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In fact the product of min ( a, 11-a , r , 11-r ) and min ( b , 11-b , r ,11-r )
(referring to #9 ,#11 and # 13 ) can further be simplified . Since 10 being an
even no. , the table in #11 is symmetric from S1 to S5 with S6 to S10
and also symmetric between k and 11-k , thus min ( a, 11-a ) may be
replaced by a . Similarly min (b ,11-b ) may be replaced by b .
Thus we may only consider the products at the left hand side and then
multiply by 2 . So the term min ( a, 11-a , r , 11-r ) may be simplified to
min ( a , r ) and min ( b , 11-b , r ,11-r ) may be simplified to min ( b , r)
Then the sum of the original products = 2 * [∑ min ( a , r ) * min ( b , r) ]
= 2 * [∑ min ( ab , ar , br , r^2 ) ]. for r from 1 to 5 .
If a = 5 and b = 3 , then the expression will be 2 * [∑ min ( 15, , 3r , r^2 ) ].
Hi thickhead ,
The range of r varies from 1 to 10 , thus 1 is the 1st term of r , and 10 corresponds to
the 1st term of 11-r .
Now for 2 variables ( bosses ) a and b , we have to find the sum of the 10 products of
min ( a , 11-a , r , 11-r ) and min ( b , 11-b , r , 11-r ) . ( where r varies from 1 to 10 .) and
then divided by ( 11-a ) * (11-b ) * 10 to yield the required probability .
E.g. let a = 5 and b = 3 , then min ( 5 , 6 , r , 11-r ) simplifies to min ( 5 , r , 11-r ) while
min ( 3 , 8 , r , 11-r ) becomes min ( 3 , r , 11-r ) .
But it seems I cannot find a formula simple enough for the total sum , it still have to
be calculated 1 by 1 for each product .
I will try whether with the form as matrix in # 8 , a formula can be obtained or not .
Referring to the table in #8 , f( k , r ) being the value at k-th row and r-th column can be found to be
min [ 3 , max ( 0 , r-k+3 ) , max ( 0 , k-r+5 ) ] where k varies from 1 to 6 while r from 1 to 8 .
This function also holds for n , the total no. of servants > 10 ,
For example , if n = 11 , then k varies from 1 to 11-5 +1 = 7 while r varies from 1 to 9 .
Thus f (1, 9 ) = min [ 3 , max( 0, 9-1+3 ) , max( 0,1-9+5) ]
= min ( 3 , 11, 0 ) = 0 ;
while f ( 7 , 1 ) = min [ 3 , max ( 0 , 1-7+3) , max ( 0 , 7-1+5 )]
= min ( 3, 0 , 11 ) = 0
and f ( 7, 9) = min [ 3, max(0, 9-7+3), max(0,7-9+5)]
= min( 3, 5, 3) = 3 .
Hi thickhead ,
It seems you have been successful to find the isomorphism of the problems
with segments and discrete objects for 3 variables But in the present time
I will be more interested to find a formula for 2 variables of this problem .
Your way of analysis is more suitable to find such a formula , therefore I
will follow your way .
In the following table with 10 rows and 10 columns , the data inside shows
the no. of times each corresponding servant receiving money from a boss
who selects k consecutive servants at a time after all the possible ways , where k varies from 1 to 10 .
k \ r S1 - S2 - S3 - S4 - S5 - S6 - S7 - S8 - S9 - S10
(1) 1 1 1 1 1 1 1 1 1 1
(2) 1 2 2 2 2 2 2 2 2 1
(3) 1 2 3 3 3 3 3 3 2 1
(4) 1 2 3 4 4 4 4 3 2 1
(5) 1 2 3 4 5 5 4 3 2 1
(6) 1 2 3 4 5 5 4 3 2 1
(7) 1 2 3 4 4 4 4 3 2 1
(8) 1 2 3 3 3 3 3 3 2 1
(9) 1 2 2 2 2 2 2 2 2 1
(10) 1 1 1 1 1 1 1 1 1 1
The data inside the table is symmetric in both x and y axis from the
middle . For total no. of servants being an even no . 10 , data at the k-th
row will be increasing from 1 to min ( k , 11-k ) till the middle , corresponding
to S 10/2 , i.e. S5 . In general the value in k-th row and r-th column ( corresponding to Sr , the r-th servant ) may be represented by min ( k , 11-k , r , 11-r ) where r varies from 1 to 10 . ( If n , the total no. of servants is odd , then it may be a bit different . )
If we only consider for r ≤ 5 , the expression can be simplified to min ( k , 11-k , r ) .
