1. ## 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.?

2. 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.

3. 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, $\displaystyle P_{1}({5}\cup{6})=2\times \frac{1}{6}=\frac{2}{6}$

For 2 rolls, your first roll must fail. To fail you must get either of these in the set {$\displaystyle 1\cup2\cup3\cup4$}, which gives $\displaystyle P_{2,1}=\frac{4}{6}$, 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 $\displaystyle X_{2,1}$

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

Same for 3 rolls, 4 rolls, 5 rolls,...,$\displaystyle n\in N$, 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.

4. 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 $\displaystyle (1/6)^4$

5. 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 $\displaystyle (1/6)^4$
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 $\displaystyle q_1=\frac{4}{6}$.

To fail the second event, the probability will depend on what appeared on $\displaystyle q_1$. 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 $\displaystyle q_2=\frac{3}{6}$.

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 $\displaystyle q_3=\frac{2}{6}$

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

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