Probability ---- consecutive individuals

bobbym · 2016-10-16 16:13:27

Hi;

mr.wong · 2016-10-16 21:52:53

Hi bobbym ,

Your results will be accepted . I wonder why you
provided 2 approximate answers instead of 2 more exact
ones as
(1) 1681/57600
(2)$ 0.43

bobbym · 2016-10-16 22:14:50

Hi;

We work in a specific way. First comes a simulation always. With my hardware I can get 2 or maybe 3 accurate digits after the decimal. Of course, this is only a good estimate. If there is a reason to I will seek an exact answer but if I fail I have the simulation answer standing by. Often, this will be enough and is much, much better than no answer at all.

It is sort of a sin, to have no answer, not even a guess.

http://www.mathisfunforum.com/viewtopic … 08#p302808

mr.wong · 2016-10-18 21:53:01

Related Problem (4)

100 soldiers formed a 10 *10 matrix . After a battle
those soldiers at one side of a diagonal ( not inclusive )
were all killed , thus remaining a triangular semi-matrix
enclosed by 10 soldiers at each side .
General A chose randomly from it a similar shape
triangular semi-matrix enclosed by 5 soldiers at each
side and gave each soldier inside 1 dollar .
General B did the same thing . ( The 2 semi - matrices
must be parallel with the big one .)
If a soldier was chosen randomly from the whole ,
find the probability that he received 2 dollars .

thickhead · 2016-10-27 03:34:34

mr.wong wrote:

Related Problem (3)
100 soldiers formed a 10 * 10 matrix . General A chose randomly a 5*5 sub-matrix contained in it
and gave each soldier inside 1 dollar . General B chose randomly a 3*3 sub-matrix contained in it
and gave each soldier inside 2 dollars . ( The 2 sub-matrices must be parallel with the big one . )
If a soldier was chosen randomly from the whole , find
(1) The probability that he received 3 dollars .
(2) The expected amount he received .

Hi mr.wong,
I tried to put it on excel.Each cell consists of submatrices through the point
5 point submatrix
1
2 2
3 4 3
4 6 6 4
5 8 9 8 5
5 10 12 12 10 5
4 8 12 13 12 8 4
3 6 9 12 12 9 6 3
2 4 6 8 10 8 6 4 2
1 2 3 4 5 5 4 3 2 1 sum=325
the probability of getting 1 dollar=325/55/21=0.281385281

3 point submatrices
1
2 2
3 4 3
3 5 5 3
3 5 6 5 3
3 5 6 6 5 3
3 5 6 6 6 5 3
3 5 6 6 6 6 5 3
2 4 5 5 5 5 5 4 2
1 2 3 3 3 3 3 3 2 1 sum=216
The probability of getting 2 dollars =216/55/36=0.109090909

product of the 2 matrices
1
4 4
9 16 9
12 30 30 12
15 40 54 40 15
15 50 72 72 50 15
12 40 72 78 72 40 12
9 30 54 72 72 54 30 9
4 16 30 40 50 40 30 16 4
1 4 9 12 15 15 12 9 4 1 sum=1533
So the probability of getting 3 dollars =1533/55/21/36=0.036868687
Expectation=0.281385281*1+0.109090909*2=0.4995671
Filling the cells is a difficult task and I hope it is without errors.

Last edited by thickhead (2016-10-27 16:16:25)

bobbym · 2016-10-27 04:21:09

Are you sure you are working on Problem 3?

thickhead wrote:

mr.wong wrote:
Related Problem (3)
100 soldiers formed a 10 * 10 matrix . General A chose randomly a 5*5 sub-matrix contained in it
and gave each soldier inside 1 dollar . General B chose randomly a 3*3 sub-matrix contained in it
and gave each soldier inside 2 dollars . ( The 2 sub-matrices must be parallel with the big one . )
If a soldier was chosen randomly from the whole , find
(1) The probability that he received 3 dollars .
(2) The expected amount he received .

thickhead · 2016-10-27 04:46:42

Sorry. I exchanged the dollars. Now it stands corrected.

