11500  Vampires
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11500  Vampires
Hello!
I Found the key to solve:
In http://en.wikipedia.org/wiki/Gambler%27s_ruin You can find the Both Formulas
The first one is to solve when AT = 3 (P2 = n1/(n1+n2))
The second one is to solve when AT != 3 (P1 = 1 (q/p)^n1... where q = (6AT)/AT ...
Good lock
I Found the key to solve:
In http://en.wikipedia.org/wiki/Gambler%27s_ruin You can find the Both Formulas
The first one is to solve when AT = 3 (P2 = n1/(n1+n2))
The second one is to solve when AT != 3 (P1 = 1 (q/p)^n1... where q = (6AT)/AT ...
Good lock

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Re: 11500  Vampires  tips
Can you please give me more details? What is the approach for solving this problem?
and please elaborate more on "Gambler ruin"
and please elaborate more on "Gambler ruin"
Sleep enough after death, it is the time to work.
Mostafa Saad
Mostafa Saad
Re: 11500  Vampires  tips
Hi Naffi and Mostafa.
You can use to D !=1 also. You must calcule n1 and n2:
n1 = EV1/D (rounded to up) page 1 from article
n2 = EV2/D (rounded to up)
for at = 3 use the formula
prob =
n1

n1 + n2
else
prob =
(1 (q/p)^n1 )

(1(q/ ***
where (q/p) = (6at)/at
the second prob formula including (***) can be found on link:
http://en.wikipedia.org/wiki/Gambler%27s_ruin ...
... in Unfair coin flipping
P1 = ...
... it is only to avoid to give here the complete answer,
good lock.
You can use to D !=1 also. You must calcule n1 and n2:
n1 = EV1/D (rounded to up) page 1 from article
n2 = EV2/D (rounded to up)
for at = 3 use the formula
prob =
n1

n1 + n2
else
prob =
(1 (q/p)^n1 )

(1(q/ ***
where (q/p) = (6at)/at
the second prob formula including (***) can be found on link:
http://en.wikipedia.org/wiki/Gambler%27s_ruin ...
... in Unfair coin flipping
P1 = ...
... it is only to avoid to give here the complete answer,
good lock.
Re: 11500  Vampires  tips
Thanks neilor for your kind elaboratin, I got AC.
But, I did not handle unfair coin flipping as you said, may be the test cases are fair.
But, I did not handle unfair coin flipping as you said, may be the test cases are fair.
Always At Your Help.

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 Joined: Wed Feb 01, 2006 12:59 pm
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Re: 11500  Vampires  tips
Hi, this is already how i did it, but last simple is not right.
Code: Select all
while(cin>>EV1>>EV2>>AT>>D && (EV1+EV2+D+AT))
{
clr(vis, 0);
cout.setf(ios::fixedios::showpoint );
cout.precision(1);
double n1 = EV1/D, n2 = EV2/D;
double p1, p2;
if(AT == 3) {
p1 = n1/(n1+n2);
p2 = n2/(n1+n2);
}
else { // won't work well if D != 1
double q = (6AT)/6.0, p = 1q;
p1 = (1.0pow(q/p, n1)) / (1.0pow(q/p, (n1+n2)));
p2 = 1p1;
}
cout<<p1*100<<" "<<p2*100<<"\n";
}
Sleep enough after death, it is the time to work.
Mostafa Saad
Mostafa Saad
Re: 11500  Vampires
My idea was to derive some equations, and then gaussian elimination.
Ami ekhono shopno dekhi...
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Re: 11500  Vampires
can anyone tell me what is the wrong with my code
&
what i hav to add in this code
(it has done in C)
&
what i hav to add in this code
(it has done in C)
Code: Select all
#include<stdio.h>
#include<math.h>
int main()
{
unsigned long ev1,ev2,at,d,n1,n2;
double pr,m;
while(scanf("%lu%lu%lu%lu",&ev1,&ev2,&at,&d)==4)
{
n1=ev1/d;
n2=ev2/d;
m=n1+n2;
if(at==3)
pr = (n1/m)*100;
else
{
double q = (6at)/6, p = 1q;
pr = (1pow(q/p, n1)) / (1pow(q/p, (n1+n2)));
}
printf("%.1lf\n",pr);
}
return 0;
}
Re: 11500  Vampires
You can also solve it without using the formula for the Gambler's Ruin... think memoization ;D
Re: 11500  Vampires
Can I solve this problem using Markov's chain?

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Re: 11500  Vampires
I simply assumed that after 1000 moves, the probability becomes so small that it is negligible. Got AC with O(20*20*1000)