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Re: [Phys-L] From a Math Prof (physics BS major) at my institution ( math challenge)

It doesn't matter, aside from efficiency.

If I put 6 numbers in a hat and draw two without replacement, then
there are 30 equally likely outcomes (preserving the order of the
numbers that are picked).

Now let's say I put 6 numbers in a hat and draw two *with*
replacement, but with the condition that if the second draw is the same
number as the first, I put it back and redo my second draw until I get
a different result. Let's calculate the probability that the two
numbers end up being "1" and "2", in that order. Well, the first draw
picks "1" with probability 1/6. The second draw picks "2" 1/6 of the
time... or "1" 1/6 of the time, in which case I redo. That redo
similarly picks "2" 1/6 of the time, or "1" 1/6 of the time which would
mean *another* redo, and so on.

The probability that the second draw ends up being "2" after as many
redos as it takes is
1/6 (zero redos) + 1/36 (one redo) + 1/216 (two redos) + ... = 1/5.
So the overall chance of picking "1" and then "2" is 1/30. Aside from
being distinct, "1" and "2" aren't special, thus all of the ordered
pairs of distinct numbers occur with equal probability 1/30.

I'm by no means proficient in matlab/octave, but it seems like all of
this choosing code could be replaced with calls to randperm(35, 5).

- Craig


On Wed, 26 Feb 2014 18:36:49 +0000
Jeffrey Schnick <JSchnick@Anselm.Edu> wrote:

-----Original Message-----
From: Phys-l [mailto:phys-l-bounces@phys-l.org] On Behalf Of brian
whatcott
Sent: Wednesday, February 26, 2014 12:41 PM
To: phys-l@phys-l.org
Subject: Re: [Phys-L] From a Math Prof (physics BS major) at my
institution ( math challenge)

If the intention is to draw five integers in the range 1..35 from a
hat, what do you do if you draw a number that's already drawn?

...
I think what Paul is saying is that after you pick a number you
shouldn't put it back in the hat before having picked all five. I
can't see how it could matter but I have created a new version of the
code, one that uses all values returned by the pseudo-random
distribution generator and it is running now with n = a million. I
have pasted it below. See any mistakes in it? If not, what is your
prediction on how the results will compare with those of the old code?

function dummy = getdist2(n)
tic
for m=1:n
a=zeros(21,5);
for i=1:21
numsLeft=0:36;
for j = 1:5
nIndex = fix(2+rand*(36-j));
a(i,j)=numsLeft(nIndex);
numsLeft=[numsLeft(1:nIndex-1),numsLeft(nIndex+1:38-j)];
end
end
a=sort(a,2);
nrounds(m)=sum(10==a(:)) + sum(20==a(:)) + sum(30==a(:));
nseq2(m)=0;
for i=1:21
b=diff(a(i,:));
nseq2(m)=nseq2(m)+sum(1==b);
end

end


for i=0:6
totalrounds(i+1)=sum(i==nrounds);
end
totalrounds=totalrounds

for i=0:6
totalseq2(i+1)=sum(i==nseq2);
end
totalseq2=totalseq2

for i=0:6
total(i+1)= sum( nseq2<=i & nrounds<=i );
end
total=total

toc
end

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