:)...
Lets say you have two players a and b
one card is distrbuted to each player
if the card with a is higher then a wins  else a loses.

probability of a winning:
total num of cards with combos where a wins a is the first player:
2-1,3-1...13-1
3-2.......13-2
..
13-12
this sum equals = (13+12+...1)*4, 4 suites are possible

therefore probability=sum/52C2.

I dont think it is half in this case.

similarly extending for 3 players with 2 cards each i dont think its
going to be 1/3 becoz of the condition given.

I might be wrong. Please correct me if so.


On Jan 19, 1:08 am, Prateek Jain <prateek10011...@gmail.com> wrote:
> ya obviously...it is 1/3.....no need of any data is required
>
> On 1/18/12, sunny agrawal <sunny816.i...@gmail.com> wrote:
>
>
>
>
>
>
>
> > isn't the answer will be 1/3, without any calculations :)
>
> > On Thu, Jan 19, 2012 at 7:10 AM, Sundi <sundi...@gmail.com> wrote:
>
> >> there are 52 cards.. there are 3 players a1,a2,a3 each player is given
> >> 2 cards each one of A=2...J=11,Q=12,K=13..a user wins if his sum of
> >> cards is greater then the other two players sum.
>
> >> find the probability of a1 being the winner?
>
> >> --
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> > --
> > Sunny Aggrawal
> > B.Tech. V year,CSI
> > Indian Institute Of Technology,Roorkee
>
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