calculating the chance to get a 6 on a dice after 6 throws

PabloDons:

Is there a way to calculate the chance to get a 6 on a dice after throwing it 6 times? the only way I can think of is setting up all possible combinations which would be insanely long, Example:
111111 111111
111111 111111
111111 111111
111111 111111
111111 222222
123456 123456 etc.
I have tried my hardest but couldn't figure out a formula.
any help is appreciated
- Pablo

Your question has more than one interpretation!

If you mean: "What is the probability of getting at least one 6 in six throws of the dice" then this is just 1 - the probability of getting no sixes.
The probability of getting no six for the first die is 5/6.
The probability of getting no six for either the first or second die is (5/6)*(5/6) or 5 ²/6 ²
The probability of getting no six for any of the first three dice is (5/6)*(5/6)*(5/6) or 5 ³/6 ³
Continuing this way, the probability of getting no sixes for any of the six dice is 5 ⁶/6 ⁶
Hence, the probability of getting at least one six is 1- 5 ⁶/6 ⁶

PabloDons:

Is there a way to calculate the chance to get a 6 on a dice after throwing it 6 times? the only way I can think of is setting up all possible combinations which would be insanely long, Example:
111111 111111
111111 111111
111111 111111
111111 111111
111111 222222
123456 123456 etc.
I have tried my hardest but couldn't figure out a formula.
any help is appreciated
- Pablo

There are two ways in which I can interpret your question, You're either asking what the odds are of throwing one six in six throws, or you're asking the odds of throwing at least one six in six throws.

The odds of throwing one six in six throws can be calculated in the following way;

We know that there is a (1/6)th chance of throwing a six with one dice and a chance of (5/6) of not throwing a six with one dice.

So the odds of throwing for example [6] [no 6] [no 6] [no 6] [no 6] [no 6] = (1/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6)

However this is only the chance of throwing a six in the first throw and no six in the next 5 throws.

Another possibility would be throwing [no 6] [6] [no 6] [no 6] [no 6] [no 6] or [no 6] [no 6] [no 6] [no 6] [no 6] [6]

All these possibilities have the same probability as the one I just calculated ( (1/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6) )

There are 6 possibilities for which in six throws you throw 1 six (in the first, second, third, fourth, fifth, or sixth throw)

So the odds of throwing 1 six in six throws is 6 * (1/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6) = 0.4019 ( = 40.19%)

These odds take into account the cases where there is only one six, and it leaves out the cases where you throw six twice, three times, or even more times

To calculate the odds where you want at least one six you need to make use of the following;

The chance of something happening = 1 - the chance of that thing NOT happening

So for example if I have 0.30 (=30%) chance of winning $1000, I have 0.70 (=70%) chance of NOT winning $1000.

Similarly, to calculate the probability of throwing at least 1 six, It is more convenient to calculate the probability of throwing less than 1 six.
Or more clearly, the probability of throwing no sixes at all.

The probability of throwing no sixes means that each dice hits the (5/6)th chance of NOT throwing a six.
Consequently the probability of NOT throwing a six = (5/6)*(5/6)*(5/6)*(5/6)*(5/6)*(5/6) = 0.3349 (or 33.49%)

Now to find the odds of throwing at least one six, we simply need to calculate 1-0.3349 = 0.6651

So there is a 66.51% chance of throwing at least one six.

As you can see, the formulation of the question is very important in calculating the right probability

Alan:

PabloDons:

Is there a way to calculate the chance to get a 6 on a dice after throwing it 6 times? the only way I can think of is setting up all possible combinations which would be insanely long, Example:
111111 111111
111111 111111
111111 111111
111111 111111
111111 222222
123456 123456 etc.
I have tried my hardest but couldn't figure out a formula.
any help is appreciated
- Pablo

Your question has more than one interpretation!

If you mean: "What is the probability of getting at least one 6 in six throws of the dice" then this is just 1 - the probability of getting no sixes.
The probability of getting no six for the first die is 5/6.
The probability of getting no six for either the first or second die is (5/6)*(5/6) or 5 ²/6 ²
The probability of getting no six for any of the first three dice is (5/6)*(5/6)*(5/6) or 5 ³/6 ³
Continuing this way, the probability of getting no sixes for any of the six dice is 5 ⁶/6 ⁶
Hence, the probability of getting at least one six is 1- 5 ⁶/6 ⁶

1-5^6/6^6

If you mean "What is the probability of getting exactly one 6 in six throws of the dice" then
The probability of one particular die getting a 6 and all the others getting no sixes is (1/6)*(5 ⁵/6 ⁵)
But there are 6 dice each of which could be the one that falls as a 6, so the probability is 6*(1/6)*(5 ⁵/6 ⁵) or 5 ⁵/6 ⁵

5^5/6^5

If you mean something else then you'll need to specify exactly what!

reinout-g:

PabloDons:

Is there a way to calculate the chance to get a 6 on a dice after throwing it 6 times? the only way I can think of is setting up all possible combinations which would be insanely long, Example:
111111 111111
111111 111111
111111 111111
111111 111111
111111 222222
123456 123456 etc.
I have tried my hardest but couldn't figure out a formula.
any help is appreciated
- Pablo

There are two ways in which I can interpret your question, You're either asking what the odds are of throwing one six in six throws, or you're asking the odds of throwing at least one six in six throws.

The odds of throwing one six in six throws can be calculated in the following way;

We know that there is a (1/6)th chance of throwing a six with one dice and a chance of (5/6) of not throwing a six with one dice.

So the odds of throwing for example [6] [no 6] [no 6] [no 6] [no 6] [no 6] = (1/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6)

However this is only the chance of throwing a six in the first throw and no six in the next 5 throws.

Another possibility would be throwing [no 6] [6] [no 6] [no 6] [no 6] [no 6] or [no 6] [no 6] [no 6] [no 6] [no 6] [6]

All these possibilities have the same probability as the one I just calculated ( (1/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6) )

There are 6 possibilities for which in six throws you throw 1 six (in the first, second, third, fourth, fifth, or sixth throw)

So the odds of throwing 1 six in six throws is 6 * (1/6) * (5/6) * (5/6) * (5/6) * (5/6) * (5/6) = 0.4019 ( = 40.19%)

These odds take into account the cases where there is only one six, and it leaves out the cases where you throw six twice, three times, or even more times

To calculate the odds where you want at least one six you need to make use of the following;

The chance of something happening = 1 - the chance of that thing NOT happening

So for example if I have 0.30 (=30%) chance of winning $1000, I have 0.70 (=70%) chance of NOT winning $1000.

Similarly, to calculate the probability of throwing at least 1 six, It is more convenient to calculate the probability of throwing less than 1 six.
Or more clearly, the probability of throwing no sixes at all.

The probability of throwing no sixes means that each dice hits the (5/6)th chance of NOT throwing a six.
Consequently the probability of NOT throwing a six = (5/6)*(5/6)*(5/6)*(5/6)*(5/6)*(5/6) = 0.3349 (or 33.49%)

Now to find the odds of throwing at least one six, we simply need to calculate 1-0.3349 = 0.6651

So there is a 66.51% chance of throwing at least one six.

As you can see, the formulation of the question is very important in calculating the right probability

Thank you very much, I was thinking of the odds to get at least 1 six out of 6 throws. Thank you rainout for the brilliant example of a the chance for something to happen = 1 - the chance of something not to happen. I didn't think of that at all.