Probability of same event happening in a row

• Jun 4th 2008, 09:13 AM
jgillett
Probability of same event happening in a row
Hi, I've been wrecking my head all day with this and I can't figure it out. I'm guessing that it is obvious but I don't really have much of an education in probability.

What I want to know is if there's a formula that you can use to work out the probability of (for example) rolling a 6 three (or more) times in a row when you roll a dice five times (so 66_66 wouldn't be a desired result despite rolling 3 or more 6s).

I've been racking my brains and have noticed some things (admittedly just with coins) but haven't been able to get a formula from it.

I would love a formula where you can change things like how many times you want to attain the result in a row, how likely that result is (e.g. 1/2 for heads, 1/6 for a number on a dice) and how many times you're going to repeat the action (e.g. how many times you're going to roll the dice or flip the coin).

Most importantly though I'd appreciate if some patient person could help guide me through the formula (or even better guide me to 'discovering' the formula myslef) so that I understand it and what it 'means' (if that makes any sense).

Cheers
• Jun 4th 2008, 11:11 AM
TheEmptySet
Quote:

Originally Posted by jgillett
Hi, I've been wrecking my head all day with this and I can't figure it out. I'm guessing that it is obvious but I don't really have much of an education in probability.

What I want to know is if there's a formula that you can use to work out the probability of (for example) rolling a 6 three (or more) times in a row when you roll a dice five times (so 66_66 wouldn't be a desired result despite rolling 3 or more 6s).

I've been racking my brains and have noticed some things (admittedly just with coins) but haven't been able to get a formula from it.

I would love a formula where you can change things like how many times you want to attain the result in a row, how likely that result is (e.g. 1/2 for heads, 1/6 for a number on a dice) and how many times you're going to repeat the action (e.g. how many times you're going to roll the dice or flip the coin).

Most importantly though I'd appreciate if some patient person could help guide me through the formula (or even better guide me to 'discovering' the formula myslef) so that I understand it and what it 'means' (if that makes any sense).

Cheers

What you are discribing is a binomial probability...

Here is the general formula the probability of x events in n trials with a probability of success P is given by

Note: this only works if there are two outcomes, sucess or failure.

$f(x)=\binom{n}{x}(P)^{x}(1-P)^{n-x}$

So in your case with rolling a 6 exactly 3 times in 5 attemps would be

$f(3)=\binom{5}{3}\left( \frac{1}{6}\right)^{3}\left( \frac{5}{6}\right)^{5-3}=\frac{125}{3888}$

Now you could do the same for 4 and 5

$f(4)=\frac{25}{7776} \\\ f(5)=\frac{1}{7776}$

$f(3)+f(4)+f(5)=\frac{23}{648}$

Read this to help with the theory behind it

Binomial probability - Wikipedia, the free encyclopedia

I hope this helps.

Good luck.(Clapping)
• Jun 4th 2008, 11:44 AM
galactus
Just remember, the binomial gives the prob. of three sixes in five rolls...period.
We just want the cases where they are consecutive.
We could have three different cases:

six,six,six,not,not
not,six,six,six,not
not,not,six,six,six

If I misunderstood, I am sorry, but the prob. of rolling 3 sixes in a row is 1/216. But the other two are not sixes so (5/6)^2.

(1/6)^3(5/6)^2. This can happen in three ways in 5 rolls. So $3(\frac{1}{6})^{3}(\frac{5}{6})^{2})=\frac{25}{259 2}\approx{.00965}$

Again, if I am misunderstanding, I apologize.
• Jun 4th 2008, 12:32 PM
Plato
Quote:

Originally Posted by jgillett
What I want to know is if there's a formula that you can use to work out the probability of (for example) rolling a 6 three (or more) times in a row when you roll a dice five times (so 66_66 wouldn't be a desired result despite rolling 3 or more 6s).

The question is about at least three 6’s in row.
Therefore, this is not binominal probability but rather a counting problem.
I suggest counting three cases: exactly three in a row, exactly four in a row and then exactly five in a row.

For the first case we could have 666XX, X666X, or XX666 (where X is not a six). Each of those can happen is 25 ways.

For the second case we could have 6666X, X6666, or X6666 (where X is not a six). Each of those can happen is 5 ways.

The third case 66666 can happen in only one way.

Thus the probability is $\frac {75 + 15 + 1} {6^5}$.

On course, if we change the question to say, “at least three in a row in seven rolls” now the counting becomes more difficult because we could have 66X666X. We could have ZZX666X, where Z could be any of the six and X is not a six. That happens in (36)(25) ways. To further complicate matters we must count 666X666 as exactly three in a row.
• Jun 4th 2008, 12:33 PM
TheEmptySet
No, Think you are right I misunderstood the question. hmmm. Now I need to think about it again.

Thanks Galactus(Bow)