Re: pattern matching

*To*: mathgroup at smc.vnet.net*Subject*: [mg128312] Re: pattern matching*From*: Dana DeLouis <dana01 at me.com>*Date*: Sat, 6 Oct 2012 01:50:32 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net

> Count[Table[RandomInteger[], {n}], 1]/n, {n, 100, 1000, 10}], x_ < .5] Hi. Here's one way. t=Table[Count[Table[RandomInteger[],{n}],1]/n,{n,10,15,1}] {7/10,4/11,1/2,7/13,3/7,8/15} Count[t, n_/; n<.5] 2 As a suggestion, note the following with 0 / 1's Mean[{1,1,0,0,0}] 2/5 Perhaps: t=Table[Mean[RandomInteger[{0,1},n]],{n,10,15,1}] {1/2,2/11,2/3,5/13,4/7,2/15} Count[t, n_/;n<.5] 3 = = = = = = = = = = HTH :>) Dana DeLouis Mac & Mathematica 8 = = = = = = = = = = On Friday, October 5, 2012 2:43:42 AM UTC-4, Daniel S wrote: > Consider the following code: > > > > Count[Table[ > > Count[Table[RandomInteger[], {n}], 1]/n, {n, 100, 1000, 10}], > > x_ < .5] > > > > I want to count the number of times there are occurrences less than .5 but some how I am not specifying the right pattern. Any suggestions on getting right?