Last edited by thickhead (2016-10-27 04:47:11)

mr.wong · 2016-10-27 15:42:06

Hi thickhead ,

I think you have mixed Related Problem (3) with Related Problem (4) .
But your answer may be for a Related Problem (5) !

thickhead · 2016-10-27 16:25:09

No. I am referring to problem(3). I have not come upto problem (4) and (5)
There seems to be mistake in my first table.
Average expectation for 1 dollar =15/55=0.272727273 whereas I got 0.281385281. I should have got 315 as total in my first table.My second table is correct.
Average expectation overall =(15+12)/55=0.490909091
as captain A distributed 15 dollars and captain B 12 dollars.

bobbym · 2016-10-27 17:16:28

There is no problem 5...

mr.wong · 2016-10-28 01:00:56

Hi thickhead ,

In Problem (3) the 2 small sub- matrices are squares of 5*5 and 3*3
respectively .
While in Problem (4) the 2 small semi - matrices are triangular , both
with 5 soldiers at each side .
Your answer may fit a proposed Problem (5) with 2 small triangular
semi-matrices with 5 and 3 soldiers at each side respectively .

Hi bobbym ,

The Problem (5) is still under proposal .

thickhead · 2016-10-28 02:18:23

Sorry,mr.wong,
I meant problem no.(4) but I don't know how I quoted your problem (3)
Evidently the triangular matrix I gave points to problem (4) Why do you think it is no.(5)?.
I fixed the error in my layout.The maximum I could get with logic was this.
1
2 2
3 4 3
4 6 6 4
5 8 9 8 5
5 9 11 11 9 5
4 8 11 13 11 8 4
3 6 9 11 11 9 6 3
2 4 6 8 9 8 6 4 2
1 2 3 4 5 5 4 3 2 1
But total is 316, one more than the correct one.Then I simulated 21 submatices in excel sheet at proper places.and superposed all.
1
2 2
3 4 3
4 6 6 4
5 8 9 8 5
5 9 11 11 9 5
4 8 11 12 11 8 4
3 6 9 11 11 9 6 3
2 4 6 8 9 8 6 4 2
1 2 3 4 5 5 4 3 2 1
This give s a total of 315.the product of 2 sub matrices is
1
4 4
9 16 9
12 30 30 12
15 40 54 40 15
15 45 66 66 45 15
12 40 66 72 66 40 12
9 30 54 66 66 54 30 9
4 16 30 40 45 40 30 16 4
1 4 9 12 15 15 12 9 4 1
with a total of 1476.
the probability of getting 3 dollars is 1476/55/21/36=0.035497835

thickhead · 2016-10-28 04:29:39

simulation for prob no.(3) yields
Captain A

1	2	3	4	5	5	4	3	2	1
2	4	6	8	10	10	8	6	4	2
3	6	9	12	15	15	12	9	6	3
4	8	12	16	20	20	16	12	8	4
5	10	15	20	25	25	20	15	10	5
5	10	15	20	25	25	20	15	10	5
4	8	12	16	20	20	16	12	8	4
3	6	9	12	15	15	12	9	6	3
2	4	6	8	10	10	8	6	4	2
1	2	3	4	5	5	4	3	2	1

Captain B

1    2    3    3    3    3    3    3    2    1
2    4    6    6    6    6    6    6    4    2
3    6    9    9    9    9    9    9    6    3
3    6    9    9    9    9    9    9    6    3
3    6    9    9    9    9    9    9    6    3
3    6    9    9    9    9    9    9    6    3
3    6    9    9    9    9    9    9    6    3
3    6    9    9    9    9    9    9    6    3
2    4    6    6    6    6    6    6    4    2
1    2    3    3    3    3    3    3    2    1

Combined

1    4    9    12    15    15    12    9    4    1
4    16    36    48    60    60    48    36    16    4
9    36    81    108    135    135    108    81    36    9
12    48    108    144    180    180    144    108    48    12
15    60    135    180    225    225    180    135    60    15
15    60    135    180    225    225    180    135    60    15
12    48    108    144    180    180    144    108    48    12
9    36    81    108    135    135    108    81    36    9
4    16    36    48    60    60    48    36    16    4
1    4    9    12    15    15    12    9    4    1

Cell total =6724
mr.wong's results perfectly match.
maximum probability of getting 3 dollars=225/576

Last edited by thickhead (2016-10-28 05:07:38)

mr.wong · 2016-10-28 16:44:04

Hi thickhead ,

Thanks much for your laborious work !
I shall reserve your result in # 37 for the proposed Problem (5) .

thickhead · 2016-10-28 18:12:10

Hi mr.wong,
O.K. I understood the problem (4) now.

