# probably wrong

• Jun 27th 2009, 02:15 PM
s_ingram
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
• Jun 27th 2009, 03:04 PM
Plato
Quote:

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.
• Jun 28th 2009, 02:18 AM
s_ingram
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?
• Jun 28th 2009, 03:28 AM
Plato
Quote:

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.
• Jun 28th 2009, 03:50 AM
s_ingram
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).
• Jun 28th 2009, 04:31 AM
Plato
Quote:

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 $\frac{{5^5 }}
{{6^7 }} + \frac{{2 \cdot 5^4 }}
{{6^7 }} + \frac{{5^3 }}
{{6^7 }} = \frac{{5^3 }}
{{6^7 }}\left( {25 + 10 + 1} \right) = \frac{{5^3 }}
{{6^5 }}$
• Jun 28th 2009, 04:41 AM
s_ingram
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!
• Jun 28th 2009, 05:18 AM
Plato
Quote:

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"