Probability Problem

alan4cult

Right, well I've been working on this problem for a while and have come up with a few answers. Can anybody give me an answer to this?

Here it is:

caseyn2008

i came up with 1/2 considering you can basically do only diagonals ill look over it some more and see if there are more possibilitys i found some more lets see if i can figure this out
here is the beginning of a long process
http://img88.imageshack.us/img88/549/15431588tt3.png
light blue dark blue black brown green cyan red yellow purple i think thats all the colors in there then you got to add the 8 diagonals
http://img88.imageshack.us/img88/549/15431588tt3.png

caseyn2008

caseyn2008 wrote: »

i came up with 1/2 considering you can basically do only diagonals ill look over it some more and see if there are more possibilitys i found some more lets see if i can figure this out
here is the beginning of a long process
http://img88.imageshack.us/img88/549/15431588tt3.png
light blue dark blue black brown green cyan red yellow purple i think thats all the colors in there then you got to add the 8 diagonals
http://img88.imageshack.us/img88/549/15431588tt3.png

tell me if that helped or im just to drugged up from being disabled im not thinking sanely

alan4cult

Well, I'll tell you what I've got:

The first coin can be placed anywhere so it has a 16/16 chance of being placed somewhere.

If we let the other coins be placed in the same row or column then the second coin has a 6/15 chance of this.

The third coin then has a 5/14

So the probability of this not happening is 1 - (16/16)(6/15)(5/14)
= 1 - 30/210 = 1 - 1/7 = 6/7

This doesn't seem right but its a start I suppose.

LeixlipRed

Hmmm, I'm tired but perhaps work out the total number of combinations (without worryin about position of coins) then subtract number of outcomes we dont want. What an input from me

EDIT: Did I see this problem on the Maths problems board in Maynooth last week?

LeixlipRed

alan4cult wrote: »

Well, I'll tell you what I've got:

The first coin can be placed anywhere so it has a 16/16 chance of being placed somewhere.

If we let the other coins be placed in the same row or column then the second coin has a 6/15 chance of this.

The third coin then has a 5/14

So the probability of this not happening is 1 - (16/16)(6/15)(5/14)
= 1 - 30/210 = 1 - 1/7 = 6/7

This doesn't seem right but its a start I suppose.

Think there's a few holes in that. For example take coin 1 and coin 2 along the top left side. Then the probability of third coin being on same row or column is 8/14

FedFrank2

well im not fully sure on this but i got 0.1285.

16/16 for the first coin, then the 2nd coin has 9/15 (as 3 from the row its in and the column cant be used) and 3/14 for the 3rd coin (as another 3 each cant be used. place a coin in the diagram above and hopefully you'll see what im on about?)

LeixlipRed

FedFrank2 wrote: »

well im not fully sure on this but i got 0.1285.

16/16 for the first coin, then the 2nd coin has 9/15 (as 3 from the row its in and the column cant be used) and 3/14 for the 3rd coin (as another 3 each cant be used. place a coin in the diagram above and hopefully you'll see what im on about?)

Does the probability of the third coin not differ depending on where you place the 2nd coin?

FedFrank2

LeixlipRed wrote: »

Does the probability of the third coin not differ depending on where you place the 2nd coin?

maybe so, having looked at it again the 3rd coin is, i think, 6/14, which gives a prob of 0.2571

LeixlipRed

FedFrank2 wrote: »

maybe so, having looked at it again the 3rd coin is, i think, 6/14, which gives a prob of 0.2571

So depending on position of the 2nd coin we either have six places to put it or four

FedFrank2

LeixlipRed wrote: »

So depending on position of the 2nd coin we either have six places to put it or four

yes it does seem that way. can we take an average in this instance? or can the answer be a range of probabilities?

alan4cult

Sorry my solution ^^^ is flawed. I was actually working on a smaller square and I'm tired. I'll retry it later.

Thomas_S_Hunterson

Ok I'm getting 0.1714286 by calculating number of possible valid combinations over number of combinations.

Total combinations shold be 16*15*14*3! = 20160

You can break it into three parts based on the location of the first coin.
1. First coin is on a corner. (4 places)
2. First coin is on the edge and not a corner. (8 places)
3. First coin is in the middle. (4 places)

For all these cases, the second coin can be on any of 9 places, and then the third, it can be on any of four places.

Combining everything, this gives us:
4*9*4*3! + 8*9*4*3! + 4*9*4*3! = 3456

So our probability then is 3456/20160 = 6/35 = 0.17142857142857

/edit: Didn't need to split it up at all, silly me:rolleyes:. That's how I worked it out on paper though.

LeixlipRed

Ah I'm an idiot. It is always 4 places for the third coin.

alan4cult

I also got 6/35 but I did it by working out all the possible combinations. I did by giving each square a letter e.g. A,B,C... and then used a program to generate the combinations. I then went through them all and the answer was indeed 6/35.

That's solved then! Thanks all!

Crumbs

I have the same answer but there's some incorrect stuff mixed in with correct stuff in this thread so I thought that I would try to tidy it up.

P(First coin can go anywhere) = 16/16
P(Second coin not in same row or column) = 9/15
P(Third coin not in same rows or columns) = 4/14

(16/16)*(9/15)*(4/14) = 0.1714


Sean_K, that should be division by 3! when working out the combinations. But because you did the same thing on both sides of the probability, they cancelled themselves out anyway to give the same answer.

i.e. Total number of combinations = 16*15*14/3! = 560
Total valid combinations = 16*9*4/3! = 96
P(E) = 96/560 = 0.1714


LeixlipRed, when working out the problem another way the case can arise of different numbers of possibilities for the second and third coins.

LeixlipRed

Crumbs wrote: »

LeixlipRed, when working out the problem another way the case can arise of different numbers of possibilities for the second and third coins.

Yeh I was majorly out of it this morning. I was confusing working our the number of unfavourable outcomes with the favourables (ie, where its 4 available places for the 3rd coin).

ramanujan

i miss probability....

anyway i like this method: 4*4*3*2/16C3 = 6/35
(choose empty col)(choose empty row)(choose spot for coin in first row)(choose spot for coin in second row) / (choose 3 spaces out of 16)


generalising the above to choose m spaces out of a nxn box, (m<=n) with same condition:
( nC(m-n) )^2 * (m)! / nCm



where xCy stands for number of ways you can choose y elements out of x.