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Subject: 
Re: Sante Fe Wish List
Newsgroups: 
lugnet.trains
Date: 
Thu, 28 Feb 2002 14:21:29 GMT
Viewed: 
1038 times
  
Scott A wrote:

In lugnet.trains, Alan Demlow writes:
Yes, I hope they don't go for a numbered Ltd Ed run of 10,000 again. The
chances of getting a match (if you buy one of each) would be ~1:10,000,000
or 1:9,900,025 if we allow for BJ taking 5 from each run! (if my Y2 stats
are correct... what are the chances of that?)

Zero.


Scott A


I can't resist this one...if you already have one Santa Fe loc and buy one
expansion pack, the chances of getting a match are 1:10,000 (or, 1:9995 to
account for the 5 taken out by BJ), not 1/100,000,000=(1/10,000)^2.  The
chances of getting any given number (say, # 5034) twice if you buy one each of
the Sante Fe and expansion pack would be 1:100,000,000, but this is different
than than the chances of getting some number t2ice, which is 1/10000 as stated.

I deserved that - Thanks Allan. I was working on the latter case, but missed
a zero in my metal calc, and then, for some reason, missed the nine at the
start of 99900025 ((10k-5)^2).

Basically, if I wanted to get two the same I'm better to put the money on
the lottery (UK odds ~1:13,000,000) and hope to win (~£4M?) and use the
proceeds to buy large amounts of 10020s (A's & B's). The chances of all that
are slim... but not zero. ;)

You're still not understanding the math. Put it this way, lets say I
have A8888, what is the chance that when I buy a B unit I get B8888? One
in 9995. The one in 99900025 chance is the chance you would have to get
both A0006 and B0006 (or A4558 and B4558 or any other specific desired
number).

Frank



Message is in Reply To:
  Re: Sante Fe Wish List
 
(...) Zero. (...) I deserved that - Thanks Allan. I was working on the latter case, but missed a zero in my metal calc, and then, for some reason, missed the nine at the start of 99900025 ((10k-5)^2). Basically, if I wanted to get two the same I'm (...) (23 years ago, 27-Feb-02, to lugnet.trains)

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