An expectation problem:

bobbym · 2012-07-26 01:46:47

A list of 1000 zeroes is created called M. A random number is picked from 1 to 1000. Each time one of these is picked, the zero in the list ( at that index ) is changed to a 1. For instance if the random number is 276 then M[276]=1. What is the average number of random numbers that must be picked to have 200 ones in the list?

This has an analytic solution given in the other thread.

http://www.mathisfunforum.com/viewtopic … 45#p227545

This thread has been created to list the various programming solutions to the problem. Please only programming here. All analytical answers can go in the other thread.

gAr · 2012-07-26 02:02:52

To check our analytic solution, we may try a simulation:

Here is an example of doing it in python

from random import randint

zeroes = [0]*1000
tries = 10000
hit = 0

for i in range(tries):
    while sum(zeroes) < 200:
        zeroes[randint(0, 999)] = 1
        hit += 1
    zeroes = [0]*1000
print hit / float(tries)

bobbym · 2012-07-26 08:28:21

This idea uses a different approach. Instead of finding the number of picks it uses the analytical answer and then uses a simulation to test if it is correct.

fubar[l_]:=Module[{h1,g1},
g1=l;
h1=RandomInteger[{1,Length[l]}];
g1[[h1]]=1;
g1]

Now run this:

Table[s=Table[0,{1000}];Count[Table[s=fubar[s],{223}]//Last,1],{100}]//Mean//N

The output is 200.03.

gAr · 2012-07-27 05:40:29

In R, we may speed up the simulation using the sample function:

sa <- c(1:1000)*0
trials <- 10000
hit <- 0
for (i in 1:trials)
{
    sa[sample(1:1000, 200, replace = TRUE)] <- 1
    hit <- hit + 200
    while (sum(sa) < 200)
    {
        sa[sample(1:1000,1)] <- 1
        hit <- hit + 1
    }
    sa <- c(1:1000)*0
}
print (hit / trials)

phrontister · 2012-07-29 02:08:39

Hi Bobby,

The Olympic Games have got my attention for now but I thought I'd post this LB attempt anyway. It's a bit scratchy, but it works...so I'm happy enough.

Some better coder would be able to improve its efficiency and speed (the 10000 cycles took just under an hour!). Compared with yours and gAr's solutions, mine seems to be stuck way back in the Dark Ages!

Results: My test run of 10000 cycles yielded an average of 223 picks per cycle.

Got some long nights of TV viewing ahead!

    FOR cycles = 1 TO 10000
        M$ = ""
        FOR zeroes = 1 TO 1000
            M$ = M$ + "0"
        NEXT zeroes

        'initial 199 picks
        FOR picks = 1 TO 199
            random = INT(RND(1) * 1000) + 1
            M$ = LEFT$(M$,random - 1) + "1" + RIGHT$(M$,1000 - random)
        NEXT picks
        picks = picks - 1

[morepicks]
        FOR count = 1 TO 1000
            sum = sum + VAL(MID$(M$,count,1))
        NEXT count
        IF sum = 200 THEN GOTO [nextcycle]
        random = INT(RND(1) * 1000) + 1
        M$ = LEFT$(M$,random - 1) + "1" + RIGHT$(M$,1000 - random)
        sum = 0
        picks = picks + 1
        GOTO [morepicks]

[nextcycle]
        totalpicks = totalpicks + picks
    NEXT cycles
    cycles = cycles - 1

    'results
    PRINT "Cycles: ";cycles
    PRINT "Total picks: ";totalpicks
    PRINT "Average picks per cycle: ";totalpicks / cycles
    END

bobbym · 2012-07-29 02:15:50

Hi phrontister;

It got the right answer! Efficiency takes second place to that.
Very good work.

anonimnystefy · 2012-07-29 02:39:15

Hi

I get the answer of 2.23058399999999E+002:approx 223.0584 with:

program Expectation;

type
  arr=array[1..1000] of boolean;
  
function count(a:arr):integer;
var
  i,cnt:integer;
begin
  cnt:=0;
  for i:=1 to 1000 do
    cnt:=cnt+ord(a[i]);
  count:=cnt;
end;

var
  a:arr;
  i,j,number:integer;
  expect:real;

begin
  randomize;
  expect:=0;
  for i:=1 to 10000 do
  begin
    for j:=1 to 1000 do
      a[j]:=false;
    number:=0;
    while count(a)<200 do
    begin
      a[random(1000)+1]:=true;
      number:=number+1;
    end;
    expect:=expect+number;
  end;
  writeln(expect/10000);
end.

