# chance on throwing a six with 6 dice

The chance to throw a 6 with one die is 1/6

And 6 times 1/6 = 1

So, if I throw with 6 dice, the chance to throw at least 1 six should be 1.

But when I throw 6 dice, I sometimes don't throw any 6 at all..

How come?

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It's a peeve of mine: "die" is singular, "dice" is plural... –  Ｊ. Ｍ. Nov 23 '10 at 1:05
Why does nobody ever consider other "dice" than D6? –  draks ... Jul 3 '12 at 13:46

With $6$ dice there are $6^6$ possible outcomes. Of these, $5^6$ don't lead to six on any of the dice. So the number of outcomes that lead to at least one six are $6^6-5^6$, so the probability of at least one six is $\frac{6^6-5^6}{6^6} \approx 0.6651$

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Thank you.. now that's why I fail at Yatzee :p –  Enrico Pallazzo Nov 23 '10 at 0:49
Doesn't Yatzee have 5 dice? :p –  Rawling Dec 21 '10 at 14:52
Yes, 5 dice. And it has an 'h'. –  Fixee Jun 3 '11 at 2:58

The expected number of dice showing $6$ is $1$ when throwing six dice. The probability to see at least one $6$ is $1 - (5/6)^6$, as explained in Timothy's answer.

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The probability of the union of independent events (unlike disjoint events) is not the sum of the individual probabilities. If $E_i$ denotes the event "$6$ is obtained on the $i$th throw", then $E_i$ are independent events, and it does not hold ${\rm P}(E_1 \cup \cdots \cup E_6 ) = {\rm P}(E_1 ) + \cdots + {\rm P}(E_6 )$. The left-hand side probability can be found as follows. The complement of the event $E_1 \cup \cdots \cup E_6$ is $E_1^c \cap \cdots \cap E_6^c$, where $E_i^c$ is the event "$6$ is not obtained on the $i$th throw". Hence, $${\rm P}(E_1 \cup \cdots \cup E_6 ) = 1 - {\rm P}(E_1^c \cap \cdots \cap E_6^c ).$$ Now, the probability of the intersection of independent events is the product of their individual probabilities. So, ${\rm P}(E_1^c \cap \cdots \cap E_6^c ) = {\rm P}(E_1^c) \cdots {\rm P}(E_6^c) = (5/6)^6$. Hence, ${\rm P}(E_1 \cup \cdots \cup E_6 ) = 1 - (5/6)^6$.

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+1,but you didn't explain "why the probabilities of union of independent events (unlike disjoint events) is not the sum of the individual probabilities ? " –  Quixotic Dec 7 '10 at 15:01
Try to figure out why, for any two events $A$ and $B$, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. So, if $A$ and $B$ are disjoint (meaning that $A \cap B = \emptyset$), we have $P(A \cup B) = P(A) + P(B)$, but if $A$ and $B$ are independent, then $P(A \cup B) = P(A) + P(B) - P(A)P(B)$. –  Shai Covo Dec 7 '10 at 16:49
If you flip a coin, there's a 50-50 chance you get heads. If you flip two coins there's a $3/4$ chance you get at least one head, since there's a $1/2 \times 1/2 = 1/4$ chance they're both tails. So similarly, there's a $5/6$ chance each die is not $6$. The chance they're all not $6$ is $(5/6) \times (5/6) \times (5/6) \times (5/6) \times (5/6) \times (5/6) = (5/6)^6$. So the chance that at least one is a six is $1 - (5/6)^6$.
You expect to get one $6$ on average. But sometimes you get more than one $6$.
Since the expected number of $6$s is one, this means that sometimes you have to get fewer than one $6$, i.e. zero $6$s.