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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. :-)

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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.

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

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  • $\begingroup$ Now I wasn't going to work that one out on my own. Thanks so much for the insightful commentary. I'll do my best to understand it and to apply to similar problems. :-) $\endgroup$ – MrGreggles Mar 14 '13 at 7:06
  • $\begingroup$ No worries, you can click on the "tick" on the answer to accept this answer. $\endgroup$ – bryanblackbee Mar 14 '13 at 7:06
  • $\begingroup$ So for rolling at least 2 dice of 2 or less (from 3 dice rolled): P(all dice 2 or lower) + P(two dice 2 or lower) = (1/3 x 1/3 x 1/3) + (1/3 x 1/3 x 3 x 2/3) = 1/27 + 2/9 = 7/27 = 0.26. Is that correct? $\endgroup$ – MrGreggles Mar 14 '13 at 7:45
  • $\begingroup$ Yes you are correct.$$P(\text{all dice 2 or lower}) + P(\text{two dice 2 or lower})\\= \frac 13 \times \frac 13 \times \frac 13 + \frac 13 \times \frac 13 \times \frac 23 \times 3\\=\frac 1 {27} + \frac{6}{27}\\=\frac {7} {27}$$ $\endgroup$ – bryanblackbee Mar 14 '13 at 7:51
  • $\begingroup$ Typically, I check my answer by calculating the probability of the complement event. They should add up to 1, which in this case happens. In this case,$$P(\text{no dice 2 or lower}) + P(\text{one die 2 or lower})\\= \frac 23 \times \frac 23 \times \frac 23 + \frac 23 \times \frac 23 \times \frac 13 \times 3\\=\frac 8 {27} + \frac{12}{27}\\=\frac {20} {27}$$ $\endgroup$ – bryanblackbee Mar 14 '13 at 7:56

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