Jamie rolls a die multiple times. find the probability that she rolls her first 5 before she rolls her second even number

Question

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.

Answer 1

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, $$p = \frac{1}{3} p + \frac{1}{2} q + \frac{1}{6} \implies p = \frac{7}{16}$$

Answer 2

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 $5$?

Answer 3

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): $$ \begin{eqnarray*} P\left(A\right)&{}={}&\sum\limits_{k=1}^{\infty} \left(P\left(k\mbox{ rolls, no even, ending in }5\right) {}+{} P\left(k\mbox{ rolls, one even, ending in }5\right)\right) \newline &{}={}&\sum\limits_{k=1}^{\infty} \left(\frac{1}{6}\left(\frac{1}{3}\right)^{k-1}{}+{}\frac{1}{6}{k-1\choose 1}\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)^{k-2}\right) \newline &{}={}&\frac{1}{6}\sum\limits_{k=1}^{\infty} \left(\left(\frac{1}{3}\right)^{k-1}{}+{}\frac{1}{2}\left(k-1\right)\left(\frac{1}{3}\right)^{k-2}\right) \newline &{}={}&\frac{1}{6}\left(\frac{1}{\left(1-\frac{1}{3}\right)}{}+{}\frac{1}{2\left(1-\frac{1}{3}\right)^2}\right)\newline &{}={}&\frac{7}{16}\,. \end{eqnarray*} $$

Answer 4

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 $$ 1-\frac9{16}=\frac7{16} $$

Answer 5

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.