Sunday, September 28, 2008

Federer v/s Roddick

Came across this question recently.

Part 1:
Earlier, the probability of Federer winning against Roddick was > 4/5.
Now it is < 4/5.

Does there necessarily have to be a point when it was exactly 4/5, during the transition from > 4/5 to < 4/5?

( The probability was determined on the past record of the players against each other )

Part 2:
Does your answer change if, earlier it was < 4/5, and later it was > 4/5, and you are asked the question:
Does there necessarily have to be a point when it was exactly 4/5, during the transition from < 4/5 to > 4/5?

7 comments:

sudeep said...

This is a question, right ?

:-)

I thought you were criticizing the media, because they never told you abt when the probability became 4/5.

Regarding the question, I guess it depends upon how many matches were played during the transition place. If the number of matches played were zero, I guess such a point never existed.

sarvagnaani said...

is this a trick question or what? (reminds me of calvin and his mom :D)

Guess the answer depends on the number of matches they have played against each other. Assume they had played x matches and fed won y of them with prob of fed wining, p(f) = y/x > 4/5 => y > .8x and x and y are such that both x and y are natural numbers. For example, y = 5 and x = 6 with p(f) = 5/6 (0.83 > 0.8).

Now, they have played some z new matches and fed won l (could be 0) of them, so the new p(f) = (y+l)/(x+z) < 4/5. Now to answer your question, i need more data, the exact value of x,z, y and l. Going back to my example, if they played only one more match and and fed lost the match, then p(f) = 5/7 (0.71) < 4/5 and there was no pt where p(f) = 4/5.

Ok, if your question is more general.. like .. are there natural numbers x,y,l,z such that x/y > 4/5 and (y+l)/(x+z) = 4/5 and l <= z?
hmm..i have to think this through :D

sarvagnaani said...

comment on my own comment :D
of course there are lots of natural numbers x,y,l,z such that x/y > 4/5 and (y+l)/(x+z) = 4/5 and l <= z?

for example
y=4,x=4, x/y=1,
they play one more match and fed does not win that..p(f) = 4/5
and one more match fed does not win again..p(f) = 4/6

now..what was your question please? :D

richie said...

Yup, it is a question :).

Haha ( criticizing the media :) ).

To rephrase the question (using similar terminology used by gnaani):


Therefore we know there exists, a (y,x) such that we have y/x > 4/5

( i.e there exists a point in time, where they have played x matches and Fed has won y of them. )


Now later, after some z new number of matches , of which Fed won l, we have:

(y+l)/((x+z) < 4/5

Now was there a point where it was exactly 4/5?

( So somewhere in the realm of the x to x+z matches, is there a point where Federer wins exactly 4/5 of them, or can u give a counterexample ).

Or can you just jump past that probability of 4/5?

More detail:
============
Instead of choosing 4/5, let's say the probability was instead 2/5.

So earlier y/x = 1/2 ( which is > 2/5 )

Now if Fed loses the next match:
it will be 1/3 ( which is < 2/5 ).

So there is an example which indicates that we can jump over 2/5.

Can u find a similar one for 4/5?

richie said...

To answer Sudeep's comment:

"Regarding the question, I guess it depends upon how many matches were played during the transition place. If the number of matches played were zero, I guess such a point never existed."

So the 2 facts that you are given:
a) there was a point in time, where it was > 4/
b) then it was < 4/5

So number of matches played during the transition phase has to be > 0 alva.

sarvagnaani said...

mama..yeradu example kottanalo..
ok..from my first comment..
y = 5 and x = 6 with p(f) = 5/6 (0.83 > 0.8).
if they played only one more match and and fed lost the match, then p(f) = 5/7 (0.71) < 4/5 and there was no pt where p(f) = 4/5.

example where it p(f) becomes exactly = 4/5 at some pt,
y=4,x=4, x/y=1 (> 0.8)
they play one more match and fed does not win that..p(f) = 4/5
and one more match fed does not win again..p(f) = 4/6

ok..now..for the last time..what was your question? :p

richie said...

Super leh :).

The true question togo eega :D.

Is the reverse possible too....i.e

earlier < 4/5, and later > 4/5
without there being a 4/5 in between.