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Old 2018-02-02, 14:29   #1
Xyzzy
 
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Default February 2018

https://researchweb.watson.ibm.com/h...ruary2018.html
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Old 2018-02-02, 16:30   #2
CRGreathouse
 
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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.
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Old 2018-02-02, 16:46   #3
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Quote:
Originally Posted by CRGreathouse View Post
I can't find a solution
Did your clock start at first throw? i.e n=1 @ t= 0?
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Old 2018-02-05, 07:47   #4
LaurV
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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 mother-freaking-lucky. 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 larger-numbered-faced 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 roly-poly, to fall always on a side, or stand up, somehow, or work in a different universe...whocares...) and I have a 2-faced die, wouldn't the game always end after n>2 minutes (by me), no matter what the roly-poly 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?

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Old 2018-02-05, 14:51   #5
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“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?




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Old 2018-02-05, 14:58   #6
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Is it me or did they update the target to 2569?
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Old 2018-02-05, 15:19   #7
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Yes, they did update!
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Old 2018-02-05, 15:23   #8
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Quote:
Originally Posted by axn View Post
Is it me or did they update the target to 2569?
Yes, they changed it -- I sent in a solution for the original problrm and they emailed me back saying the problem had changed.
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Old 2018-02-07, 07:12   #9
LaurV
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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.

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Old 2018-02-09, 23:03   #10
uau
 
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Anyone have an idea what the '*' marks in answer list mean? There doesn't seem to be any visible bonus objective...
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Old 2018-02-10, 04:23   #11
VBCurtis
 
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Quote:
Originally Posted by LaurV View Post
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?
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.
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