ISI INTERVIEW PREP

[Mr. Nobody]

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[anurag]

by 13c2 n 13c3 are u nt selecting 3 n 2 diffrnt denominatns instead f card of same denominations? each denomination contqins 4 cards ryt??

[neha]

With ref. to
 Prove that there is a unique function f : R2 → R satisfying both
the properties below:
(i) f(x,y) is non-decreasing in x and non-decreasing in y;
(ii) f(x,x) = c for all x ∈ R, where c is a constant.

I was just thinking cant we take f(x,y)= 3 or something for all the x,y belonging to R? This function is also satisfying both the properties.

[deepak]

I believe Anurag is right.
The question here is to choose a '3 of a kind' and a '2 of a kind' (like a cool poker hand). So in a 'kind', there are 4 cards (ie, 4 queens, 4 jacks etc). Getting a 2 of a kind is possible in 4C2 ways and a 3 of a kind is possible in 4C3 ways. Also, there are a total of 13 'kinds'. So for the 1st 'set' (of either 2 cards or 3 cards) we will have 13 kinds to choose from and for the second 'set' we will have 12.
So Number of ways = 13*12*4C2*4C3

[anurag]

@neha - m confused  whether f(x,y) = 3 is a constant , or a wil it be a level curve (nd in case of level curve how wil u ensure non-decreasing in x n y) . pls help

[anurag]

@deepak- Ya exactly..
@ram- dont confuse denomination wid suite..

[anurag]

RE1 2011
Q.2a .wil d answer be (x-y) / n^2 , and 0 as n-> infinity

[deepak]

It's an equation of a plan. And looking at it trivially, I can't see why it doesn't satisfy the 2 properties, Neha. Perhaps this is an acceptable answer?

[Mr. Nobody]

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[neha]

@anurag:

I am nt getting this answer [(x-y) / n] -y,, and apparently as n approaches infinity,, it will be -y.

[neha]

I mean my answer is [(x-y) / n] -y

[Mr. Nobody]

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[deepak]

Ram, let's suppose f(x,y) = c. This does satisfy property 1. Now on to property 2. f(x,x) will also be 'c' right? The functional value is a constant irrespective of what values of x or y are considered. So this does satisfy what is asked?

[Mr. Nobody]

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[anurag]

@neha- won't it be like prob assigned to whether winning x or losing y d same i.e 1/n^2

[anurag]

sry i mean 1/n whether chosing same no. or different on 2nd attempt?

[anurag]

@DEEPAK - a straight line is both non-increasing as well as non-decreasing?

[Amit Goyal]

The proof of uniqueness is attached.
RE1.11.05sol.png

[anurag]

thnku sir

[deepak]

Neha and Anurag, P(a=b) is 1/n and P(a!=b) = 1 - 1/n and the pay offs are x and (-y) respectively, right?
So E = 1/n* (x) + (1-1/n)*(-y) = (x+y)/n - y.
n-> infinity will, like you say, result in a loss of y. But for finite numbers, i think the numerator of the first term is x+y.