Probability for 5 dice (high d6 result +1 per additional die 4 or higher)

Question

I think I am calculating these odds properly. Hopefully someone can confirm that I am, or tell me where I am going wrong if I'm not.

When rolling 5 regular six-sided dice, the total result I am looking for is the value of the highest die +1 for every additional die that comes up 4+.

I have a 59.81% chance of one of the 5 dice coming up a 6. Of the remaining 4 dice there is a 93.75% chance that one of them is 4+, so $59.81\% \times 93.75\%$ gives me a 55.89% chance of getting a 7, or 6+1. Of the remaining 3 dice there is a 87.5% chance of getting a 4+. So the odds of getting a 6 on 1 of 5 dice (0.5981) times the odds of getting a 4+ on 4 dice (0.9375) times the odds of getting a 4+ on 3 dice (0.875) gives me a 49.06% chance of getting 6+2.

Is this correct?

Answer 1

You get $6 + 1$ if you throw one six, one $4$ or $5$ and three dice lower than $4$, or if you throw two sixes and three dice lower than $4$. The probability of the former equals:

$$\frac{{5 \choose 1}1^4{4 \choose 1}2^1{3 \choose 3}3^3}{6^5} = \frac{1080}{7776}$$

The probability of the latter equals:

$$\frac{{5 \choose 2}1^2{3 \choose 3}3^3}{6^5} = \frac{270}{7776}$$

As such, the probability of getting $6 + 1$ equals:

$$\frac{1080+270}{7776} = \frac{1350}{7776} \approx 0.174$$

A similar approach can be used for $5 + 2$, resulting in a probability of:

$$\frac{{5 \choose 1}1^1{4 \choose 2}1^2{2 \choose 2}3^2 + {5 \choose 2}1^2{3 \choose 1}1^1{2 \choose 2}3^2 + {5 \choose 3}1^3{2 \choose 2}3^2}{6^5} = \frac{270 + 270 + 90}{7776} = \frac{630}{7776}$$

Again, the same approach can be used for $4 + 3$, resulting in a probability of:

$$\frac{{5 \choose 4}1^4{1 \choose 1}3^1}{6^5} = \frac{15}{7776}$$

As such, the probability of getting $7$ equals:

$$\frac{1350 + 630 + 15}{7776} \approx 0.257$$

Answer 2

Too long for a comment on jvdhooft's answer: of the $6^5=7776$ equally likely events, the outcomes look like

       extrafourplus
                0    1    2    3    4
maxdice
      1         1    0    0    0    0
      2        31    0    0    0    0
      3       211    0    0    0    0
      4       405  270   90   15    1
      5       405  810  630  225   31
      6       405 1350 1710  975  211

and (dividing by $7776$) probabilities of

       extrafourplus
                   0       1       2       3       4
maxdice
      1      0.00013 0.00000 0.00000 0.00000 0.00000
      2      0.00399 0.00000 0.00000 0.00000 0.00000
      3      0.02713 0.00000 0.00000 0.00000 0.00000
      4      0.05208 0.03472 0.01157 0.00193 0.00013
      5      0.05208 0.10417 0.08102 0.02894 0.00399
      6      0.05208 0.17361 0.21991 0.12539 0.02713

which does not give your probabilities in the original question, even with Guy's re-interpretation.

This gives combined scores of

   1    2    3    4    5    6    7    8    9   10 
   1   31  211  405  675 1305 1995 1936 1006  211

and so probabilities of

      1       2       3       4       5       6       7       8       9      10 
0.00013 0.00399 0.02713 0.05208 0.08681 0.16782 0.25656 0.24897 0.12937 0.02713