<s>[INST] Probability for 5 dice (high d6 result +1 per additional die 4 or higher) 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?
 [/INST] 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:
$$\frac{1350 + 630 + 15}{7776} \approx 0.257$$
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