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I never studied probability at school and this problem has been bothering me for a long time:

Let's say I have a perfectly fair die. If I roll it, the odds of it landing on $6$ are $\frac{1}{6}$. If I roll two dice, the odds of at least one of them landing on 6 are $\frac{1}{6}\times 2 =\frac{1}{3}$.

But what about if I roll six dice? What are the odds that one will land on $6$? Based on the previous reasoning, it should be: $$\frac{1}{6}\times 6 = 1$$

But that can't be true. It's actually possible that I roll six dice and none of them land on 6. What about if I roll $100$ dice? It's still possible that none of the land on 6. So what are the odds that at least one will?

eje211
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  • @Avatar: no, as others have shown, it is $\frac {11}{36}$ – Ross Millikan Jul 16 '13 at 17:34
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    The probabilities add only if the events are independent of eachother... While rolling two dice seem like two independent events, requesting "at least one $6$" makes them dependent: the outcome of the first die influences what you expect from the second (if you get a 6 in the first try, second doesn't matter, while if you don't it matters)... – N. S. Jul 16 '13 at 17:37

6 Answers6

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If I roll two dice, the odds of at least one of them landing on 6 are $\frac16×2=\frac13$

Not true - these are easily proved by thinking about it the other way - what are the odds that neither die lands on 6?

That is simply $$\frac56 \cdot \frac56 = \frac{25}{36}$$

So the odds that either of them lands on 6 is $$1- \frac{25}{36} = \frac{11}{36}$$ which is slightly less than $\frac13$.

The next example works the same way, the odds that none of the 6 dice land on 6 is

$$\frac56 \cdot \frac56 \cdot \frac56 \cdot \frac56 \cdot \frac56 \cdot \frac56 = \frac{5^6}{6^6} \approx0.33489$$

So the odds that any of the dice lands on 6 is the complement: about 66.5%.

MJD
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D Stanley
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The probability of rolling at least one 6 is $1$ - the probability of rolling no sixes. This is then

$$1-\left (\frac{5}{6}\right)^6 \approx 0.665$$

This works for any number of dice.

Ron Gordon
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    As pre the title, the OP seems also to be interested in the question for $n$ independent rolls of $n$-sided dice, not just the standard case $n=6$. The method generalizes and gives us $$1-\left(1-\frac1n\right)^n,$$which for $n\to\infty$ converges to $1-\frac1e\approx0.63212$. – Hagen von Eitzen Jul 16 '13 at 17:53
  • @HagenvonEitzen: I think the OP rolled out the title in haste. But thanks for the nifty result. – Ron Gordon Jul 16 '13 at 17:57
  • You're right, I was in a bit of a hurry. I was arguing a point about how doing something that can result in deadly injury 1 in 200 times twenty times meant a high likelihood of death, but I realized I didn't have my math right. – eje211 Jul 16 '13 at 22:26
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Let's say I have a perfectly fair die. If I roll it, the odds of it landing on 6 are 1/6. If I roll two dice, the odds of at least one of them landing on 6 are 1/6×2=1/3.

The chance of rolling the first 6 is 1/6, including the following cases (1/36 chance each):

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

The chance of rolling the second 6 is also 1/6, which counts the following cases:

(1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (6, 6)

Using this method, you've actually counted the odds of rolling two 6's twice. This overcounting gets worse with three+ dice: (6, 6, 2) (6, 4, 6) (3, 6, 6), (6, 6, 6), etc.

The easiest way is to find the complement probability (odds of rolling no 6's) and subtracting that from 1 to find your answer with 6 dice:

$$1-(\frac{5}{6})^6 = 1 - \frac{15625}{46656}= \frac{31031}{46656} \approx 66.5\%$$

Or you could count the cases of rolling a 6, two 6's, three 6's, ... six 6's. But that's too much casework.

James Zhu
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As you have discovered, probability of independent events is not additive. If it was, then by rolling 10 dice, you'd have a probability of getting a 6 as $10\cdot \frac{1}{6} > 1$. Clearly this is false, since you could get $\{1,1,1,1,1,1,1,1,1,1\}$.

There are two ways of looking at this: either you count the number of events where no 6s come up and subtract this from the total number of events, or you work the variables as separate events:

$$P(\textrm{at least one 6}) = P(\textrm{first die is a 6}) + P(\textrm{first die is not a six AND second die is a six}) \\ = \frac{1}{6} + \frac{5}{6}\cdot \frac{1}{6} = \frac{6}{36}+\frac{5}{36} = \frac{11}{36}.$$

Emily
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  • Your answer is the only one so far that also works with coins. I thought of looking at probability of not getting a six, but it didn't seem to work for coins. The probability of getting at least two heads for one coin would be the same as getting at least one: 1/2 * 1/2. But with your second example, it would be: 1/2 + 1/2 * 1/2, or 3/4. That looks correct to me. If I'm correct, the general solution for a coin would be, for k attempts, the sum 1 to k of: 0.5 * (0.5)^(k-1). Correct? – eje211 Jul 16 '13 at 18:00
  • Getting "two heads for one coin" doesn't really make sense, unless you mean two separate coin flips. The probability of getting at least one head in two coin flips can be found by enumerating all possibilities: ${ HH, HT, TH, TT }$, so it's $\frac34$. The general solution of any "at least one" type of problem is always "one minus the probability of none" -- the opposite of "at least one" is "exactly none". – Emily Jul 16 '13 at 18:04
  • Sorry, things got mixed up in my mind. I meant at least one head for two coins. – eje211 Jul 16 '13 at 22:22
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Bernoulli's trials is the best solution. For an event occurring $n$ times , if '$p$' gives the probability of a success and '$q$' gives the probability of a failure, then the probability of the event occurring '$r$' times is given by

$P(r) = \binom{n}{r}p^rq^{n-r}$

rurouniwallace
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Suraj M S
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If you list the $36$ possibilities and count, you will find the chance of getting at least one $6$ in two rolls is $\frac {11}{36}$. Your calculation of $2 \cdot \frac 16$ gives the expected number of sixes, but you get two of them at once when you roll $12$, so only eleven different rolls have at least one six. Similarly, when rolling six dice, on average you will get one six. Sometimes you will get more, sometimes less.

Ross Millikan
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