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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?

  • 2
    It's a peeve of mine: "die" is singular, "dice" is plural...2010-11-23
  • 2
    Why does nobody ever consider other "dice" than D6?2012-07-03

8 Answers 8

9

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 $\frac{6^6-5^6}{6^6} \approx 0.6651$

  • 0
    Thank you.. now that's why I fail at Yatzee :p2010-11-23
  • 0
    Doesn't Yatzee have 5 dice? :p2010-12-21
  • 2
    Yes, 5 dice. And it has an 'h'.2011-06-03
7

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.

5

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, ${\rm P}(E_1 \cup \cdots \cup E_6 ) = 1 - (5/6)^6$.

  • 0
    +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 ? "2010-12-07
  • 0
    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 $P(A \cup B) = P(A) + P(B) - P(A)P(B)$.2010-12-07
2

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 $1 - (5/6)^6$.

1

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.

1

The classic method to calculate this is considering the probability of not throwing a six which equals 5/6 , six times and then subtracting that probability from the probability of all possible outcomes, 1, which gives the probability of having thrown at least one six

= 1 - (5/6)6

= 0.665

But consider another method, the probability of the first die being a 6, which is 1/6, then consider that there is an additional probability contribution of a 6 from the next die from the remaining possible events which are now 5/6 of the total, which is 5/6*1/6, and so on until one gets the following sum

= 1/6 + (5/6) / 6 + (5/6)2 / 6 + (5/6)3 / 6 + (5/6)4 / 6 + (5/6)5 / 6

= 0.665 or

= 67%

0

The probability of rolling a six on a dice is 1/6.If you roll a dice the chances of you getting a 6 will only be once

  • 0
    This is really not an answer at all.2017-02-17
0

You can get the probability of rolling at least one six with a single line or R. Try.

sum(rowSums(rolldie(6) == 6) >= 1) / nrow(rolldie(6))