# Sixes

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(R6 &W6&B6) = P(R6) x P(W6) x P(B6) = 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(~R6&~W6&~B6) = 1 – (5/6 x 5/6 x 5/6) = 1 – (125/216) = 91/216 = 0.421.
• 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((R6&~W6&~B6) v (~R6&W6&~B6) v (~R6&~W6&B6)) = P(R6&~W6&~B6) + P(~R6&W6&~B6) + P(~R6&~W6&B6) = (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.
• 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.