Probability ---- consecutive individuals

mr.wong · 2016-08-13 16:32:30

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 .

thickhead · 2016-08-13 21:51:21

thickhead · 2016-08-14 02:40:07

bobbym · 2016-08-14 06:28:09

Hi mr.wong;

thickhead · 2016-08-14 15:28:50

mr.wong · 2016-08-14 16:41:02

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 .

bobbym · 2016-08-14 16:49:15

Hi;

You mean a 6 x 8 matrix?

mr.wong · 2016-08-15 17:02:57

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 .

thickhead · 2016-08-15 18:22:01

I would like to show the number of combinations in which the servant in the queue is present like below.
(1,2,3,4,5,5,4,3,2,1) for boss A.
(1,2,3,3,3,3,3,3,2,1) for boss B. They are also relative probabilities.The probability of winning both tips is proportional to

Summing all and dividing by (6*8*10) gives the probability. If third boss C appears and distibutes tips to 4 consecutive servants we can show it by
(1,2,3,4,4,4,4,3,2,1) and probability can be found by multiplication of all 3.

Probability of getting all 3 tips is found by the sum of the terms divided by 6*8*7*10

Hi mr. wong,
You might have noticed that conjugate concept works here also.

Last edited by thickhead (2016-08-15 18:22:56)

thickhead · 2016-08-15 20:25:03

We can extend the concept to n servants (n is even)with 3 bosses giving prizes to consecutive na,nb and nc servants.a<b<c<1/2. If we prepare the matrix for the number of selections each servant is involved we get a similar pattern. For A the number goes on increasing from 1 to na , then becomes constant at na till we reach halfway.we can leave the further portion as it is symmetrical. For B it increases from 1 to nb then remains constant at nb ,similarly for C upto nc. Beyond nc it remains constant for all. the probability can be calculated as follows.

Last edited by thickhead (2016-08-15 20:57:57)

mr.wong · 2016-08-17 17:00:37

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 ) .

.

Last edited by mr.wong (2016-08-18 14:53:12)

thickhead · 2016-08-18 16:54:33

Did not understand min ( 1,10 , k , 11-k ) 1 is always minimum Isn't it? Or is 1..10 the range?

mr.wong · 2016-08-19 16:55:42

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 .

mr.wong · 2016-08-23 21:33:10

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 ) ].

mr.wong · 2016-08-24 22:51:01

In general , let n be the total no. of servants being an
even no ., sum of the corresponding products will be
2* ∑[ min ( ab , ar , br , r^2 ) ] or 2* ∑[ min ( ab , min (a,b) r , r^2 ) ]
for r from 1 to n/2 .
Then the required probability will be
2* ∑[ min ( ab , min (a,b) r , r^2 ) ] / n * (n-a+1) *( n-b +1)
However , the value of min ( ab , ar , br , r^2 ) still have to be
calculated 1 by 1 for each value of r .

For n to be an odd no. ,say 11 , the value in k-th row and r-th
column is still min ( k , n+1-k , r , n+1-r ) , i.e. min ( k , 12-k , r , 12-r ) . For k = n+1-k = r = n+1-r , the expression will = n+1/ 2 being the unique greatest value in the table .
In this case the whole table will be symmetric vertically for k from 1 to [ n/2 ] ( the greatest integer contained in n/2 ) ( which = n-1 /2 ) with k from n+1 - [ n/2 ] ( = [ n/2 ] + 2 ) to n . For k = [ n/2 ] + 1 it will be unique .
The table will also be symmetric horizontally for r from 1 to [ n/2] ( or
n-1 /2 ) with r from [ n/2 ]+2 to n . For r = [ n/2 ] + 1 it will be unique .
Then the sum of product of min ( a, n+1-a , r , n+1-r ) and
min ( b , n+1-b , r ,n+1-r ) , may be simplified to ∑ min ( a, r ,n+1-r ) * min ( b , r ,n+1-r ) . where r varies from 1 to n .
It may also be simplified to 2 * {∑ min ( a , r ) * min ( b , r) } + a*b ( where r varies from 1 to n-1 /2 .) The term min ( a , r ) * min ( b , r) may also be expressed as
min ( ab , ar , br , r^2 )

Last edited by mr.wong (2016-08-25 15:55:06)

mr.wong · 2016-08-26 16:16:22

Thus if n is even :
Sum of corresponding products = 2* {∑ min ( ab , min (a,b) r , r^2 ) }
for r from 1 to n/2 .

If n is odd :
Sum of corresponding products = 2* {∑ min ( ab , min (a,b) r , r^2 ) } + ab
for r from 1 to [ n/2] .

No matter n is even or odd , a general formula for the sum may be expressed as :
2* {∑ min ( ab , min (a,b) r , r^2 ) } + 2 { n/2 - [ n/2] } * ab
for r from 1 to [ n/2] .
as the term 2 { n/2 - [ n/2] } will = 0 if n is even , and = 1 if n is odd since then n/2 = [ n/2] + 1/2 .