Last edited by thickhead (2016-10-28 20:33:30)

thickhead · 2016-10-28 20:44:57

Problem No.4
1
4 4
9 16 9
16 36 36 16
25 64 81 64 25
25 81 121 121 81 25
16 64 121 144 121 64 16
9 36 81 121 121 81 36 9
4 16 36 64 81 64 36 16 4
1 4 9 16 25 25 16 9 4 1
Total=2331
probability of getting 2 dollars=2331/(55*21*21)=111/1155=0.096103896

Last edited by thickhead (2016-10-29 00:23:34)

mr.wong · 2016-10-28 23:02:21

Hi thickhead ,

Have you omitted a " 25 " at the 5th row , a " 25" at the
6th row and a " 16 " at the 7th row in the table at # 41 ?
Also for the probability should the denominator be 55*21*21 ?

thickhead · 2016-10-29 00:25:35

I don't know how I missed them but the total is not affected. As for 36 I was still in the mood of 3x3 submatrix. Now it stands corrected.

mr.wong · 2016-10-29 16:05:33

Thanks thickhead ,

Now it seems if the no. of points of the big triangle with the 2 half - sized
small triangles trend to ∞ , then the probability will trend to 1/10 .

mr.wong · 2016-10-31 20:20:39

It may be ambiguous to define what is half - sized for a
matrix ( no matter a square matrix or a triangular semi-matrix )
with n dots at each side . ( Assuming that the distance between
the dots are fixed and the dimension of the dots may be neglected . )
For n to be even , readily half - sized means the one with n / 2 dots .
But for n to be odd , quite probably we shall mean a sub - matrix
with [ n/2 ] + 1 = ( n + 1 ) / 2 dots to be half - sized so that its length
between the end points will be 1/2 of that of the big one .
For example for a matrix with 3 dots at each side , a half - sized
sub - matrix will be referred to one with 2 dots at each side .

Math Is Fun Forum

#26 2016-10-16 16:13:27

Re: Probability ---- consecutive individuals

#27 2016-10-16 21:52:53

Re: Probability ---- consecutive individuals

#28 2016-10-16 22:14:50

Re: Probability ---- consecutive individuals

#29 2016-10-18 21:53:01

Re: Probability ---- consecutive individuals

#30 2016-10-27 03:34:34

Re: Probability ---- consecutive individuals

#31 2016-10-27 04:21:09

Re: Probability ---- consecutive individuals

#32 2016-10-27 04:46:42

Re: Probability ---- consecutive individuals

#33 2016-10-27 15:42:06

Re: Probability ---- consecutive individuals

#34 2016-10-27 16:25:09

Re: Probability ---- consecutive individuals

#35 2016-10-27 17:16:28

Re: Probability ---- consecutive individuals

#36 2016-10-28 01:00:56

Re: Probability ---- consecutive individuals

#37 2016-10-28 02:18:23

Re: Probability ---- consecutive individuals

#38 2016-10-28 04:29:39

Re: Probability ---- consecutive individuals

#39 2016-10-28 16:44:04

Re: Probability ---- consecutive individuals

#40 2016-10-28 18:12:10

Re: Probability ---- consecutive individuals

#41 2016-10-28 20:44:57

Re: Probability ---- consecutive individuals

#42 2016-10-28 23:02:21

Re: Probability ---- consecutive individuals

#43 2016-10-29 00:25:35

Re: Probability ---- consecutive individuals

#44 2016-10-29 16:05:33

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#45 2016-10-31 20:20:39

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