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Jamie rolls her fair 6-sided die multiple times.

Find the probability that she rolls her first 5 before she rolls her second (not necessarily distinct) even number?

This is what I have so far...

the probability that she rolls a 5 is $\frac{1}{6}$ the probability that she rolls an even number if $\frac{1}{2}$

Now I'm stuck...can anyone help me get to the final product? please be clear and concise.

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  • $\begingroup$ Look into the geometric distribution. This models the number of attempts before a "success" that has a fixed probabilty. For example, the number of times you need to roll a die before a 5 comes up is represented by a geometric distribution with $p=1/6$. $\endgroup$ – KSmarts Dec 19 '14 at 20:17
  • $\begingroup$ @KSmarts i don't believe i've done geometric distribution before. algebraic distribution, yes. geometric distribution, no. is there another method? $\endgroup$ – Math is Life Dec 19 '14 at 20:20
  • $\begingroup$ Geometric distribution and algebraic distribution are actually two completely different things. The distributive property used in algebra is used in multiplying polynomials and the like, while the geometric distribution is a probability distribution, which represents the possible values of a random variable. $\endgroup$ – KSmarts Dec 19 '14 at 20:24
  • $\begingroup$ @KSmarts I see. Can you provide an example? $\endgroup$ – Math is Life Dec 19 '14 at 20:27
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First, we compute the probability that Jamie rolls a 5 before she rolls an even number. Let this probability equal $q$.

We can write down an equation involving $q$ by considering the different things that could happen on Jamie's first roll. If she rolls a $5$ on her first roll, the game ends and she "wins"; this happens with probability $1/6$. If she rolls an even number on her first roll, the game ends and she "loses"; this happens with probability $1/2$. Otherwise, the game restarts. Thus, $$q = \frac{1}{6} \cdot 1 + \frac{1}{2} \cdot 0 + \frac{1}{3} q \implies q = \frac{1}{4}$$

Now, let $p$ be the probability that Jamie rolls a 5 before she rolls her second even number. Again, we look at cases. If Jamie rolls a 1 or 3, the game resets; this happens with probability $1/3$. If she rolls an even number, then we're reduced to the game that Jamie wins with probability $q$; this happens $1/2$ of the time. And last, if she rolls a 5, then she just wins, and the game ends. So, $$p = \frac{1}{3} p + \frac{1}{2} q + \frac{1}{6} \implies p = \frac{7}{16}$$

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Hints:

It's probably easier to find the probability of the complement: two even numbers before a $5$.

Consider rolling until she gets an even number or a $5$ (so ignore any $1$s or $3$s). Since there are three possible even numbers and only one $5$, what is the probability that she gets an even before a $5$?

Then what's the probability that she gets two evens before a $5$?

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  • $\begingroup$ probability that she gets an even before a 5 is 1/3*1/6 = 1/18? and probability that she gets TWO evens before a 5 is..1/3*1/3*1/6 = 1/54? $\endgroup$ – Math is Life Dec 19 '14 at 20:36
  • $\begingroup$ You're going to keep rolling until you get an element of the set $\{2,4,5,6\}$. What's the probability that you get an even when you pick one element from that set? $\endgroup$ – paw88789 Dec 19 '14 at 20:41
  • $\begingroup$ 3/4. probability that she gets an even before a 5 i 3/4*1/4 = 3/16. probability that she gets TWO even before a 5 is 3/4*3/4*1/4 = 9/64. 3/16+9/64 = 21/64. 1-21/64 = 23/64? $\endgroup$ – Math is Life Dec 19 '14 at 20:42
  • $\begingroup$ @paw88789: I just saw that your hint leads to my answer. I should probably delete it. The only thing is that the OP accepted another answer; I don't know if they will want to think to see how nicely this hint works. Perhaps I should leave my answer up for just that reason. $\endgroup$ – robjohn Dec 19 '14 at 22:36
  • $\begingroup$ @robjohn, I'd say not to worry about it. :-) But thanks for checking $\endgroup$ – paw88789 Dec 19 '14 at 23:34
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You roll the dice until you roll an even number or a five. Then you write down the letter E or a 5. You continue until you roll another even number or a five. Again, you write down the letter E or a 5.

