Probabilities of Dice Pools

Question

Given 5 fair, 6-sided dice, calculate the probability of an equal or greater number of dice showing 1 appearing than the number of dice showing 4,5, and 6 combined.

For example: if 5 dice are rolled and show:

1,1,1,4,5: this means 3 dice showing 1's > 2 dice showing 4+ = fulfills condition

1,1,2,5,6: this means 2 dice showing 1's = 2 dice showing 4+ = fulfills condition

5,2,1,3,5: this means 1 dice showing 1 < 2 dice showing 4+ = fails condition

Edit:

I began by listing combinations that fulfill the condition from most number of 1's rolled to least:

1,1,1,1,1 = 1

1,1,1,1,2+ = 5 possibilities * 5 positions = 25

1,1,1,2+,2+ = (5*5) possibilities * (5+5) positions = 250

1,1,(2,3),2+,2+ = (5*5*2) possibilities * (5*5) positions = 1250

1,(2,3),(2,3),(2,3),2+ = (5*2*2*2) possibilities * (5*4) positions = 800

So Probability = (1+25+250+1250+800)/5^6

which is: 0.1488

This seems high!

Answer 1

Your $1,1,1,2+,2+$ should have $5 \choose 2$ positions, which happens to be equal to $5+5$. Your $1,1,(2,3),2+,2+$ includes some cases like $1,1,2,5,6$ which fail. It should be $1,1,(2,3),(2,3),2+$. It also triple counts cases like $1,1,2,2,3$, because each of $2,2,3$ can be the $2+$ Both are true in the next line as well. Finally, the denominator should be $6^5$, not $5^6$

Answer 2

Let $a$ represent ones, $b$ represent two and three, and $c$ represent four, five, and six.

$(a+b+c)^5 = a^5+5 a^4 b+5 a^4 c+10 a^3 b^2+20 a^3 b c+10 a^3 c^2+10 a^2 b^3+30 a^2 b^2 c+30 a^2 b c^2+10 a^2 c^3+5 a b^4+20 a b^3 c+30 a b^2 c^2+20 a b c^3+5 a c^4+b^5+5 b^4 c+10 b^3 c^2+10 b^2 c^3+5 b c^4+c^5$

provides the representation of possible outcomes, where the coefficient is the number of ways some outcome can happen, and the exponents are the number of each of the three types, e.g., $10a^2b^2c$ shows there are ten ways that the five rolls can be two of type $a$ and two of type $b$ and one of type $c$.

By inspection, remove the terms where the exponent of $c$ is greater than that of $a$, leaving

$a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5 + 5 a^4 c + 20 a^3 b c + 30 a^2 b^2 c + 20 a b^3 c + 10 a^3 c^2 + 30 a^2 b c^2$

Substitute the probabilities of each type ($1/6$ for $a$, $2/6$ for b and $1/2$ for $c$), resulting in a probability of $\frac{77}{324}$ ~ $0.24$.

If the requirement exludes the case where neither ones or 4/5/6 are present, simply remove the isolated $b$ term, resulting in a probability of $\frac{227}{972}$ ~ $0.23$