For any 1 boss choosing k consecutive servants at a time (there are 11-k such cases ), the total no. of times for the 10 servants receiving money , denoted by x , =
min ( 1,10 , k , 11-k ) + min ( 2 , 9 , k , 11-k ) +.... min ( r ,11-r , k , 11-k ) + .....
+ min ( 10, 1 , k , 11-k )
Since the table is symmetric between S1 to S5 and S6 to S10 , therefore x can also be expressed as
2*[ min ( 1, k , 11-k ) + min ( 2 , k , 11-k ) + min ( 3 , k , 11-k ) + ..... ] ( totally 5 terms )
Thus the probability of a servant chosen randomly from the 10 who has received money from the boss will be x / (11-k) * 10 .
Since the probability can also be found directly to be k / 10 ,
therefore x = k * (11-k ) .
.
Hi bobbym ,
Perhaps matrix will be a more correct term .
Let the 10 servants be labelled from S1 to S10 . In the following table ( matrix )
the 1st row denotes the case for Boss A to choose S1 to S5 , the 2nd row for
S2 to S6 , and so on . Thus there are 10-5+1 = 6 rows . In each row the 1st
term denotes the no of servants receiving 3 dollars in the case for Boss B to choose S1 to S3 , and there are min (5,3) = 3 such servants . The 2nd term also = 3 being the case for Boss B to choose S2 to S4 , and so on . Thus there are
10-3+ 1 = 8 terms in each row .
(1) (3,3,3,2,1,0,0,0) (totally 12 servants receiving $3 in the 8 cases )
(2) (2,3,3,3,2,1,0,0) (totally 14 servants receiving $3 in the 8 cases )
(3) (1,2,3,3,3,2,1,0) (totally 15 servants receiving $3 in the 8 cases )
(4) (0,1,2,3,3,3,2,1) (totally 15 servants receiving $3 in the 8 cases )
(5) (0,0,1,2,3,3,3,2) (totally 14 servants receiving $3 in the 8 cases )
(6) (0,0,0,1,2,3,3,3) (totally 12 servants receiving $3 in the 8 cases )
The grand totality of servants receiving $3 = 82 in the 6*8 = 48 combinations ,
thus in average there are 82/ 48 = 41/24 such servants in each case within the 10 servants . Therefore for 1 servant chosen randomly from the 10 the probability he received $3 = 41/24 * 1/10 = 41/240 .
Hi thickhead and bobbym ,
Both of you are right ! The 2nd answer seems quite intuitive , it will be the same even if the servants chosen by the bosses were not consecutive .
For the 1st question , I find it convenient to use a 6*8 determinant (?) to denote the no. of servants receiving 3 dollars in various combinations . In general if Boss A chooses ' a ' consecutive servants while Boss B chooses ' b ' consecutive servants , a general formula for no. of servants chosen simultaneously could be derived .
10 servants lined up a queue . Boss A chose randomly 5 consecutive
( neighbouring ) of them and paid each 1 dollar . Boss B chose randomly
3 consecutive of them and paid each 2 dollars . If a servant was chosen
randomly from the 10 , find
(1) The probability that he received 3 dollars .
(2) The expected amount he received .
Hi thickhead ,
For (1) should it be a < 1/3 ?
Hi thickhead ,
Will the property of conjugates also applies to triangles ?
For example , if the length of the corresponding side of
B is changed to 4 units , will the corresponding probability be double ?
or something like that ?
Related problem :
Let E denotes a triangle PQR with PQ = QR = 6 units and
angle Q = 90 degree . Inside E there are 2 similar triangles
A and B both being parallel with E and can move freely and
uniformly inside E but must keep parallel with E in moving .
The lengths of the corresponding sides of A and B are
3 units and 2 units respectively .
If a point is chosen randomly on E , find the probability that
the point also lies inside A and B at the same time .
This property also applies for variables with numerator
other than 1 . For example : P( 1/3 , 2/5 ) = 137/810 , while
P( 1/3 , 3/5 ) = 137/ 540 = 3/2 * 137/810 .
In general , for 0 ≤ a ≤ 1 and 0 ≤ b ≤ 1 ,
this formula may be true : P( 1-a , b) = (1-a / a) * P( a , b) .
Thus P(1-a,1-b) = [(1-a)(1-b)/ab] * P(a,b)
Hi thickhead ,
The property of conjugates can also be applied to problems
involving 2 variables . For example : P ( 1/3 , 1/4 ) = 29/ 288 ,
while P( 1/3 , 1- 1/4) = P( 1/3 , 3/4 ) = 29/ 96 = 3* 29/ 288 ;
similarly P( 1-1/3, 1/4) = P( 2/3, 1/4) = 29/144 = 2* 29/288 ;
and P( 1-1/3 , 1-1/4 ) = P( 2/3, 3/4 ) = 29/ 48 = 6* 29/288 .
