Basic Probability Question

[thomas reid]

I've just begun the study probability and I have a very basic question.

If I roll a fair dice I know I have a one in six chance of getting a "1".

What is my chance of getting a "1" if I roll it twice?  (It could be either on the first roll or the second roll or both.  I am just concerned with getting a "1" at some point.)

What is my chance of getting a "1" if I roll it ten times?  (Same thing: I'm just concerned with getting a "1" (could be more than one "1") at some point.)

I realize this may be too basic of a question for this forum, but I'm hoping someone will have the time to help me out with this really simple question.

Thank you.

[studiot]

No problem.

 

The easy way to approach this is to realise that rolling one die twice or two dice together is the same in probability theory.

Please note the singular and plural of the word die.

So it we roll two dice together and list all the possibilities we can see that they are

1 plus 1 , 2 , 3 , 4 , 5 or 6 that is 6 possibilities

2 plus 1 , 2 , 3 , 4 , 5 or 6 that is 6 possibilities

3 plus 1 , 2 , 3 , 4 , 5 or 6 that is 6 possibilities

4 plus 1 , 2 , 3 , 4 , 5 or 6 that is 6 possibilities

5 plus 1 , 2 , 3 , 4 , 5 or 6 that is 6 possibilities

6 plus 1 , 2 , 3 , 4 , 5 or 6 that is 6 possibilities

Adding these up we see that there are a grand total of 36 or 6x6 possibilities (or outcomes) though not all include a 1.

 

Now this is where we need to become more precise in our wording.

Because there are two situations where a 1 is rolled.

The roll can produce at least one 1  (two 1s in this scenario)

or

The roll can produce exactly one 1

 

In the first of the above lines each of the six possibilities produces at least one 1, making 6 ways to get a 1
In the last five of the above lines one of the 6 possibilities is produces at least one 1,  adding a further 5 ways to get at least one 1

So there are a total of (6 + 5) = 11 ways to get at least one 1 out of a grand total of 36 possibilities.

So the probability of getting at least on 1 is 11/36.

 

In the first of the above lines, five of the six ways will produce exactly one 1 and the sixth will produce two 1s, making 5 ways to get exactly one 1.

In the last five of the above lines one of the 6 possibilities is produces exactly one 1; there are no ways to produce more than one 1.
So this adds a further 5 ways to get exactly one 1.

So there are a total of (5 + 5) = 10 ways to get exactly one 1 out of a grand total of 36 possibilities.

So  the probability of getting exactly one 1 is 10/36

 

This analysis is meant to help understanding and provide something to fall back on.

You can always use formula, which is quicker,  once you have the understanding.

 

Does this help?

 

 

Edited January 31, 2018 by studiot

[thomas reid]

Yes ... very helpful!

Thank you for taking the time to spell all out so clearly.

Where can I find the formula (for 10 rolls or so)?

Again, thank you.

[studiot]

Beware there are some incorrect formulae rattling around the net.

This is the time to introduce a couple of new concepts.

Firstly the stopping or termination criterion.

I assume you will roll the die until you get a 1 and then stop (after n throws)

Getting a 1 is the stopping criterion.

This is equivalent to getting at least one 1 in n throws.

Which brings us to the second concept.

The probability of something happening and not happening equals 1

So if the probability of getting a 1 in one throw is 1/6 then the probability of not getting a 1 is (1 - 1/6) = 5/6

I am not going to produce a formal proof of the formula (you can have it if you really want it)

I am just going to show that it works.

The probability of getting a 1 in n throws is

[math]{P_{1in\,n}} = \left\{ {1 - {{\left( {{P_{not\,n}}} \right)}^n}} \right\} = \left\{ {1 - {{\left( {\frac{5}{6}} \right)}^n}} \right\}[/math]

So for n = 0  (no throws) the simplest possible there is zero probability of getting a 1

[math]{P_{n = 0}} = \left\{ {1 - {{\left( {\frac{5}{6}} \right)}^0}} \right\} = \left\{ {1 - 1} \right\} = 0[/math]

For n = 1 we have what we already know that the probability is 1/6

[math]{P_{n = 1}} = \left\{ {1 - {{\left( {\frac{5}{6}} \right)}^1}} \right\} = \left\{ {1 - \left( {\frac{5}{6}} \right)} \right\} = \frac{1}{6}[/math]

For n = 2 we have what we worked out above that the probability is 11/36

[math]{P_{n = 2}} = \left\{ {1 - {{\left( {\frac{5}{6}} \right)}^2}} \right\} = \left\{ {1 - \left( {\frac{{25}}{{36}}} \right)} \right\} = \frac{{11}}{{36}}[/math]

and for n = 3 we have

[math]{P_{n = 3}} = \left\{ {1 - {{\left( {\frac{5}{6}} \right)}^3}} \right\} = \left\{ {1 - \left( {\frac{{125}}{{216}}} \right)} \right\} = \frac{{91}}{{216}}[/math]

 

 

I will leave you to do some work now for n = 10.

 

 

[thomas reid]

 

Very helpful ... thanks for the formula ... exactly what I was looking for!

 

Edited February 1, 2018 by thomas reid

[thomas reid]

oops ... I wrote "0.84 %" instead of "0.84" or "84 %"

 

[studiot]

A bit of algebra, especially useful in exams.

 

[math]1 - {\left( {\frac{5}{6}} \right)^{10}} = 1 - \frac{{{{\left( 5 \right)}^{10}}}}{{{{\left( 6 \right)}^{10}}}} = \frac{{{{\left( 6 \right)}^{10}} - {{\left( 5 \right)}^{10}}}}{{{{\left( 6 \right)}^{10}}}} = \frac{{60466176 - 9765625}}{{60466176}} = 0.84[/math]

 

[MigL]

" we don't need no stinkin' algeebra, we gots calcumalators"

Seriously, excellent explanation