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Re: average of the consecutive coin tosses

  • To: mathgroup at smc.vnet.net
  • Subject: [mg121949] Re: average of the consecutive coin tosses
  • From: Michael Stern <nycstern at gmail.com>
  • Date: Fri, 7 Oct 2011 04:46:30 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201110060818.EAA22381@smc.vnet.net>

Your way:

Timing[(flips = 1; (#/flips++) & /@
      Accumulate[Table[RandomInteger[], {10000}]]) // N;]

{0.078, Null}



My way:

Timing[Accumulate[RandomInteger[{0, 1}, 10000]]/Range[10000] // N;]

{0.032, Null}




On 10/6/2011 4:18 AM, michael partensky wrote:
> Hi.
>
> What are the better ways of doing what this guy does very inefficiently for
> the average of the consecutive coin tosses:
>
> tr = (flips = 1; (#/flips++)&  /@ Accumulate[Table[toss, {1000}]]) // N);
>
> where
>
> toss= RandomInteger[];
>
> Thanks
> Michael



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