# fair die probability

• Oct 23rd 2009, 08:37 AM
mhitch03
fair die probability
A fair die is rolled over and over until the sum of all the numbers obtained is greater than 4. What is the probability that this takes 1 roll or 2 rolls or 3 rolls or 4 rolls or 5 rolls, etc.?
• Oct 23rd 2009, 10:24 AM
jalko
To complete task in one roll you have to get "5" or "6", so there are 2 possible outcomes that are correct, but there are total of 6 possible outcomes. So probability have to be some combination of these numbers, you can roll "5", so one number from six and also roll "6", so again one number from six. In total you can roll 2/6 numbers to be successful.

In two rolls you can get 6*6 possible outcomes, how many times you can be succesful? just write it down and you will see the answer. Also for 3 and 4 rolls. Maybe start with 5 rolls and you will see it imediately.
• Oct 23rd 2009, 10:53 AM
novice
Quote:

Originally Posted by mhitch03
A fair die is rolled over and over until the sum of all the numbers obtained is greater than 4. What is the probability that this takes 1 roll or 2 rolls or 3 rolls or 4 rolls or 5 rolls, etc.?

For 1 roll you will need a 5 or 6, $P_{1}({5}\cup{6})=2\times \frac{1}{6}=\frac{2}{6}$

For 2 rolls, your first roll must fail. To fail you must get either of these in the set { $1\cup2\cup3\cup4$}, which gives $P_{2,1}=\frac{4}{6}$, given that it happened before proceeding to the subsequent roll. For the subsequent roll immediately to this, you will get the probability given that the first roll was X=1, 2, 3, or 4. This seems to be and open ended question for we do not know not what comes first; we only know the probability only one has already occured. The following is the list of all four possible first numbers denoted as $X_{2,1}$

$\begin{array}{ccc}X_{2,1}&S&p_{2,2}(X)\\
1&{4,5,6}&{3/6}\\
2&{3,4,5,6}&{4/6}\\
3&{2,3,4,5,6}&{5/6}\\
4&{1,2,3,4,5,6,}&1\end{array}$

Same for 3 rolls, 4 rolls, 5 rolls,..., $n\in N$, repeat the process.

In your case, at worse, you will need 5 rolls, at which you get {1,1,1,1,1}

Note: I have re-evaluated this post many times and concluded that it is labor intensive to answer this question. It appeared rather innocent when I first attempted, but as I comtemplated, I found it more of a busy work than brain work.
• Oct 23rd 2009, 03:25 PM
matheagle
This is off a bit...........
In your case, at worse, you will need 5 rolls, at which you get {1,1,1,1,1}

In the case of five rolls you must have ...{1,1,1,1,anything}

that would be $(1/6)^4$
• Oct 23rd 2009, 05:32 PM
novice
Quote:

Originally Posted by matheagle
This is off a bit...........
In your case, at worse, you will need 5 rolls, at which you get {1,1,1,1,1}

In the case of five rolls you must have ...{1,1,1,1,anything}

that would be $(1/6)^4$

Matheagle, may I strike a bargain with you? Here is my take:

The reason for taking five rolls to succeed is that the first 4 events failed. Let q denotes the probability of failure, and p the probability of success.

To fail the first, we need 1 or 2 or 3 or 4, from which $q_1=\frac{4}{6}$.

To fail the second event, the probability will depend on what appeared on $q_1$. If indeed "1" appeared, 4 is the smallest number we must get subsequently. To fail, we need to get 1, 2, or 3, from which $q_2=\frac{3}{6}$.

To fail the third event, if indeed "1" had just appear the second round, this time we need 1, or 2 to give us failure, from which $q_3=\frac{2}{6}$

Segue, to fail the forth, $q_4=\frac{1}{6}$, and of course we will succeed on the fifth attemp, at which $p_5=\frac{6}{6}$.

Consequently, were we to take five attempts to succeed, proveded that we have got a string of 1's in the last 4 attempts, $p=q_1q_2q_3q_4p_5$, where $p=\left(\frac{4}{6}\right)\left(\frac{3}{6}\right) \left(\frac{2}{6}\right)\left(\frac{1}{6}\right)\l eft(\frac{6}{6}\right)=\frac{1}{324}$