@Don,
I think i misunderstood the question again.. :)..
One of major things that i went wrong with was that for me the deck
consisted of 13*3=39 cards..( basically an assumption made based on
the way i understood the question)
Thanks for the explanation..
On Jan 23, 9:17 pm, Don wrote:
> Close.
Close. You actually have to be sure that all 6 cards dealt to the
players are unique.
For instance, if I get 3 points, if you don't require that all the
cards dealth in the game are unique, you would conclude that there is
a very small, but positive probability that I will win. In reality, 3
points
@Don and Sundi..
As Don pointed out, all we are looking for is:
sum of a1 > sum of a2
sum of a1 > sum of a3
Assumption:
1) The 2 cards picked for a particular player are unique.
2) Cards are numbered : 1,..., 12, 13.
Hence, the following code should give the answer for the a1's
probability to
@Don..
Yup, it seems I misread it ... :) .. Thanks
On Jan 23, 9:17 am, Don wrote:
> I think that you are misreading the problem. A1 wins if his sum is
> larger than A2's sum and larger than A3's sum. A1's sum doesn't have
> to be larger than A2+A3.
> Don
>
> On Jan 22, 5:18 pm, Lucifer wrote:
>
I think that you are misreading the problem. A1 wins if his sum is
larger than A2's sum and larger than A3's sum. A1's sum doesn't have
to be larger than A2+A3.
Don
On Jan 22, 5:18 pm, Lucifer wrote:
> @sundi..
>
> Lets put is this way..
>
> Probability of (a1 wins + a1 draws + a1 losses) = 1,
>
@sundi..
Lets put is this way..
Probability of (a1 wins + a1 draws + a1 losses) = 1,
Now, sample count a1 wins = 46298 ( using the above given code)
Hence, the probability (win) = 46298/474552 = .097561
[ @ Don - as i mentioned in my previous post that i had initially
missed a f
Hi Lucifer,
Have you checked the sum of probability of (a winning + b
winning + c winning + draw)==1 ?
On Jan 22, 2:38 pm, Lucifer wrote:
> @above
>
> editing mistake.. (btw the working code covers it)
> /*
> int j =*1*;
> for(int i = 0; i < 12 ; i+=2)
> {
> A[i] = A[i+1] = A[22-i]
@above
editing mistake.. (btw the working code covers it)
/*
int j =*1*;
for(int i = 0; i < 12 ; i+=2)
{
A[i] = A[i+1] = A[22-i] = A[21-i] = j;
++j;
}
*/
On Jan 22, 6:53 pm, Lucifer wrote:
> @Don..
>
> Well i will explain the approach that i took to arrive at the
> probability..
> Well ye
@Don..
Well i will explain the approach that i took to arrive at the
probability..
Well yes u are correct in saying that it doesn't make a lot of sense
but then the no. of wins by a1 keeping in mind that a1 > a2 + a3 is
much less than a1 <= a2 + a3..
Or may be I have gone wrong in calculating the
You are saying that a1 wins roughly 1 in 20 times? How does that make
any sence?
Don
On Jan 19, 2:35 pm, Lucifer wrote:
> @correction:
>
> Probalilty (a1 wins) = 24575/474552 = .051786
>
> On Jan 20, 1:30 am, Lucifer wrote:
>
>
>
> > hoping that the cards are numbered 1,2,3,,13..
>
> > Proba
@correction:
Probalilty (a1 wins) = 24575/474552 = .051786
On Jan 20, 1:30 am, Lucifer wrote:
> hoping that the cards are numbered 1,2,3,,13..
>
> Probalilty (a1 wins) = 21723/474552 = .045776
>
> On Jan 20, 12:47 am, Don wrote:
>
>
>
>
>
>
>
> > P= 8800/28561 ~= 0.308112461...
>
> > On Ja
hoping that the cards are numbered 1,2,3,,13..
Probalilty (a1 wins) = 21723/474552 = .045776
On Jan 20, 12:47 am, Don wrote:
> P= 8800/28561 ~= 0.308112461...
>
> On Jan 18, 7:40 pm, Sundi wrote:
>
>
>
>
>
>
>
> > there are 52 cards.. there are 3 players a1,a2,a3 each player is given
> > 2
P= 8800/28561 ~= 0.308112461...
On Jan 18, 7:40 pm, Sundi 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 t
@Sunny: The probability of a1 being the winner is not 1/3 because of
ties. I.e., if a1 = a2 > a3, then a1 and a2 are tied and there is no
winner. What we can say with no calculations is that P(a1 winning) =
(1 - P(no winner)) / 3.
Dave
On Jan 18, 10:52 pm, sunny agrawal wrote:
> isn't the answer
:)...
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+...
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