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Re: average of the consecutive coin tosses

  • To: mathgroup at smc.vnet.net
  • Subject: [mg121942] Re: average of the consecutive coin tosses
  • From: Heike Gramberg <heike.gramberg at gmail.com>
  • Date: Fri, 7 Oct 2011 04:45:13 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201110060818.EAA22381@smc.vnet.net>

You could do something like

tr = With[{list = RandomInteger[{0, 1}, 10000]},  Accumulate[list]/Range[Length[list]]];

On my machine this is about a factor 2.5 faster than the original code.

Heike.

On 6 Oct 2011, at 10:18, 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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