Back to Probability Theory

### Problem

- Three dice are tossed. What’s the probability of rolling:
- Three sixes
- At least one six
- Exactly one six

- Assume the dice are red, white, and blue.

### Solution

###### Three sixes

- Six possibilities exist for the red die: 1, 2, 3, 4, 5, and 6. For each of these, the same six possibilities exist for the white die. Thus between the red and white dice 36 possibilities exist, 6 x 6. For each of these 36 possibilities, six possibilities exist for the blue die. There are thus 216 possibilities in all, 6 x 6 x 6. The possibilities being equally likely, the probability of three sixes is 1/216.
- The
*Special Conjunction Rule*yields the same result:**P(R**_{6 }&W_{6}&B_{6}) = P(R_{6}) x P(W_{6}) x P(B_{6}) = 1/6 x 1/6 x 1/6 = 1/216.

###### At least one six

- Computing the probability of at least one six is more challenging. It’s tempting to add the 1/6 probabilities rather than multiply them, getting ½. The problem is that the same logic means the probability of getting at least one six tossing six dice is 1.0.
- Though cumbersome, you could run through all 216 possibilities, tallying those with at least one six. For example, one of the eight rows of the following truncated Possibility Table qualifies:

- The best approach is using the Negation Rule, the key being that the negation of at least one six is no sixes. Thus:
- P(at least one six) = 1 – P(no sixes) = 1 – P(~R
_{6}&~W_{6}&~B_{6}) = 1 – (5/6 x 5/6 x 5/6) = 1 – (125/216) = 91/216 = 0.421.

- P(at least one six) = 1 – P(no sixes) = 1 – P(~R
- In general, to calculate the probability of
*at least one*thing of a kind, calculate the probability there’s nothing of that kind and subtract from one.

###### Exactly one six

- Rolling exactly one six can happen in three ways:
- the red die six, the others not;
- the white die six, the others not;
- the blue die six, the others not.

- Combining the
*Special Conjunction*and*Disjunction Rules*:- P(exactly one six) = P((R
_{6}&~W_{6}&~B_{6}) v (~R_{6}&W_{6}&~B_{6}) v (~R_{6}&~W_{6}&B_{6})) = P(R_{6}&~W_{6}&~B_{6}) + P(~R_{6}&W_{6}&~B_{6}) + P(~R_{6}&~W_{6}&B_{6}) = (1/6 x 5/6 x 5/6) + (5/6 x 1/6 x 5/6) + (5/6 x 5/6 x 1/6) = 25/216 + 25/216 + 25/216 = 75/216.

- P(exactly one six) = P((R
- For each die there are 25 ways it can be a six and the others not. With three dice, there are 3 x 25 ways a die can be the lone six. The chance of exactly one six is therefore 75 of 216.