1. ## dice probability question

This is a revision question for a phase test that i could sure use some help with!

A dice game consitsts of throwing 2 six sided dice and noting the score on them. If the score is greater than 9 or at least one dice shows a 6 then you score 3 points, otherwise you lose 1 point. Calculate the probability of gaining 3 points on a single throw. What is the average (or expected) number of points you will gain per throw.

i make the probability of scoreing 3 points as 12 over 36, but i dont know how to do the rest.

any suggestions?

Cheers

2. Originally Posted by marciano1976
This is a revision question for a phase test that i could sure use some help with!

A dice game consitsts of throwing 2 six sided dice and noting the score on them. If the score is greater than 9 or at least one dice shows a 6 then you score 3 points, otherwise you lose 1 point. Calculate the probability of gaining 3 points on a single throw. What is the average (or expected) number of points you will gain per throw.

i make the probability of scoreing 3 points as 12 over 36, but i dont know how to do the rest.

any suggestions?

Cheers
Colour the die red and blue, then the combinations that win are (r6,b4), (r6,b5), (r6,b6), (r5,b5), (r5,b6), (r4,b6), (r6,b1), (r6,b2), (r6,b3), (r1,b6), (r2,b6), (r3,b6). That is 12 winning configurations out of a total of 36.

p(3)=12/36=1/3

p(1)=1-p(3)=2/3

So the expected number of points is:

(-1)p(1)+(3)p(3)=(-2/3)+(3/3)=1/3

RonL

3. Originally Posted by CaptainBlack
So the expected number of points is:
(1)p(1)+(3)p(3)=(2/3)+(3/3)=5/3
Surely that should be (-1)p(1)+(3)p(3)=(-2/3)+(3/3)=1/3

4. Hello, marciano1976!

A dice game consists of throwing two six-sided dice and noting the score on them.
If the score is greater than 9 or at least one dice shows a 6 then you score 3 points,
otherwise you lose 1 point.

(a)Calculate the probability of gaining 3 points on a single throw.

(b) What is the average (or expected) number of points you will gain per throw?

There are 36 possible outcomes . . .

$\displaystyle \begin{array}{cccccc}(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & {\color{blue}(1,6)} \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & {\color{blue}(2,6)} \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & {\color{blue}(3,6)} \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & {\color{blue}(4,6)} \\ (5,1) & (5,2) & (5,3) & (5,4) & {\color{blue}(5,5)} & {\color{blue}(5,6)} \end{array}$
$\displaystyle \begin{array}{cccccc}{\color{blue}(6,1)} & {\color{blue}(6,2)} & {\color{blue}(6,3)} & {\color{blue}(6,4)} & {\color{blue}(6,5)} & {\color{blue}(6,6)} \end{array}$

There are 12 in which the sum is greater than 9 or there is at least one 6.

Therefore: .$\displaystyle (a)\;\;P(\text{gain 3 points}) \:=\:\frac{12}{36} \:=\:\frac{1}{3}$

Hence: .$\displaystyle \begin{array}{ccc}P(\text{win 3 points}) &=& \frac{1}{3} \\ P(\text{lose 1 point}) & = & \frac{2}{3} \end{array}$

And: .$\displaystyle (b)\;\;\text{Average} \;=\;(+3)\left(\frac{1}{3}\right) + (-1)\left(\frac{2}{3}\right) \:=\:\frac{1}{3}$

You can expect to win an average of $\displaystyle \frac{1}{3}$ of a point per throw.

. . . as Plato already pointed out.
.

5. Originally Posted by Plato
Surely that should be (-1)p(1)+(3)p(3)=(-2/3)+(3/3)=1/3
Of course it should, I misread the question (again)

RonL