Last edited by anonimnystefy (2012-07-29 03:30:03)

bobbym · 2012-07-29 02:42:15

Hi;

That is very good!

anonimnystefy · 2012-07-29 02:45:34

What is the actual number, again?

bobbym · 2012-07-29 02:53:41

anonimnystefy · 2012-07-29 02:57:37

And in the decimal form?

bobbym · 2012-07-29 02:58:47

Hi;

anonimnystefy · 2012-07-29 03:02:29

I ran the simulation once again and got the same number as before.

bobbym · 2012-07-29 03:05:27

The exact same number? Then you have a bug in the code.

See you in a bit. Chore time!

anonimnystefy · 2012-07-29 03:31:49

Changed my code a little, but now it always gives output less then, but close to the actual answer. Haven't gotten a number over 223 with the new code.

bobbym · 2012-07-29 06:27:19

What is the output of a couple of runs?

anonimnystefy · 2012-07-29 07:25:46

One was something aroung 222.8 and another one was around 222.9. Let me get the actual figures.

bobbym · 2012-07-29 07:52:39

That seems reasonable. Over many runs you will average out to around 223.

anonimnystefy · 2012-07-29 07:59:06

Here is what I got now:

 2.23019100000000E+002
 2.23065900000000E+002
 2.23091500000000E+002
 2.23027800000000E+002
 2.23021100000000E+002
 2.22971200000000E+002
 2.23048500000000E+002
 2.23036800000000E+002
 2.22945700000000E+002
 2.23067400000000E+002
 2.23025500000000E+002
 2.23024000000000E+002

Last edited by anonimnystefy (2012-07-29 08:04:28)

bobbym · 2012-07-29 08:02:29

That is about right.

anonimnystefy · 2012-07-29 08:03:19

I will add a few more values in a minute.

bobbym · 2012-07-29 08:08:11

The answer is a little more than 223, I think.

anonimnystefy · 2012-07-29 08:11:00

Did you see the edited post 19?

bobbym · 2012-07-29 08:13:17

Yes, it looks fine to me. The sd is very small on this problem.

anonimnystefy · 2012-07-29 08:14:45

Standard deviation?

Math Is Fun Forum

#1 2012-07-26 01:46:47

An expectation problem:

#2 2012-07-26 02:02:52

Re: An expectation problem:

#3 2012-07-26 08:28:21

Re: An expectation problem:

#4 2012-07-27 05:40:29

Re: An expectation problem:

#5 2012-07-29 02:08:39

Re: An expectation problem:

#6 2012-07-29 02:15:50

Re: An expectation problem:

#7 2012-07-29 02:39:15

Re: An expectation problem:

#8 2012-07-29 02:42:15

Re: An expectation problem:

#9 2012-07-29 02:45:34

Re: An expectation problem:

#10 2012-07-29 02:53:41

Re: An expectation problem:

#11 2012-07-29 02:57:37

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#12 2012-07-29 02:58:47

Re: An expectation problem:

#13 2012-07-29 03:02:29

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#14 2012-07-29 03:05:27

Re: An expectation problem:

#15 2012-07-29 03:31:49

Re: An expectation problem:

#16 2012-07-29 06:27:19

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#17 2012-07-29 07:25:46

Re: An expectation problem:

#18 2012-07-29 07:52:39

Re: An expectation problem:

#19 2012-07-29 07:59:06

Re: An expectation problem:

#20 2012-07-29 08:02:29

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#21 2012-07-29 08:03:19

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#22 2012-07-29 08:08:11

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#23 2012-07-29 08:11:00

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#24 2012-07-29 08:13:17

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#25 2012-07-29 08:14:45

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