Divided the sum by ( n-a+1 ) * (n-b+1 ) * n , the required probability will be obtained

mr.wong · 2016-08-27 22:40:27

If a graph of y = min { ab , min (a , b) r , r^2 } is drawn it can be found
that the curve can be divided into 3 parts .
The 1st part will be for r from 1 to min (a , b) where y = r^2 since when
r = min (a,b) , y = min { min (a,b) * max(a,b) , min (a,b) min(a,b) ,min(a,b)^2 }
= r^2 = 1^2 + 2^2 .....+ min(a,b)^2 . ( totally min(a,b) terms)
( Is there any formula to calculate the sum of squares ? )

The 2nd part will be for r from min(a,b) + 1 to max (a,b) where
y = min(a,b) * r since when r = max(a,b) ,
y = min { min(a,b) * max(a,b) , min(a,b) * max(a,b) , max(a,b) ^2 } =
min(a,b) * r = min(a,b)*[ min(a,b) +1] + ... min(a,b)* max(a,b)
(totally max(a,b)- min(a,b) terms )
Thus the sum of the 2nd part will be
min(a,b)* (a+b+1)/2 * {max(a,b)- min(a,b)}

The 3rd part will be for r from max(a,b) +1 to [ n/2] where
y= ab ( being a constant ) and there are totally [ n/2] -max(a,b)
terms . Thus the sum of the 3rd part will be
ab * {[ n/2] -max(a,b) }

The above 3 parts will all exist only if max(a,b) +1 ≤ [ n/2] ,
if max(a,b) ≥[ n/2] ≥ min(a,b) , then only part 1 and part 2 will exist . If [ n/2] ≤ min(a,b) , then only part 1 will exist .

For example , let n = 11 , a = 5 and b = 3 ; then the total sum
= 2 * { 1+4+9 + 27 + 0 } + 15
= 2 * 41 + 15
= 97
Since ( n-a+1 ) * (n-b+1 ) * n = 7 * 9 * 11 ,
therefore P = 97/ 693 ( about 0.14 )

Last edited by mr.wong (2016-09-02 20:44:11)

mr.wong · 2016-08-29 22:25:10

( continued from # 8 )

The value of the term in k-th row and r-th column of the table in
# 8 can be expressed as min [ 3 , max ( 0 , r-k+3 ) , max ( 0, k-r+5)].
= min { 3 , max [ 0 , min ( r-k+3 , k-r+5 ) ] }
where k varies from 1 to 6 and r varies from 1 to 8 .

Let us turn the table shown in # 8 anti-clockwisely by 45 ° with
the point ( 1, 8 ) at the top and the point (6, 1) at the bottom ,
then we separate the table into 3 parts . ( This is only true for
min (a,b) , i.e 3 to be odd . )
The 1st part is the triangular shape with ( 1,8 ) as vertex and the
base will be the line joining (1,3) and (6,8) .
The 2nd part consists of only 1 row by joining (1,2) and ( 6,7) .
( For min (a,b) to be an even no . , the 2nd part will be void . )
The 3rd part is the inverse triangle with the base formed by
joining (1,1) and (6,6) while with (6,1) as vertex . The items
inside are exactly the same with the 1st part .
There are min ( 10-5+1 , 10-3+1 ) = 6 items in each side of the
upper triangle . ( also for the lower inverse one .) The base ( lowest
row ) are filled with the value 3 and there are 6 such 3 ' s .
The upper row are all 2 and there are 5 such 2's . The values are
descending by 1 for each upper row , until the row are all 1's and
there are 12 - 5 - 3 = 4 of them . ( The other rows filled with all
0 's will be neglected . )
Thus the sum of the all values in each triangle will be
( 3 * 6 + 2 * 5 + 1 * 4 ) = 32 , and 2 * 32 = 64 for the 2 triangles , together with the row between them with sum = 3 * 6 = 18 ,
the total sum in the table = 82 .

Divided 82 by 10 * 6 * 8 , we get P = 41/240 .

mr.wong · 2016-09-03 23:01:20

In general , let n be the total no. of servants while a and b be the nos of servants chosen by Boss A and Boss B respectively . A table with n-a+1 rows and n-b+1 columns in the form like the one in # 8 may be constructed with the value of the term in k-th row and r-th column of the table expressed as
min { a , b , max [ 0 , min ( r-k+min (a,b) , k-r+max(a,b) ] }
where k varies from 1 to n-a+1 and r varies from 1 to n-b+1 .
( This is for the case that a ≥b , the result will be the same for
a ≤b just by exchanging the rows with columns .)

