Puzzles - try this

1

Philip Calvert ·2009-01-03 07:16:01

........no one??
it may take time but
atleast give it a try

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11

Anirudh Narayanan ·2009-01-03 07:18:12

What do attacking positions mean? Sorry, I don't know anything about chess [7]

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1

Philip Calvert ·2009-01-03 07:19:37

if any two bishops are present on the same diagonal
they can attack each other

b4 posting you can check your ans for foll. data

n=8 k=6
5599888

5 2
240

4 4
260

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62

Lokesh Verma ·2009-01-11 11:08:28

I am not sure if a formula for this exists....
why? (I dont know!)

Well i could only say that the maximum no of bishops that cna be placed is 2n-1... that too may not be attainable!! (But this is the upper limit)

I for one could only place 2n-2 for any n.... (other than 1)

I will be eagerly waiting for rahul to post his solution...

I dont see that answer which he has given.. cos it is a fraction!!!

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1

Philip Calvert ·2009-01-12 04:52:30

this is definitely solvable and a combinatorical solution does exist though i don't know anything more than that and definitely Rahul's answer isn't correct. i have the feeling the answer would be more complicated
otherwise i don't see why this kind of a question can't come in JEE
looks quite within the syllabus doesn't it?
so plz "try this" everyone "try this"..

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62

Lokesh Verma ·2009-01-12 06:13:29

so where is k rahul..

ur approach is good..

but how do we generalise...

is it necessary for us to place the bishop at the boundary?

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1

rahul1993 Duggal ·2009-01-18 18:49:47

let us assume 3 cases

CASE 1

i have chosen the door with a prize. then the host can pick at any of the other 2 doors randomley

CASE 2

i have chosen a door with nothing in it. now the host will obviously pick the one without a prize

CASE 3

as there are 2 doors without a prize, i again pick a door without a prize, the host will pick the other one without a prize.

so i can conclude probability of me doing as done in CASE 2 and CASE 3 is higher than me doing as in CASE1 as there are 2 doors without a prize and 1 with a prize. so probability of CASE 1 for me is 1/3 whereas for CASE 2 and CASE 3 is 2/3 which is higher

thus i would switch the door to win the prize as in CASE 2 and CASE 3

let us assume the winning door to be B

as in case 1 if i chose door B with a probability of 1/3 and then the host opened door A and i changed my coice to door C with a probability of 1/2 only to lose. so losing probability is 1/6

similarly if initially i chose door B with a probability of 1/3 and the host opened door C and i changed my choice to door A with a probability of 1/2. again i would lo lose with a probability of 1/6

so total losing probability is 1/6+1/6=1/3

now as in case 2 if i chose door A with a probability of 2/3 and the host opened door C and i changed my choice to door B with a probability of 1/2, i would win with a probability of 1/3

similarly if initially i had chosen door C with a probability of 2/3 and after the host had opened door A i would change my choice to door B with a probility of 1/2, i would win with a probability of 1/3

so total chances of me winning with changing my choice would be 1/3+1/3=2/3 and losing would be 1/3.

thus i would prefer to change my choice after ythe host opened a losing door

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62

Lokesh Verma ·2009-01-18 21:41:46

posted in the wrong thread :)

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1

rahul1993 Duggal ·2009-01-18 23:12:29

oh sorry. actually i was deleting my previous 2 posts on this thread as they were not mine. Actually a person named abhijit had solved the question posted by me on the site goiit http://www.goiit.com/posts/list/algebra-probability-bishops-on-a-chessboard-80169.htm#388399