Conference rusure::math

>    Parachutists have an equal chance of landing on the south half of a
>    small island as on the north half.

I think you also need some information about the prob. density function, 
and about the actual prob's to answer this. If the island is sufficiently 
small, EACH prob. may be zero...

Lynn Yarbrough 

>    If 6 parachutists jump at one time, what is the ratio of probability
>    that four land on the north half and two land on the south half, to the
>    probability that three land on each half.

     Assuming the chance is not zero then I think the answer is 3 to 4.
    
    Peter

.1 is a red herring, unless the probabilities are exactly zero, in which case 
the ratio is undefined.  Otherwise, the ratio is independent of the actual
probabilities, we have complete information, and we can assume for convenience
a large island so that 

	P( landing on the island ) = 1

The unstated assumptions in .0 is that the probabilities are stationary and
independent, both of which may be contrary to reality.  The motion of the plane 
and changing winds will ensure that the probabilites are not stationary.  The
control the parachutists have over their descent and their motivation to 
disperse or stick together will ensure that the probabilities are not 
independent.

However, assuming for the sake of the problem that the probabilities are
stationary and independent...

There are 2^6 or 64 possible outcomes, ranging from NNNNNN to SSSSSS.  If you 
just count the relevant cases, you will see that there are 15 cases with 4 Ns,
and 20 cases with 3Ns.  So the requested ratio is just 3/4.

Or you can notice that this is just a selection of 2 objects out of 6 or 3
objects out of 6, without regard to order.  These numbers are given by the
rule for combinations or binomial coefficient

	nCr = n!/(r!(n-r)!)

and this will give you the same answer.

Or you could notice that these assumptions define a Bernoulli process with equal 
probabilities and look up the binomial probabilities and you get the same 
answer.

RE: .3

>Or you can notice that this is just a selection of 2 objects out of 6 or 3
>objects out of 6, without regard to order.  These numbers are given by the
>rule for combinations or binomial coefficient
>
>	nCr = n!/(r!(n-r)!)
>
>and this will give you the same answer.
>
>Or you could notice that these assumptions define a Bernoulli process with equal 
>probabilities and look up the binomial probabilities and you get the same 
>answer.

    Specifically in the latter case you would find that the formula is:

		   r	    (n-r)
(1)	    nCr * p  * (1-p)

    When p = 1/2, p = (1-p) so it becomes:

		       n
(2)	    nCr * (1/2)
				      n
    When the ratio is formed the (1/2)  term is the same top and bottom so
    it is simply the ratio of the nCx terms, and so the same.

    Note that if the two halves of the island are not equal then the simple
    ratio of nCx's by itself could not be used, and one would have to
    either use the explicit enumeration or the binomial probability
    formula.

					Topher

I think there are still some unstated assumptions about the prob. density 
function. Take, as a limiting case, the following:

Assume the skill level is such that P(landing on the surface)=1 for one 
chutist. Assume even that if there is space on the island for even one more 
chutist, any of them can land in it. However, if there is space for only 
three in each side of the island, then P(3-3)=1 and P(4-2)=P(2-4)=0.

If the island is large enough so that the chutists can be idealized into 
points (the problem statement does not say this), then the underlying prob. 
density may be taken to be uniform, and the equations in previous notes
hold. If the area is smaller relative to the size of the chutists then the 
presence of previous occupants affects the likelihood (and, given that 
skill enters into the action, the choice) of subsequent chutist's sites.

BTW, if one looks only at one side of the island (say N) then the 2-4 and 4
-2 cases are equally likely and it is their sum that has the ratio 4/3 with 
the 3-3 case. Each individually is in the ratio 2/3 with the 3-3 case.

BTW, if we completely ignore skill for the chutists, the we need to take 
into account the probability that some of them will miss the island 
altogether. In that event, P(2-4)=P(3-3)=P(4-2) = 0.    B^)

If the probability of landing on the North side is p, and q = 1-p, then
Prob( 4 north & 2 south ) / Prob( 3 north & 3 south ) is (3p)/(4q).


If there are P+Q independent jumps, and the probability of landing on the
North side is p, and q = 1-p, then
							    p (Q+1)
Prob( P north & Q south ) / Prob( P-1 north & Q+1 south ) = -------
							    (1-p) P


This suggests a technique for measuring the ratio between the areas of the
two parts of the island.

RE: .5

    We do need to assume some specific probability of each 'chutist hitting
    the island -- with certainty being the most convenient assumption.
    "Stationary and independent" would seem to preclude the big feet
    scenario, however.

				    Topher

RE: .6

>This suggests a technique for measuring the ratio between the areas of the
>two parts of the island.

    ... which seems similar in principle but rather clumsier than standard
    Monte Carlo integration (i.e., you drop 'chutists on the the island with
    independent, uniform distributions and the ratio of 'chutists in
    the north to 'chutists to the south is an unbiased estimator of the
    ratio of areas).

					    Topher

        If you assume p is the prob. of landing on the north
        half, then p is also the prob. of landing on the south
        half.  Let q = 1 - 2p be the prob. of missing the island
        altogether.  The result of a trial of all six
        parachutists jumping at one time is (n1,n2,n3) where n1
        is how many land on the north half, n2 is how many land
        on the south half, and n3 miss the island.  We have 0 <=
        ni <= 6 and n1 + n2 + n3 = 6.  The probability of
        (n1,n2,n3) is (6! / (n1! n2! n3!)) p^n1 p^n2 q^n3.
        
        If p =/= 0, then the probability of (4,2,0) is 15 p^6 and
        the probability of (3,3,0) is 20 p^6 and the ratio
        between them is 3/4.
        
        As an earlier reply already indicated.
        
        Dan

>BTW, if one looks only at one side of the island (say N) then the 2-4 and 4
>-2 cases are equally likely and it is their sum that has the ratio 4/3 with 
>the 3-3 case. Each individually is in the ratio 2/3 with the 3-3 case.

Drat, I got that backwards. C(6,3)=20, C(6,4)=C(4,2)=15, so it's the 
(3,3)/(4,2) ratio that's 4/3 and the (3,3)/((4,2)+(2,4)) that's 2/3. Can't 
do these things in my head anymore.

RE: .9 (Dan)

    Yes, quite right, sorry, there is no need to know the proportion of
    'chutists who land on the island -- I slipped into analyzing a
    different problem.  If you know how many land north and how many south
    (equivalently, how many land north and how many land on the island)
    then you may make your calculation.  If you know how many land north
    and how many left the plane *then* you need to know what proportion
    you could expect to land on the island.

					Topher