# Thread: Probability of throwing two dice

1. ## Probability of throwing two dice

• Consider throwing two dice

What is the probability an 8 will be rolled before a 7 is rolled?
What is the probability the player will either: roll a 7 or 11; or (if not) roll a 4, 5, 6, 8, 9 or 10 AND THEN repeat that same number before a 7 is rolled?
Any help will be appreciated.
Thank you,

Consider

2. hi August80,

if you mean "what is the probability of rolling a combined 8 followed immediately by rolling a combined 7",

then the probability of rolling an 8 is $\frac{5}{36}$

while the probabilty of rolling a 7 is $\frac{6}{36}$

as there are 5 ways to roll an 8 ....... 62 53 44 35 26
and 6 ways to roll a 7 .................... 61 52 43 34 25 16

multiply the probabities to find the probability of an 8 before a 7 (or a 7 before an 8)

since the sequences are 6261 6252 6243 6234 6225 6216 and similarly for the other ways starting 5, 4, 3 and 2.

That's 6(5)=30 ways out of $6^4$

which means the probability is $\frac{30}{6^4}=\frac{5}{36}\ \frac{6}{36}=\frac{5}{6^3}$

3. Thank you very much Archie!

Best,
August80

4. Originally Posted by August80
• Consider throwing two dice

What is the probability the player will either: roll a 7 or 11; or (if not) roll a 4, 5, 6, 8, 9 or 10 AND THEN repeat that same number before a 7 is rolled?
Any help will be appreciated.
Thank you,

Consider
In rolling a 7 or 11, we sum their individual probabilities.

7 = 16, 25, 34, 43, 52, 61
11=65, 56

So the probability of rolling a 7 or 11 is $\frac{8}{36}=\frac{2}{9}$

For the last one, you are seeking the probability of the following sequences arising...

4-4-7, 5-5-7, 6-6-7, 8-8-7, 9-9-7, 10-10-7

For these the denominator will be $6^6$

As an example, 10 can arise in the following ways....
64, 55, 46

10-10 can arise in the following ways...
6464, 6455, 6446,
5564, 5555, 5546,
4664, 4655, 4646

Since 7 can arise in 6 ways, there are 6(3)(3)=54 ways to get 10-10-7

so it's probabilty is $\frac{54}{6^6}$ or $\frac{9}{6^5}$

4 can occur as follows....
13, 22, 31

so you could try the rest as it's the same procedure and let us know how you get on if you like.

5. Thanks Again Archie,

I definetely will work on this and I'll let you know!

6. Originally Posted by August80
• Consider throwing two dice

What is the probability an 8 will be rolled before a 7 is rolled?
What is the probability the player will either: roll a 7 or 11; or (if not) roll a 4, 5, 6, 8, 9 or 10 AND THEN repeat that same number before a 7 is rolled?
Any help will be appreciated.
Thank you,

Consider
The probability of rolling an 8 is 5/36; the probability of rolling a 7 is 6/36; and the probability of rolling something else (neither a 7 nor an 8) is 25/36.

In order to roll an 8 before a 7, you must roll a sequence of zero or more "others" (neither 7s nor 8s), followed by an 8.

The probability that the final 8 occurs on the first roll is 5/36.

The probability that the final 8 occurs on the second roll is (25/36) (5/36).

The probability that the final 8 occurs on the third roll is (25/36) (25/36) (5/36).

And in general, the probability that the final 8 occurs on the nth roll is
$(25/36)^{n-1} \cdot (5/36)$.

Summing over all the possibilities for the final 8 to occur (there are infinitely many of them), the total probability of rolling an 8 before a 7 is

$(5/36) + (25/36) (5/36) + (25/36)^2 (5/36) + (25/36)^3 (5/36) + \dots$
$= (5/36) \; [1 + 25/36 + (25/36)^2 + (25/36)^3 + \dots]$
$=\frac{5}{36}\cdot \frac{1}{1 - 25/36}$
$= \frac{5}{11}$

where we have used the formula for the sum of an infinite geometric series.

7. Thank you very much awkward,

Best,
August80

8. Hello, August80!

I totally agree with awkward.
Here's my approach . . .

Consider throwing two dice
What is the probability an 8 will be rolled before a 7 is rolled?
There are: . $6^2 = 36$ possible outcomes with a pair of dice.

There are 6 ways to roll a "7": . $P(7) \,=\,\frac{6}{36}$
. . There are 5 ways to roll an "8": . $P(8) \,=\,\frac{5}{36}$
. . There are 25 ways for any Other sum: . $P(\text{Other}) \:=\:\frac{25}{36}$

He could get an "8" on his first roll.
. . $P(\text{8 on 1st roll}) \:=\:\frac{5}{36}$

He could get an "8" on his second roll.
His first roll must be an Other.
. . $P(\text{8 on 2nd roll}) \:=\:\left(\frac{25}{36}\right)\cdot\frac{5}{36}$

He could get an "8" on his third roll.
His first two rolls must be Others.
. . $P(\text{8 on 3rd roll}) \:=\:\left(\frac{25}{36}\right)^2\cdot\frac{5}{36}$

He could get an "8" on his fourth roll.
His first three rolls must be Others.
. . $P(\text{8 on 4th roll}) \:=\:\left(\frac{25}{36}\right)^3\cdot\frac{5}{36}$

And so on . . . we see the pattern, right?

Hence: . $P(\text{8 before 7}) \;=\;\frac{5}{36} + \left(\frac{25}{36}\right) + \frac{5}{36}\left(\frac{25}{36}\right)^2 + \frac{5}{36}\left(\frac{25}{36}\right)^3 + \hdots$

. . . . . . . . . . . . . . . $=\;\frac{5}{36}\underbrace{\bigg[1 + \frac{25}{36} + \left(\frac{25}{36}\right)^2 + \left(\frac{25}{36}\right)^3 + \hdots \bigg]}_{\text{geometric series}}$

The geometric series has the sum: . $\frac{1}{1-\frac{25}{36}} \:=\:\frac{1}{\frac{11}{36}} \:=\:\frac{36}{11}$

Therefore: . $P(\text{8 before 7}) \;=\;\frac{5}{36}\cdot\frac{36}{11} \;=\;\frac{5}{11}$

9. Thanks Soroban for your approach!

I appreciate it!