The table may be divided into 3 portions , the 1st portion will be the
parallelogram in the middle , from r = k to r = k + max(a, b ) - min( a, b)
which are all filled with the value min (a , b) .

For r = k , value of each term in the corresponding inclined " diagonal " will be fixed to be min { a , b , max[ 0 , min ( min (a,b) , max (a,b)) ] } = min (a ,b ) and there are [n - max(a,b) +1 ] terms .

For r = k + max(a, b ) - min( a, b) ,
each term will be min { a, b, max [ 0 , min ( max(a,b) , min(a,b) )]}
also = min (a b) and there are also [n - max(a,b) +1 ] terms .
Thus there are totally [ max(a, b ) - min( a, b) + 1 ] *[ n - max(a,b) +1]
such terms with value min(a,b) .

The remaining portion of the table consists of 2 identical triangles
with base all with value [ min(a,b) - 1 ], and there are [ n- max(a,b) ] terms .
Both values will be descending towards the vertex of the triangle .
It will be stopped at either one value = 1 .

Thus the sum of the total values in the 1st portion ( the parallelogram )
will be min (a,b) * [ max(a, b ) - min( a, b) + 1 ] * [ n - max(a,b) +1] .

While the sum of values in the 2 triangles will be
2 * { [ min(a,b) - 1 ] * [ n- max(a,b) ]
+ [ min(a,b) - 2 ] * [ n- max(a,b) - 1]

+ .........................* ......................... } ( stop when either 1 side = 1 )

Thus the total sum of values in the whole table will be the sum of
the 2 portions .
Divided this value by n * (n-a+1) * (n-b+1 ) , the require probability
will be obtained .

mr.wong · 2016-09-04 17:20:36

Hi thickhead ,

It seems that the rule of conjugates really holds when the value a
is replaced by n-a+1 , then the corresponding P will be multiplied by
n-a+1 / a .

mr.wong · 2016-09-06 16:21:43

It can be observed ( from limited no. of examples ) that
(1) If n is even , and a = b = n/2 , then P = n+1 / 3 (n+1) + 3 ;
(2) If n is odd , and a = ( n-1 ) / 2 , while b = ( n+1 ) / 2 , ( a and b being conjugate ) , then P = n-1 / 3 ( n-1 ) + 3 = n-1 / 3n .

For example , let n = 20 , a = b = 10 , then P = 21 / 63 + 3 = 21 / 66 = 7/ 22 .
If the answer is correct , then by multiplied 7/ 22 by n*(n-a+1)* (n-b+1)
= 7 / 22 * 20 * 11 * 11
= 770 which should = 10 * 1 * 11 +
2 * { 9 * 10 + 8 * 9 + 7 * 8 ...... + 1 * 2 } ,
i.e. 9 * 10 + 8 * 9 + 7 * 8 ...... + 1 * 2 = ( 770 - 110 ) / 2 = 330 .

Thus the sum of a series of x * (x + 1 ) ( and then x^2 ) can be found in this way without adding the products 1 by 1 .

mr.wong · 2016-09-07 20:50:13

If n is even and a = b = n/2 , from # 9 it can be found that
P = [1^2 + 2^ 2 + ...( n/2 )^2] * 2 / [ n *( n/2 +1 ) * ( n/2 +1 ) ]

In fact there is already a formula to calculate the sum of
consecutive squares . ( Square pyramid numbers ) which states that :
1^2 + 2^2 + ... + n^2 = n*(n+1)*(2*n+1)/6.

Thus P = n/2 * ( n/2 +1 ) * ( n + 1) / [3 * n *( n/2 +1 ) * ( n/2 +1 ) ]
= ( n+1) / 3 * 2 * ( n/2 +1 )
= ( n+1) / 3 ( n + 2 ) = ( n+1) / 3 ( n + 1) + 3
as in # 21 .

mr.wong · 2016-09-10 16:54:13

Related Problem (1)
Let n = 100 , a = 50 , b = 30 , find P .

Solution :
(1st method ) : ( according to # 8 and # 19 )
Let a ∩ b denotes the total no. of times all the servants receiving money from both Boss A and Boss B at the same time .
Then a ∩ b consists of 2 parts :

(Part 1) min ( a,b) * [ max(a, b ) - min( a, b) + 1 ] * [ n - max(a,b) +1]
= 30 * 21 * 51 = 32130

(Part 2) 2 * { 29 * 50 = 29 * ( 29 + 21 ) = 29 ^2 + 29 * 21
+ 28 * 49 = 28 * ( 28 + 21) = 28 ^2 + 28 * 21
+ 27 * 48 = 27 * ( 27 + 21) = 27 ^2 + 27 * 21
+ ................= ...........................= ...........................
+ 1 * 22 = 1 * ( 1 + 21) = 1 ^2 + 1 * 21 }

Since 29^2 + 28^2 + 27^2 + .............. + 1^2
= 29 * 30 * 59 / 6 = 8555 .
while 29* 21 + 28* 21 + 27* 21 + ............+ 1 * 21
= 30 * 21 * 29 / 2 = 9135 .