Hi thickhead ,
It is quite interesting for your property of conjugates .
But it seems only valid if the original variables are
1/2 , 1/3 and 1/4 . Also if c is changed from 1/4 to
2/4 ( = 1/2) , P will not be double . ( I got P = 25/ 162 )
Can this property be generalized ?
Hi thickhead ,
We have found that for 3 variables , if a = 1/2 ,
b =1/3 and c = 1/4 , then P = 709/10368 by P(3)
directly .
If we treat a and b firstly with P(2) , we get an
intermediate value d = 23/108 , if we want to find
the value of corresponding k , perhaps we have to
solve the equation :
min(k,1/4) - [min ( 1-k, 3/4)^3 - max(0, 3/4- k)^3 ] / 9/4*(1-k)
= 709/10368
In fact I really don't know how to solve it !
So it seems this is not a feasible way to find P(3)
by P(2) indirectly .
Hi thickhead ,
Perhaps we may do it in another way .
As we got the intermediate product D , the common
portion of A and B being 1/3 unit , can we adjust
its length to a value k so that after substituted it to
the original formula P(2) with C being 1/2 un . we can
get the final correct result 1/4 .
Can we find the rule for such adjustment ?
Hi thickhead ,
Since I cannot derive a formula for P(3) readily , I have
tried to solve problem involving 3 moving segments with
P(2) step by step . For example , in problem with segment E
being 1 unit long and 3 moving segments A , B and C each being
1/2 unit , I found the expected length of common portion of A and B firstly .
The formula P(2) gives the common portion of A and B ( denoted
by D , should also be considered as a segment .) to be 1/3 un . Then
I substituted D and C into P(2) again , but I got a wrong answer other than
the correct answer .( the expected length of common portion of A and B and C
should be 1/4 un . and the corresponding probability = 1/4 .)
This may be because the segment D so obtained will no longer be
movable . It is fixed at certain position(s) in E . Thus P(2) should
not be applied again . Instead , another formula involving 1 movable
segment and 1 fixed segment with appointed position should be used .
Can you derive such a formula ?
Thanks thickhead ,
The formula should be valid . But unless it can be simplified ,
in practical use the lmn formula will be much faster .
Hi thickhead ,
For a = 1/4 , b = 1/3 and c = 1/2 , what will be the results for
both formulas ?
Hi thickhead ,
Please check your formula in #61 carefully .
Hi thickhead ,
Your formula in #85 seems not symmetric in x and y .
If P = 1/9 , I think there should be 4 answers for ( x, y) .
In fact the aim of this related problem is to find whether in the
problem involving 2 moving squares , 1 of them can be replaced
by a fixed square with same length and yields the same answer .
( of course the location of the fixed square cannot be random .)
Hi thickhead ,
I need time to analyse your formula for P(3) .
Hi thickhead ,
Why P is not symmetric in x and y ?
For x = y can the results be expressed in fractions ?
Should there be totally 4 answers ( or more ? ) for (x , y ) ?
We know that for (x , y ) = ( 0,0) , ( 1,0) , (0,1) or (1,1)
P will have smallest value while for (x,y) = (1/2,1/2)
P will have greatest value . ( This is the case when B is
located at the centre of E .) 1/9 should be the average value
of P .
Related problem :
Inside a square E with co-ordinates ( 0,0) , (2,0), (2,2) and
(0,2) there are 2 smaller squares A and B both with length of sides being 1 unit and parallel with E . A can move freely and uniformly inside E but keep parallel with E during moving ; while B stays fixed in E with co-ordinate of its south-west vertex being ( x ,y ) . A point is chosen randomly on E .
(1) Find the probability that the point lies inside A and B
at the same time .
(2) If the probability = 1/9 , solve x and y .
Hi thickhead ,
Once you have assumed a < b , then a = min (a,b) and
1-b = min ( 1-a,1-b ) ; thus formula (3) can be re-written
as : min(a,b)- min( 1-a,1-b) ^3 - max(0,1-a-b) ^3/ 3(1-a)(1-b) .
If you have assumed a > b , the formula will remain the
same since then b= min(a,b) and 1-a = min ( 1-a,1-b) .
Hi thickhead ,
The aim I re-wrote your original formula to the one
in # 47 is to eliminate the restriction of the order of
the original m and n ( i.e. K65 and K66 ) defined by
you . This is always the benefit of using functions of
min and max though it looks a bit too prolix . I think
there is no need to emphasize too much in this point .
Hi thickhead ,
Will you state more and give some examples ?