Distribution Of Chips.

[Patrick D. Rockwell]

Here is a probability problem that I read in a couple of books.

You have a room in which there are a men and b women. Another
person comes in (whose gender is irrelevant) and distributes
a number of chips to random people.

After n chips are distributed, what is the probability that
the men have an even number of chips?

The solution would be

P(n)=(b/(a+b))*P(n-1)+(a/(a+b))*(1-P(n-1)) where

P(n)=1 if n=0.

I thought about this and came up with a generalization.
Let's say that you have the same problem, but you want to
know the probability that after n chips are distributed,
the men will have a number of chips equal to (x*m+r).

In the above problem, m=2 and r=0.

My generalization would be the following recursive formula.

P(n,r)=(b/(a+b))*P(n-1,r)+(a/(a+b))*P(n-1,(r-1) mod m)

where P(n,r)=1 if n=0 and r=0,
P(n,r)=0 if r>n
(r-1) mod m = (m-1) if r=0

I've tested this out on EXCEL, and it seems to work.
I've also seen that as n gets larger, the probability
gets closer to 1/m.

Now this is where it gets interesting. I'd like to
generalize it further. Let's say that instead of men
and women, (which are two groups) you have say, 3
groups. A room of 10 people.

2 policemen
3 firemen
5 paramedics.

After distributing n chips randomly to these people,
what is the probability that the 2 policemen will
get a number of chips equal to 5x+1 (m=5, r=1), the
3 firemen will get a number of chips equal to 5x+3
(m=5, r=3)?

I can do it for each group seperately. For the police,
I'd set a=2, b=3+5. For the firemen, I'd set a=3, b=2+5.
For the paramedics, I'd set a=5, b=2+3 and use the
above generalized formula. But what about doing the
calculations for 2 or more goups at one time when the
chips are finished being distributed?

Thanks in advance.

[Nick Wedd]

In message
<10ddaaba-085f-4254...@w4g2000prd.googlegroups.com>,
Patrick D. Rockwell <proc...@thegrid.net> writes

>Here is a probability problem that I read in a couple of books.
>
>You have a room in which there are a men and b women. Another
>person comes in (whose gender is irrelevant) and distributes
>a number of chips to random people.

I'm not sure what this means.
The newcomer arbitrarily selects some people, then gives them each the
same number of chips?
The newcomer arbitrarily selects some people, then gives them each an
arbitrary number of chips?

>After n chips are distributed, what is the probability that
>the men have an even number of chips?

Does "the men have" mean "each of the men has" or "the total given to
the men is"?
And does "even" mean "uniform", or "not odd"?

--
Nick Wedd ni...@maproom.co.uk

[Patrick D. Rockwell]

On May 15, 2:51 pm, Nick Wedd <n...@maproom.co.uk> wrote:
> In message
> <10ddaaba-085f-4254-97ea-13de121a2...@w4g2000prd.googlegroups.com>,
> Patrick D. Rockwell <prockw...@thegrid.net> writes

>
> >Here is a probability problem that I read in a couple of books.
>
> >You have a room in which there are a men and b women. Another
> >person comes in  (whose gender is irrelevant) and distributes
> >a number of chips to random people.
>
> I'm not sure what this means.
> The newcomer arbitrarily selects some people, then gives them each the
> same number of chips?
> The newcomer arbitrarily selects some people, then gives them each an
> arbitrary number of chips?

No, the newcomer pulls a chip out of his bag, and then hands it to a
random person. He does this a n times.

> >After n chips are distributed, what is the probability that
> >the men have an even number of chips?
>
> Does "the men have" mean "each of the men has" or "the total given to
> the men is"?

The total given to the men.

> And does "even" mean "uniform", or "not odd"?
>

It means NOT ODD.

> Nick Wedd    n...@maproom.co.uk- Hide quoted text -
>
> - Show quoted text -

[The Qurqirish Dragon]

On May 15, 5:15 pm, "Patrick D. Rockwell" <prockw...@thegrid.net>
wrote:

A change-of-state matrix is probably the best way to go for an
arbitrary setup.

Consider for the original problem (a men, b women)
let [ x (n), y (n) ] be the (row) vector where x(n) is the probability
that with n chips distributed, the men have an even number of chips,
and y (n) is the probability that the men have an odd number of chips
(note: y(n), obviously, is simply 1 - x(n))

Then, [ x(0), y (0) ] = [ 1 , 0 ], and
[ x (n + 1), y (n + 1) ] =
[ x (n), y (n) ] * A

where A is the following matrix, with p = a/(a +b) as the probability
that one chip is given to a man. (and q = 1 - p)
[ q p]
[ p q]

And so [ x (n), y (n) ] = [ 1 , 0] * A^n.

For more complicated problems (such as your later example), you may
need an initial-state matrix rather than a vector (for your latter
example, 2-rows and 5 columns may work- one row for police, one for
firemen, and 5 columns for the quantity modulo 5) and the change of
state matrix will be larger (5x5 in this case), but the same principle
works.

[Patrick D. Rockwell]

> works.- Hide quoted text -

>
> - Show quoted text -

So, the matrix and vector would look like this?

1 0 0 0 0 .8 .2 .0 .0 .0
1 0 0 0 0 AND .0 .8 .2 .0 .0
.0 .0 .8 .2 .0
.0 .0 .0 .8 .2
.2 .0 .0 .0 .8

Did I understand you correctly? In not, please let me know.

Assuming that I did understand you, my problem is this. The
above matrix would be valid for policemen because they make up
20% of the people in the room, but Firemen make up 30% and so,
the matrix for them would be the above except with all of the
.8's and .2's replaced by .7's and .3's respectively.

My problem seems to be this. If we're going to compute the
probability that out of 10 people, the 2 policemen have
5x+1 chips between them, while simultaneiousely, the 3 firemen
have 5x+3 chips between them, the two aren't independant,
otherwise I could compute F(n,1) with a=2, b=8, and F(n,3)
with a=3, b=7, then multiply the two probabilities.

If they aren't independant, then how about this approach.

There are 2 policemen, and 3 firemen. 5 people out of 10 in
all. For the police, the r value we have is 1, and for the
firemen, it is 3 so, we want the probability that ALL of the
NON-paramedics have a number of chips equal to 5x+(1+3)=5x+4.
We do this for a specific value of n, and let that v=that
probability.

Then we calculate the probability that the police have
5x+1 chips. and let u = that probability.

So, then our quesion is, what is the probability that
the police have 5x+1 chips given that all of the
non-paramedics in the room have 5x+4. If this is a
conditional probability we divid u/v.

And so, the probability that the firemen have 5x+4
chips is (1-(u/v)).

Any comments? Ideas? Am I right or wrong here? I'm
still not sure, but it's my best idea to date.