 at 07:50 
20180205, 14:51  #5 
Oct 2017
11^{2} Posts 
“The gamewill just end when the guy with the larger die exhausted all possibilities...”
That iscorrect only when the other guys have dices with 1 face (that could besimulated by a “normal” die with a 1 on each face). In allother cases there are probabilities <>0 for all dices and every number ofthrows and these probabilities have an effect on the expected value of the time. Oneexample: The probability for all cases: “The guy with the larger die has exhausted all possibilities, but the guys with the smaller dices have not yet finished”is not zero. Myproblem is: for computing the expected time I have to summarize an infinitenumber of summands – decreasing, but never =0. How do I know when to finish? Howdo I know if this infinite sum is convergent? 
20180205, 14:58  #6 
Jun 2003
1010001001001_{2} Posts 
Is it me or did they update the target to 2569?

20180205, 15:19  #7 
Oct 2017
11^{2} Posts 
Yes, they did update!

20180205, 15:23  #8 
Aug 2006
3×1,993 Posts 

20180207, 07:12  #9 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2^{4}·613 Posts 
When I first accessed the site, at the time of my first post, it was as it is now. So the update was before that. I was wondering from where CRG got those many thousands in the black boxes.
Last fiddled with by LaurV on 20180207 at 07:14 
20180209, 23:03  #10 
Jan 2017
11^{2} Posts 
Anyone have an idea what the '*' marks in answer list mean? There doesn't seem to be any visible bonus objective...

20180210, 04:23  #11 
"Curtis"
Feb 2005
Riverside, CA
11·461 Posts 
Consider a 6sided die for you, and an 8sided 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 conditionalprobability calculation.

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