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