# Rolling two dice over and over

• July 11th 2010, 04:48 AM
Hitman6267
Rolling two dice over and over
We have one 4 sided die and a 6 sided one. We are tossing them together and observing their sum.
X: sum of the dice
From a previous part
P(4)= 3/24
P(7)= 4/24

Now we are throwing the dice over and over what is the probability that a sum of 4 appears before a sum of 7 ?
How do I tackle this kind of a problem ?
• July 11th 2010, 05:53 AM
SpringFan25
There are 3 outcomes at every trial, 4,7 and something else. I will denote "something else" by 0. It has a probability of 17/24.

The only way that a 4 can appear before a 7 is if you roll it on the first go; or you roll 0 a few times and then get a 4. So you are interested in the followign events

P(4)
+P(0,4)
+P(0,0,4)
+P(0,0,0,4)
.....

which have probabilities
$P(4) = \frac{3}{24}$

$P(0,4) = \frac{17}{24} \times \frac{3}{24}$

$P(0,0,4) = \left(\frac{17}{24})^2 \times \frac{3}{24}$

$P(0,0,0,4) = \left(\frac{17}{24})^3 \times \frac{3}{24}$
...

You can recognise the above as a geometric progression, with first term $\frac{3}{24}$ ; and common ratio $\frac{17}{24}$. use the standard formula for the sum of an infinite geometric progression

$P(4~before~7) =\frac{\frac {3}{24}}{1-\frac{17}{24}} = \frac{3}{7}$

The fact that the answer is 3/7 may not be surprising. Can you see why?

Spoiler:

3/7 is the relative likelihood of 4 compared to 7.

$
P(4|4~or~7) = \frac{\frac{3}{24}}{\frac{3}{24} + \frac{4}{24}} = \frac{3}{7}
$

so, at whatever point we stop rolling 0 we have a 3/7 chance of getting a 4.
• July 11th 2010, 06:04 AM
Soroban
Hello, Hitman6267!

Same solution as Springfan25 . . . slightly different wording.

[After all this typing, I flatly refuse to delete this.]

Quote:

We have one 4-sided die and a 6-sided die.
We toss them together and observe their sum.

From a previous part

. . $\begin{array}{ccc}P(\text{sum of 4}) &=& \frac{3}{24} \\ \\[-3mm] P(\text{sum of 7}) &=& \frac{4}{24} \end{array}$

We throw the dice repeatedly.
What is the probability that a sum of 4 appears before a sum of 7 ?

We have these probabilities: . $\begin{array}{ccc}P(\text{sum of 4}) &=& \frac{1}{8} \\ \\[-4mm]
P(\text{sum of 7}) &=& \frac{1}{6} \\ \\[-4mm]
P(\text{other}) &=& \frac{17}{24} \end{array}$

A sum of 4 could appear on the first roll:
. . $P(\text{4 on 1st roll}) \:=\:\frac{1}{8}$

A sum of 4 could appear on the second roll:
. . $P(\text{4 on 2nd roll}) \:=\:P(\text{other, 4}) \:=\:\left(\frac{17}{24}\right)\left(\frac{1}{8}\r ight)$

A sum of 4 could appear on the third roll:
. . ${(\text{4 on 3rd roll}) \:=\:P(\text{other, other, 4}) \:=\:\left(\frac{17}{24}\right)^2\left(\frac{1}{8} \right)$

A sum of 4 could appear on the fourth roll:
. . $P(\text{4 on 4th roll}) \:=\:P(\text{other, other, other, 4}) \:=\:\left(\frac{17}{24}\right)^3\left(\frac{1}{8} \right)$

. . . and so on.

Hence: . $P(\text{4 appears before 7}) \;=\;\frac{1}{8} + \frac{1}{8}\left(\frac{17}{24}\right) + \frac{1}{8}\left(\frac{17}{24}\right)^2 + \frac{1}{8}\left(\frac{17}{24}\right)^3 + \hdots$

. . . . . . . . $=\;\frac{1}{8}\underbrace{\bigg[1 + \tfrac{17}{24} + \left(\tfrac{17}{24}\right)^2 + \left(\tfrac{17}{24}\right)^3 + \hdots \bigg]}_{\text{geometric series}}$

The sum of the geometric series is: . $\frac{1}{1-\frac{17}{24}} \:=\:\dfrac{1}{\frac{7}{24}} \:=\:\frac{24}{7}$

Therefore: . $P(\text{4 before 7}) \:=\:\dfrac{1}{8}\cdot\dfrac{24}{7} \;=\;\dfrac{3}{7}$

• July 11th 2010, 06:09 AM
Hitman6267
Thank you :)