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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

A general solution to:

http://www.mathisfunforum.com/viewtopic.php?id=13145

using the formula in #2027 from http://www.mathisfunforum.com/viewtopic … 30#p299130

```
#include <stdio.h>
unsigned long bin[53][53];
void buildbin(){
long i,j;
for(i=0;i<53;i++){
for(j=0;j<53;j++){
if (j==0 || j==i)
bin[i][j]=1;
else if(j>i)
bin[i][j]=0;
else
bin[i][j]=bin[i-1][j]+bin[i-1][j-1];
}
}
}
int main(){
unsigned long sum=0;
unsigned long n[]={0,4,4,4,4,4,4,4,4,4,16}, i, j, k, h, m, sel=8; // sel: no. of cards selected
buildbin();
for(i=1;i<=10;i++)
for(j=i;j<=10;j++)
for(k=j;k<=10;k++) {
h=0;
if (i==j && i==k){
for(m=1;m<=i;m++)
h += n[m];
h=52-h;
for(m=3;m<=sel;m++)
sum += (i+j+k)*bin[n[i]][m]*bin[52-n[i]-h][sel-m];
}
else if (j!=k && i!=j){
for(m=1;m<=i;m++)
h += n[m];
h=52-h;
for(m=1;m<=n[i];m++)
sum += (i+j+k)*bin[n[k]][1]*bin[n[j]][1]*bin[n[i]][m]*bin[52-n[i]-h][sel-2-m];
}
else if (j==k && i!=j){
for(m=1;m<=i;m++)
h += n[m];
h=52-h;
for(m=1;m<=n[i];m++)
sum += (i+j+k)*bin[n[j]][2]*bin[n[i]][m]*bin[52-n[i]-h][sel-2-m];
}
else{
for(m=1;m<=i;m++)
h += n[m];
h=52-h;
for(m=2;m<=n[i];m++)
sum += (i+j+k)*bin[n[k]][1]*bin[n[i]][m]*bin[52-n[i]-h][sel-1-m];
}
}
printf("%.15f\n",sum/(double)bin[52][sel]);
return 0;
}
```

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 89,048

Hi gAr;

Thanks for including that and for the whole amazing method of solving the card problem. Wunderbar!

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Hi bobbym,

You're welcome!

It's one of the common methods in probability, I don't think there's something special in what I attempted.

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 89,048

Hi gAr;

Sometimes using the simplest techniques to do a problem is best. I gave up on the approach until John on another site convinced me it was tedious but doable. You eliminated the tedious part and that was good work.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

Hi,

I have improved the code to eliminate the conditions for the formula.

But couldn't get rid of the loops though, sigh!

This is the code for expected sum of highest 5 numbers from 8 randomly selected numbers:

```
#include <stdio.h>
unsigned long bin[53][53];
void buildbin(){
long i,j;
for(i=0;i<53;i++){
for(j=0;j<53;j++){
if (j==0 || j==i)
bin[i][j]=1;
else if(j>i)
bin[i][j]=0;
else
bin[i][j]=bin[i-1][j]+bin[i-1][j-1];
}
}
}
void init(unsigned long *t){
int i;
for(i=1;i<=10;i++)
t[i]=0;
}
int main(){
unsigned long sum=0, prod=1,x,t[11], cnt=0;
unsigned long n[]={0,4,4,4,4,4,4,4,4,4,16}, i, j, k, l, h, m, o, sel=8;
buildbin();
for(i=1;i<=10;i++)
for(j=i;j<=10;j++)
for(k=j;k<=10;k++)
for(l=k;l<=10;l++)
for(o=l;o<=10;o++)
{
h=0;
cnt=0;
prod=1;
init(t);
for(m=1;m<=i;m++)
h += n[m];
h=52-h;
t[i]++; // t[x] means the no. of times x appears in top 5
t[j]++;
t[k]++;
t[l]++;
t[o]++;
for(x=i+1;x<=10;x++){
cnt += t[x];
if (t[x] != 0)
prod *= bin[n[x]][t[x]];
}
for(m=t[i];m<=n[i];m++)
sum += (i+j+k+l+o)*prod*bin[n[i]][m]*bin[52-n[i]-h][sel-cnt-m];
}
printf("%.15f\n",sum/(double)bin[52][sel]);
return 0;
}
```

which is close to the simulation.

*Last edited by gAr (2014-02-03 00:29:05)*

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 89,048

Hi gAr;

Looks like you only have one more challenge with these type problems. n highest cards in m picked.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

Offline

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