Find The Rainbow

There are $4!$ ways to permute the four colors. The probability that green is heavier than blue is 1/2, the probability that red is heavier than yellow is 1/2, and these events are independent, so $4!(1/2)(1/2) = 6$ of the permutations satisfy the givens.

Now, if red is not heavier than blue, then the order must be green > blue > red > yellow, so there is only one such permutation out of the six.

Therefore, the probability of the complement, which is what we want, is 5/6.

Letting $g,b,r,y$ being the respective weights of the green, blue, red and yellow balls, and given that $g \gt b$ and $r \gt y$ , there are $6$ possible sequences of $g,b,r,y$ embedded among the rankings of the weights of the seven balls written in descending order, namely

• $g \gt b \gt r \gt y$

• $g \gt r \gt b \gt y$

• $g \gt r \gt y \gt b$

• $r \gt g \gt b \gt y$

• $r \gt g \gt y \gt b$

• $r \gt y \gt g \gt b$

Each of these sequences is equally likely, since they each occur in $\binom{7}{4} = 35$ possible rankings of the weights of the seven balls. In $5$ of these we have $r \gt b$ , so the desired probability is $\dfrac{5}{6}$ , and so $a + b = 5 + 6 = \boxed{11}$ .

This is, in fact, the same explanation as given elsewhere. I am making the lists explicit.

Ignoring the orange, indigo and violet ball as distractors.

Assigning the remaining balls indices in the permutations to be generated.

$\{r,y,g,b\}=\{1,2,3,4\}$

Generating the $4!=24$ permutations of 4 things take 4 at a time. The numbers in each permutation are the weights of the respective balls.

$\text{perms}=\text{Permutations}[\text{Range}[4]] \Longrightarrow \\ \left( \begin{array}{cccc} 1 & 2 & 3 & 4 \\ 1 & 2 & 4 & 3 \\ 1 & 3 & 2 & 4 \\ 1 & 3 & 4 & 2 \\ 1 & 4 & 2 & 3 \\ 1 & 4 & 3 & 2 \\ 2 & 1 & 3 & 4 \\ 2 & 1 & 4 & 3 \\ 2 & 3 & 1 & 4 \\ 2 & 3 & 4 & 1 \\ 2 & 4 & 1 & 3 \\ 2 & 4 & 3 & 1 \\ 3 & 1 & 2 & 4 \\ 3 & 1 & 4 & 2 \\ 3 & 2 & 1 & 4 \\ 3 & 2 & 4 & 1 \\ 3 & 4 & 1 & 2 \\ 3 & 4 & 2 & 1 \\ 4 & 1 & 2 & 3 \\ 4 & 1 & 3 & 2 \\ 4 & 2 & 1 & 3 \\ 4 & 2 & 3 & 1 \\ 4 & 3 & 1 & 2 \\ 4 & 3 & 2 & 1 \\ \end{array} \right)$

Keeping the permutations where the green ball is heavier than the blue ball.

$\text{perms}=\text{Select}[\text{perms},\text{\$\#\$1}[[g]]>\text{\$\#\$1}[[b]]\&] \Longrightarrow \\ \left( \begin{array}{cccc} 1 & 2 & 4 & 3 \\ 1 & 3 & 4 & 2 \\ 1 & 4 & 3 & 2 \\ 2 & 1 & 4 & 3 \\ 2 & 3 & 4 & 1 \\ 2 & 4 & 3 & 1 \\ 3 & 1 & 4 & 2 \\ 3 & 2 & 4 & 1 \\ 3 & 4 & 2 & 1 \\ 4 & 1 & 3 & 2 \\ 4 & 2 & 3 & 1 \\ 4 & 3 & 2 & 1 \\ \end{array} \right)$

Keeping the permutations where the red ball is heavier than the yellow ball.

$\text{perms}=\text{Select}[\text{perms},\text{\$\#\$1}[[r]]>\text{\$\#\$1}[[y]]\&] \Longrightarrow \\ \left( \begin{array}{cccc} 2 & 1 & 4 & 3 \\ 3 & 1 & 4 & 2 \\ 3 & 2 & 4 & 1 \\ 4 & 1 & 3 & 2 \\ 4 & 2 & 3 & 1 \\ 4 & 3 & 2 & 1 \\ \end{array} \right)$

Selecting the permutations where the red ball was heavier than the blue.

$\text{heavier}=\text{Select}[\text{perms},\text{\$\#\$1}[[r]]>\text{\$\#\$1}[[b]]\&] \Longrightarrow \\ \left( \begin{array}{cccc} 3 & 1 & 4 & 2 \\ 3 & 2 & 4 & 1 \\ 4 & 1 & 3 & 2 \\ 4 & 2 & 3 & 1 \\ 4 & 3 & 2 & 1 \\ \end{array} \right)$

Computing the fraction.

$\frac{\text{Length}[\text{heavier}]}{\text{Length}[\text{perms}]} \Longrightarrow \frac{5}{6}$

Computing the answer.

$\text{Denominator}[\text{fraction}]+\text{Numerator}[\text{fraction}] \Longrightarrow 11$

Just lie down the balls along a line with 1st being the largest and last the smallest. Total ways is 7!. But given that green > blue and red>yellow which makes total case to be 7!/2.2 (since now there is only one way instead of 2 to arrange green and blue among themselves and similarly for red and yellow). Now if red>blue then out of 4! cases possible to arrange B,G,Y and R among themselves only 5 cases (GRBY,GRYB,RGBY,RGYB,RYGB) are favorable. Hence total favorable ways are 5.7!/4!. Just taking the ratio gives 5/6. So the answer is 5+6=11

Heres some ugly php that did it for me

<?php
//build all permutations of 0-6
$arr = range(0,6);
do{
$newarr = array();
foreach($arr as $a){
for($i=0;$i<=6;$i++){
if(!preg_match('@'.$i.'@',$a)){
$newarr[] = $a.$i;
}
}
}
$arr = $newarr;
} while(strlen($arr[0])<7);

//eliminate any instances where $a[3]<$a[4] or where $a[0]<$a[2]
$c = 0;
foreach($arr as $key=>$val){
$s = str_split($val);
if($s[3]<$s[4] || $s[0]<$s[2]){
unset($arr[$key]);
} elseif($s[0]>$s[4]){
$c++;
}
}

//count remaining combos, save for later (1240)
$rc = count($arr);

//treat $c as numerator, $rc as denominator, reduce fraction
$f = simplify($c,$rc);

//print answer
echo array_sum($f);

function simplify($num,$den) {
$g = gcd($num,$den);
return array($num/$g,$den/$g);
}

function gcd($a,$b) {
$a = abs($a); $b = abs($b);
if( $a < $b) list($b,$a) = array($a,$b);
if( $b == 0) return $a;
$r = $a % $b;
while($r > 0) {
$a = $b;
$b = $r;
$r = $a % $b;
}
return $b;
}
?>

5+6=11 bcuz 5/6 was the prob