Thus sum of Part 2 = 2 * { 8555 + 9135 } = 2 * 17690 = 35380

Thus a ∩ b = 32130 + 35380 = 67510

So P = 67510 / 100 * 51 * 71
= 6751 / 36210 ( about 0.18 )
quite near to 41 / 240 ( about 0.17 )

( 2nd method ) ( according to # 9 , # 11 and # 13 )
Since the value in k-th row and r-th column in the table may be represented by min ( k , n-k+1 , r , n-r+1 ) where r varies from 1 to n .

Thus a ∩ b = ∑ { min ( 50 , 51 , r , 101-r )
* min ( 30 , 71 , r , 101-r ) }
= ∑ { min ( r , 101-r )
* min ( 30 , r , 101-r ) }
where r varies from 1 to 100 .

(1) For r ≤ 30 ,sum of the products will be ∑ r * r =∑ r^2 .
(2) For r from 31 to 50 , sum of the products will be
∑ r * 30
(3) For r from 51 to 100 (i.e. 101 - r will be from 50 to 1), sum of the products will be ∑ (101 - r) * min ( 30, 101-r )
= ∑ min [ (101 - r) * 30 , (101 - r) ^2 ]

For (1) ∑ r^2 = 1^2 + 2^2 + ... + 30^2
= 30* 31* 61 / 6 = 9455
For (2) ∑ r * 30 = 31 * 30 + 32*30 + ...+ 50*30
= 81 * 30 * 20 / 2
= 24300

For (3) , 30 = 101-r ⇒ r = 71 ,
( i ) for r from 51 to 70 ,( i.e. 101-r from 50 to 31 )
min (30, 101-r ) = 30 ,
Sum of this portion will be ∑ [ (101 - r) * 30 ]
= 50*30 + 49*30 + .... + 31*30 = 81 * 30 * 20 / 2
= 24300

( ii ) For r from 71 to 100 ,( i.e. 101-r from 30 to 1 ) ,
min ( 30, 101-r ) = 101 - r ,
Sum of this portion will be ∑ (101 - r) ^2
= 30^2 + 29^2 + ... + 1^2
= 30 * 31 * 61 / 6 = 9455

Thus sum of ( 3 ) = 24300 + 9455 = 33755

Thus a ∩ b = 9455 + 24300 + 24300 + 9455
= 67510

So P = 67510 / 100 * 51 * 71
= 6751 / 36210 ( same as 1st method )

mr.wong · 2016-09-12 20:00:39

Math Is Fun Forum

#1 2016-08-13 16:32:30

Probability ---- consecutive individuals

#2 2016-08-13 21:51:21

Re: Probability ---- consecutive individuals

#3 2016-08-14 02:40:07

Re: Probability ---- consecutive individuals

#4 2016-08-14 06:28:09

Re: Probability ---- consecutive individuals

#5 2016-08-14 15:28:50

Re: Probability ---- consecutive individuals

#6 2016-08-14 16:41:02

Re: Probability ---- consecutive individuals

#7 2016-08-14 16:49:15

Re: Probability ---- consecutive individuals

#8 2016-08-15 17:02:57

Re: Probability ---- consecutive individuals

#9 2016-08-15 18:22:01

Re: Probability ---- consecutive individuals

#10 2016-08-15 20:25:03

Re: Probability ---- consecutive individuals

#11 2016-08-17 17:00:37

Re: Probability ---- consecutive individuals

#12 2016-08-18 16:54:33

Re: Probability ---- consecutive individuals

#13 2016-08-19 16:55:42

Re: Probability ---- consecutive individuals

#14 2016-08-23 21:33:10

Re: Probability ---- consecutive individuals

#15 2016-08-24 22:51:01

Re: Probability ---- consecutive individuals

#16 2016-08-26 16:16:22

Re: Probability ---- consecutive individuals

#17 2016-08-27 22:40:27

Re: Probability ---- consecutive individuals

#18 2016-08-29 22:25:10

Re: Probability ---- consecutive individuals

#19 2016-09-03 23:01:20

Re: Probability ---- consecutive individuals

#20 2016-09-04 17:20:36

Re: Probability ---- consecutive individuals

#21 2016-09-06 16:21:43

Re: Probability ---- consecutive individuals

#22 2016-09-07 20:50:13

Re: Probability ---- consecutive individuals

#23 2016-09-10 16:54:13

Re: Probability ---- consecutive individuals

#24 2016-09-12 20:00:39

Re: Probability ---- consecutive individuals

#25 2016-10-14 16:40:11

Re: Probability ---- consecutive individuals

Board footer