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Could someone help me with how the following is calculated:

What is the probability of rolling a die seven times and getting at least one six?

My instinct told me it would be $1/6+\cdots + 1/6$ but this ends up being $7/6$.

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I've posted an answer that not only gives a correct method, but explains why the method proposed in the question is incorrect even when one throws the die only twice, so that the number purporting to be the probability is not more than $1$. –  Michael Hardy Aug 21 '13 at 19:13
    
And, as Chris Cutler's answer points out, the average number of $1$s that you get when you throw a die seven times is indeed $7/6$. –  Michael Hardy Aug 21 '13 at 19:15

5 Answers 5

up vote 6 down vote accepted

HINT:

The probability of getting no six $p=\left(\frac56\right)^7$

The probability of getting at least one six $=1-p$

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What I don't get is, the probability of getting at least one six across all the dice is independent per dice- so why is it not 1/6 + 1/6.... etc (besides the fact that sum is greater than 1) –  user997112 Aug 21 '13 at 18:48
    
@user997112, what is the probability of exactly one six or exactly two sixes? –  lab bhattacharjee Aug 21 '13 at 18:51

Consider a slightly different question:

With two fair coins what's the probability of at least one head.

By your reasoning the probability would be $\frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1$.

That means it's certain to happen: bet your house against a penny. You can't lose! right?

However, for this example we can simply list all the possibilities:

HH, HT, TH, TT

All these outcomes are equally likely so the real probability is $\frac{3}{4}$

I can calculate this by noting that I win every time except TT and the probability of TT is:

$$\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$

So the probability of at least one head is:

$$1 - \frac{1}{4} = \frac{3}{4}$$

As we got by listing all possibilities.

Now to return to the original question the probability of getting at lest one 6 from 7 dice is:

One minus the probability of getting no 6.

$$Answer = 1 - {\left( \frac{5}{6} \right)}^7 = \frac{6^7 - 5^7}{6^7} = \frac{279936 - 78125}{279936} = \frac{201811}{279936} \approx 0.7209$$

If instead you wanted the probability of exactly one six from seven dice then consider the specific case first dice is a six and in the next 6 rolls there is no six.

The probability of this would be:

$$\frac{1}{6} \times {\left( \frac{5}{6} \right)}^6$$

And since we don't care about the order of these rolls the six could be on any of the seven rolls. Giving a probability of:

$$7 \times \frac{1}{6} \times {\left( \frac{5}{6} \right)}^6 \approx 0.3907$$

Hope that helps clear up any misunderstanding.

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The quickest way to obtain this probability is as follows:

Subtract from $1$ the probability of getting NO rolls of a $6$: The probability of getting at least one six is equal to $$1 - \left(\frac 56\right)^7$$

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What I don't get is, the probability of getting at least one six across all the dice is independent per dice- so why is it not 1/6 + 1/6.... etc (besides the fact that sum is greater than 1) –  user997112 Aug 21 '13 at 18:48
    
Think about it for the case of two dice, instead of being $\frac{2}{6} = \frac{12}{36}$, the probability is only $\frac{11}{36}$. That's because we don't want to count the case where both dice are sixes twice. –  Dan Brumleve Aug 21 '13 at 18:49
    
Think about a similar problem with coins...or using fewer dice. –  amWhy Aug 21 '13 at 18:50
    
If you want to add probabilities, you need to add "getting exactly one six with 7 dice" + "getting exactly two sixes using 7 dice" + ... + "getting exactly seven sixes when rolling 7 dice". This will not total nor exceed a probability of $1$, but requires a lot more work than if you simply compute the probability of obtaining zero 6's when rolling seven dice, and subtracting that total from one. –  amWhy Aug 21 '13 at 18:56
    
Thanks, @Amzoti! Hope your work day has ended!? –  amWhy Aug 22 '13 at 0:29

Your instinctual answer, $\frac16+\cdots+\frac16=\frac76$, is the expected number of sixes rolled. The reason why we can add up all the fractions to calculate the expectation is called the linearity of expectation. In fact, linearity doesn't even require that the events are independent!

What you've proved is that the expected number of times an event happens is different from the probability that it happens at least once.

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Is the probability of getting at least one "$1$" when the die is thrown twice equal to $1/6+1/6$?

There the answer is no.

One time in six you get a $1$ the first time you throw the die.

One time in six you get a $1$ the second time you throw the die.

But sometime you get a $1$ both times, and that must be taken into account.

You get 36 possible outcomes: $$ \begin{array}{c|cccc} 11 & 12 & 13 & \cdots & 16 \\ \hline 21 & 22 & 23 & \cdots & 26 \\ 31 & 32 & 33 & \cdots & 36 \\ \vdots & \vdots & \vdots & & \vdots \\ 61 & 62 & 63 & \cdots & 66 \end{array} $$ The first horizontal row contains the six cases where the first trial gives you a "$1$". There are six of those.

The first vertical column contains the six cases where the second trial gives you a "$1$". There are six of those.

There's one case in which both trials give you a $1$.

If you just say the number of trials where you get a $1$ is $6+6$, that one gets counted twice. So it's $6+6-1$. The probability of getting at least one "$1$" is therefore $$ \frac{6+6-1}{36} = 0.305555\ldots $$

The rule that $\Pr(A\text{ or }B)=\Pr(A)+\Pr(B)$ is valid only if the events $A$ and $B$ are mutually exclusive, i.e. they cannot both happen. But the event of getting a $1$ on the first trial and the event of getting a $1$ on the second trial are not mutually exclusive: they can both happen.

Now do it by a different method:

The probability of getting no $1$ the first time is $5/6$.

The probability of getting no $1$ the second time is $5/6$.

The probability of getting no $1$s in the sequence of two trials is therefore $$\frac56\cdot\frac56 = 0.69444444\ldots$$

The probability of getting at least one $1$ in those two trials is therefore $$ 1-\left(\frac56\cdot\frac56\right) = 1-\frac{25}{36} = \frac{11}{36} = 0.30555555\ldots $$

When you write $$ \frac16+\frac16+\frac16+\frac16+\frac16+\frac16+\frac16, $$ you neglect the possibility of getting a $1$ more than once. Therefore this method will give you a wrong answer.

When you write $$ 1-\left(\frac56\cdot\frac56\cdot\frac56\cdot\frac56\cdot\frac56\cdot\frac56\cdot\frac56\right), $$ you don't neglect anything and you get a right answer.

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