# Math Help - probably wrong

1. ## probably wrong

Hi folks,

This probability question has me puzzled!

In order to start a game of chance a player throws a dice until he obtains a 6. He then records whatever score he obtains on subsequent throws. For example: Throws of 2,4,3,6,4,6,2,5 give recorded scores of 4,6,2,5.

Calculate the probability that
a) the first score recorded is that of the player's fourth throw
b) the player does not record a score in his first 5 throws

The player has 7 throws. Calculate the probability that he will have recorded

c) exactly three 5s and a 3

d) a total score of 3

So:The first bit is ok. If the first score recorded is on the 4th throw, then the 3rd throw must have been a six. For this to happen the first throw and second throw must not be six, so

P(6 on 3rd throw) = 5/6. 5/6. 1/6 = 25/216

The second bit is where I get stuck. The problem is with the wording. Does this mean that I am to assume a six is thrown on the 5th try, so that scores can be recorded from the 6th throw onwards or do I just assume that the first 5 throws do not produce a six?

i.e. P(no 6 in 5 throws) = 5/6.5/6.5/6.5/6.1/6.

or = 5/6.5/6.5/6.5/6.5/6

in fact neither give the correct answer! Any suggestions here?

For part c we have to calculate the probability of throwing:
x.x.6.3.5.5.5. where x is between 1 and 5
x.x.6.5.3.5.5.
x.x.6.5.5.3.5
x.x.6.5.5.5.3

P(score of 3) = 5/6.5/6.1/6.1/6.1/6.1/6.1/6.
since selecting any number on a dice is a 1/6. Then multiply by 4 to cover all the permutations. This gives the correct answer but now I am stuck on d!

To score 3, I reckon we have the following possibilities:

x.x.x.x.x.6.3 i.e. 5/6.5/6.5/6.5/6.5/6.1/6.1/6.

x.x.x.x.6.2.1 ditto

x.x.x.x.6.1.2 ditto

x.x.x.6.1.1.1 ditto.

Then add up the probabilities.This gives the wrong answer and makes me doubt whether my approach in part c is correct.

Any help with this would be appreciated!

best regards

2. Originally Posted by s_ingram
In order to start a game of chance a player throws a dice until he obtains a 6. He then records whatever score he obtains on subsequent throws. For example: Throws of 2,4,3,6,4,6,2,5 give recorded scores of 4,6,2,5.

Calculate the probability that
a) the first score recorded is that of the player's fourth throw
b) the player does not record a score in his first 5 throws
The correct amswer is $\left( {\frac{5}{6}} \right)^4$.
Notice that none of the first four is a six.
Even if the fifth throw is a six it still is not recorded.

3. Thanks Plato, that is the correct answer. I felt I had to account for the 5th dice. Now I see that I don't!

As you saw I got the correct answer for part c but I am still stuck on part d. Any idea how to do that bit?

4. Originally Posted by s_ingram
I am still stuck on part d. Any idea how to do that bit?
For the part (d), there are only three cases to consider.
1) He throws a first six on the fourth throw; then throws three ones.
2) He throws a first six on the fifth throw; then throws a one and a two or two and one.
3) He throws a first six on the sixth throw; then throws a three.

5. Exactly! And from my post you can see that I lined up the cases and showed how I calculated the probabilities:

So, P(x.x.x.x.x.6.3) = 5^5/6^7 (the x means anything except 6)

i.e. 5/6 for each x and 1/6 for a six or a 3 or a 5. Sorry about the notation here I haven't got round to LaTex yet!

Then P(x.x.x.x.6.1.2) = P(x.x.x.x.6.2.1) = 2 x 5^4 / 6^7

And P(x.x.x.6.1.1.1) = 3. 5^3 / 6^7.

My answer is 5^5 / 6^7 + 2.5^4 / 6^7 + 3.5^3 / 6^7

And like I say, I don't get the right answer (which is 5^3 / 6^5).

6. Originally Posted by s_ingram
And like I say, I don't get the right answer (which is 5^3 / 6^5).
First $P(xxx6111) = \frac{{5^3 }}{{6^7 }}$.

Then $\frac{{5^5 }}
{{6^7 }} + \frac{{2 \cdot 5^4 }}
{{6^7 }} + \frac{{5^3 }}
{{6^7 }} = \frac{{5^3 }}
{{6^7 }}\left( {25 + 10 + 1} \right) = \frac{{5^3 }}
{{6^5 }}$

7. Ahh!!! Now I see. Thanks, Plato.
Sorry just noticed something! In part c
P(x.x.6.3.5.5.5) = 4C3 5/6. 5/6. 1/6. 1/6. 1/6. 1/6. 1/6.
The 4C3 is because there are three 5s. For this reason I did the same with the three 1s when calculating P(x.x.x.6.1.1.1) that's why I was getting the wrong answer. So, may I ask you dear Plato, why is the 4C3 included in the first case but not in the second. It's details like this that make all the difference!

8. Originally Posted by s_ingram
Ahh!!! In part c P(x.x.6.3.5.5.5) = 4C3 5/6. 5/6. 1/6. 1/6. 1/6. 1/6. 1/6. The 4C3 is because there are three 5s. For this reason I did the same with the three 1s when calculating P(x.x.x.6.1.1.1) that's why I was getting the wrong answer. Why is the 4C3 included in the first case but not in the second.
The string "3555" can be arranged in $\frac{4!}{3!}$ ways.
There is only one way to arrange "111"