Re: average of the consecutive coin tosses
- To: mathgroup at smc.vnet.net
- Subject: [mg122001] Re: average of the consecutive coin tosses
- From: "Dr. Wolfgang Hintze" <weh at snafu.de>
- Date: Sat, 8 Oct 2011 05:35:23 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <j6jobf$lv7$1@smc.vnet.net>
Hi,
a = Table[2*Random[Integer, {0, 1}] - 1, {10^3}];
b = FoldList[Plus, a[[1]], Drop[a, 1]];
N[Mean[b]]
... and say hello to "this guy" ;-)
Wolfgang
"michael partensky" <partensky at gmail.com> schrieb im Newsbeitrag
news:j6jobf$lv7$1 at smc.vnet.net...
> 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