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Re: average of the consecutive coin tosses
*To*: mathgroup at smc.vnet.net
*Subject*: [mg121952] Re: average of the consecutive coin tosses
*From*: michael partensky <partensky at gmail.com>
*Date*: Fri, 7 Oct 2011 04:47:03 -0400 (EDT)
*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com
*References*: <201110060818.EAA22381@smc.vnet.net>
On Thu, Oct 6, 2011 at 10:28 AM, Sseziwa Mukasa <mukasa at gmail.com> wrote:
>
> On Oct 6, 2011, at 9:32 AM, michael partensky wrote:
>
> Thanks, Sseziwa.
> This example is designed for students, to compare the Mathematica style
> with the "For" loop.
> I just wanted to make it more elegant.
>
> For instance, I did not find a nice way of dividing the total value of the
> random table corresponding to each step, by the corresponding number of
> trials. That forced me to multiply # by the flipCount
> before dividing it by ++flipCount. It does not look neat.
>
> Any ideas?
>
>
> The only options I could thing of were carrying the trial number in the
> NestList argument:
>
> coinFlipNestNoGlobal[n_]:=Rest[NestList[Block[{iter=#[[2]],mean=#[[1]]},{(iter
> mean + RandomInteger[])/(iter+1),iter+1}]&,{0.,0},n]][[All,1]]
>
> If the number of trials is small, you can use FoldList:
>
> coinFlipFold[n_]:=Rest[FoldList[((#2-1) #1 +
> RandomInteger[])/#2&,0.,Range[n]]
>
> Another approach is work backward and use Reap and Sow
>
>
> coinFlipReap[n_]:=Reverse[Last[Reap[Nest[Block[{},Sow[Mean[#]];Rest[#]]&,RandomInteger[{0,1},{n}],n]]][[1]]]
>
> None of these are completely satisfying to me, what you need, and I've
> often wanted it in Mathematica, is a form of nest where you can tell how
> deep in the nest you are in the callback function.
>
*This is exactly what I wanted. Why isn't it there? Would be very natural to
have. Well, how would I know ?! :)*
>
> Regards,
> Sseziwa
>
> PS - For unrelated reasons I can't test these expressions in Mathematica
> now so there may be typos in the expressions above but I think you get the
> gist.
>
> Best wishes,
> and thanks again for your help.
> MP
>
> On Thu, Oct 6, 2011 at 9:17 AM, Sseziwa Mukasa <mukasa at gmail.com> wrote:
>
>> Michael,
>>
>> I misunderstood the intent of your previous post, the code example you
>> sent is almost correct you should probably wrap the NestList in a Rest to
>> remove the initial 0. If you use 0. (note the period) as your initial value
>> you don't need the //N either.
>>
>> Regards,
>> Sseziwa
>>
>> On Oct 6, 2011, at 8:22 AM, michael partensky wrote:
>>
>> Thanks, Sseziwa.
>> My example showed evolution of mean value with the increasing number of
>> trials.
>> This would be another take, but I am looking for something more elegant
>> and robust.
>>
>> coinFlipNest[n_]:=Module[{flipCount=0},NestList[((flipCount # +
>> RandomInteger[])/(++flipCount)//N)&,0,n]]
>>
>> Michael
>>
>> On Thu, Oct 6, 2011 at 6:37 AM, Sseziwa Mukasa <mukasa at gmail.com> wrote:
>>
>>> Mean[RandomInteger[{0,1},{1000}]
>>>
>>> Regards,
>>> Ssezi
>>>
>>> On Oct 6, 2011, at 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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