Everytime you write something down, the probability is 3/4 that you write an E, and 1/4 that you write a 5. The chance that you wrote down two Es is 9/16. The probability that you didn't write two Es is 7/16, and that's what we were asked about.

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  • $\begingroup$ Sorry about the edit. The things on the screen were moving about and I hit the wrong post to edit. I have rolled back the changes. $\endgroup$ – robjohn Dec 19 '14 at 22:41
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I just noticed that this follows directly from paw88789's excellent hint. However, just to get attention to that hint, perhaps we need to see how nicely it works:

Given that the only rolls that count are $2,4,5,6$, what is the probability that two evens are rolled first? $\left(\frac34\right)^2=\frac9{16}$. Thus, the probability that the $5$ comes before the second even face is $$ 1-\frac9{16}=\frac7{16} $$

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  • $\begingroup$ Very close to paw88789's hint $\endgroup$ – robjohn Dec 19 '14 at 22:32
  • $\begingroup$ Why is 5's 1/6 probability not included? $\endgroup$ – Math is Life Dec 19 '14 at 22:44
  • $\begingroup$ @MathisLife: the only faces we are considering are $2,4,5,6$ and ignoring the others. Each of those faces have the same probability, so they each show up with probability $\frac14$. By ignoring the $1,3$, it is like we are rolling a 4 sided die. $\endgroup$ – robjohn Dec 19 '14 at 22:49
  • $\begingroup$ Ok. I get it now...if the only faces are 2, 4, 5, 6, and the number is not even, it is a given that that face is 5. $\endgroup$ – Math is Life Dec 19 '14 at 23:39
  • $\begingroup$ @MathisLife: yes; given that a $2,4,5,$ or $6$ comes up there is a $\frac34$ chance that it is a $2,4,$ or $6$ and a $\frac14$ chance that it is a $5$. $\endgroup$ – robjohn Dec 19 '14 at 23:43
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Consider the following two events:

  1. $A:=\left\{\mbox{sequence of rolls containing one or no even number, and ending in a 5}\right\}$
  2. $B:=\left\{\mbox{sequence of rolls containing some even number}\right\}$

The probability of interest, in a sequence of independent dice rolls, is $$ P\left(A \mbox{ followed by } B\right){}={}P\left(B\, |\, A\right)P\left(A\right)\,. $$

Note that, because the rolls are independent and the event $B$ is an almost-sure event (that is, the probability of it not occurring is zero), we have $$ P\left( B \, | \, A\right){}={}P\left( B \right){}={} 1\,. $$

So, $$ P\left(A \mbox{ followed by } B\right){}={}P\left(A\right)\,. $$

We proceed, therefore, to compute the event $A$ as follows (making liberal use of independence): $$ \begin{eqnarray*} P\left(A\right)&{}={}&\sum\limits_{k=1}^{\infty} \left(P\left(k\mbox{ rolls, no even, ending in }5\right) {}+{} P\left(k\mbox{ rolls, one even, ending in }5\right)\right) \newline &{}={}&\sum\limits_{k=1}^{\infty} \left(\frac{1}{6}\left(\frac{1}{3}\right)^{k-1}{}+{}\frac{1}{6}{k-1\choose 1}\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)^{k-2}\right) \newline &{}={}&\frac{1}{6}\sum\limits_{k=1}^{\infty} \left(\left(\frac{1}{3}\right)^{k-1}{}+{}\frac{1}{2}\left(k-1\right)\left(\frac{1}{3}\right)^{k-2}\right) \newline &{}={}&\frac{1}{6}\left(\frac{1}{\left(1-\frac{1}{3}\right)}{}+{}\frac{1}{2\left(1-\frac{1}{3}\right)^2}\right)\newline &{}={}&\frac{7}{16}\,. \end{eqnarray*} $$

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