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The approach in the text is using the "not" rule, where we first find the probability of getting zero number of 6's. For one dice this probability is $\frac{5}{6}$ and for three die it becomes $\frac{5}{6}^3$. Hence using the not rule the probability of getting at least one 6 is $1-\frac{125}{216} = \frac{91}{216}$. But I have a different approach which leads to a different answer - Let's say we have one dice then we have just 1 way to get atleast one 6. For two die, it becomes 6 ways and for three die the answer comes to be 36 ways. Hence the probability turns out to be $\frac{36}{216}$. What is wrong with my approach?

StubbornAtom
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Chetan Naik
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  • For three dice you could have $6XX, X6X, XX6$ where $X$ stands for any non-six. that's already $3\times 5^2=75$ possibilities and this is not the full list. Where did you get $36$? – lulu Jun 02 '20 at 18:49
  • The correct answer is $91/216$. Note that $36/216$ is a fancy way to write $1/6$: this is the probability that the first die is $6$. But the event "at least one six" is larger than that. – Crostul Jun 02 '20 at 18:50
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    I know that my answer is wrong but what is wrong with my answer? I want atleast one 6 so I will fix one die with 6 and there are total 36 ways for the other two die – Chetan Naik Jun 02 '20 at 18:54
  • You come out with 36 because you fix one die, say die 1 and assign its value to 6. This means that dice 2 and 3 can each assume 6 values. Then in the denominator you consider cases where each die can assume 6 values. So in the numerator you are not considering cases where for example die 1 is 3, die 2 is 4, die 3 is 6 – Peanut Jun 02 '20 at 18:58
  • "For two die (sic), it becomes 6 ways..." looks like where you went wrong. There are 11 ways: (1,6) (2,6) (3,6) (4,6) (5,6) (6,6) (6,5) (6,4) (6,3) (6,2) (6,1) – The Chaz 2.0 Jun 02 '20 at 19:25
  • Before you go and say that the dice are indistinguishable and so $(1,6)$ and $(6,1)$ are effectively the same... I'm going to stop you right there and say that this is an incorrect train of logic. Even if the dice were in fact indistinguishable, because we prefer working in equiprobable sample spaces it is preferable to pretend the dice were in fact distinguishable as this allows us to calculate more easily. Even barring that, although there are indeed six different recognizable outcomes in the case they were two indistinguishable dice that doesn't mean there are $36$ for three dice... – JMoravitz Jun 02 '20 at 19:29
  • I recommend reading my answer here with regards to the incorrect argument that the dice being indistinguishable should lead to a different answer. If that wasn't your argument, then I can only imagine you came up with your numbers then by incorrectly mistaking the problem of "The probability at least one die is a six" with "The probability the first die is a six." – JMoravitz Jun 02 '20 at 19:31
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    One 'die', several 'dice' for gambling. 'Dies' for manufacturing. – BruceET Jun 02 '20 at 19:43

2 Answers2

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"Let's say we have one die then we have just 1 way to get at least one 6": yes, $1$ out of $6$ outcomes.

"For two dice, it becomes 6 ways". Nope, you have $1$ way to get a double 6, and $10$ ways to get only one 6 (first die: 6, second die: 1-5; first die= 1-5, second die: 6). $1+10=11$ out of $36$ outcomes.

"for three dice the answer comes to be 36 ways". Nope, you have $1$ way to get three 6s, $15$ ways to get two 6s (three times five ways: 6, 6, 1-5; 6, 1-5, 6; 1-5, 6, 6), $75$ ways to get only one 6 (three times twentyfive ways: 6, 1-5, 1-5; 1-5, 6, 1-5; 1-5, 1-5, 6). $1+15+75=91$ out of $216$ outcomes.

Hence even not using the (easier) not rule the probability of getting at least one 6 is $\frac{91}{216}$ :)

Sergio
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That approach is wrong. Although your answer for only one die is correct you are missing some possibilities incase of two and three dies. Let me tell you what you are missing.

  1. In case of second die you are missing one of the two possibilities eg (1,6),(6,1).

  2. Actually there are 11 cases.

    $$1*5 + 5*1 + 1 =11$$

  3. When three die appear there is chance of 6 appears only on one or two or on all dies. You need to add all the cases. If you just multply you will miss $(*,*,6)$. (* Represent any number other than 6.)

Now for three dies in your way if you agree with corrected case of two dies there will be $$6*11=66$$ ways. Which is wrong But it could be corrected by adding missing cases

Let us add missing case

$(*,*,6)$ have 5x5 = 25 cases

Now total cases will be $$66+25=91$$

Now probability will be $$\frac{91}{216}$$

You may want to note that the case such as $(6,6,6),(6,*,6),(*,6,6)$ had included in your way of multiplying the two die case with six.

Community
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Harikrishnan S
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