[Math] Dice question: Probability of rolling at least 2 of 3 dice with score 3 or less

combinatoricsdiceprobability

my first post here.

I've recently gotten back to boardgaming (Labyrinth: The War on Terror in this case) and would like to get clear on the probability of various actions. I've learnt about 'basic' dice probability but I'm not sure about this one:

Rolling 3 dice, I need to get at least 2 rolls of 3 or lower. What's the probability of a successful result?

Thanks for an help or links that will point me in the right direction. 🙂

Best Answer

In this answer, dice is the plural form of die. I assume that you are using regular 6-sided dice.

For one die, the probability of rolling $3$ or lower is $\frac 12$. For three dice, you want at least two out of three with $3$ or lower.

$P\,(\text{all dice 3 or lower}) = \frac 12 \times \frac 12 \times \frac 12 = \frac 18$

$P\,(\text{two dice 3 or lower}) = \frac 12 \times \frac 12 \times \frac 12 \times 3 = \frac 38$

This is calculated as follows: For each die, there is a probability of $\frac 12$ that it will get $3$ points or lower. In fact, the last $\frac 12$ actually represents the probability that the die turns out $4$ or larger. The last term $3$ means that there are $3$ arrangements for this combination to occur $-$ the "$4$ or larger" die appears on the 1st, 2nd or 3rd die.

$\text{Required Probability} = \frac 18 + \frac 38 = \frac 48 = \frac 12$

Related Question