1. ## die rolling problem

hey guys,
just having trouble with this question.

if 2 people play a dice game and each rolls a dice, if A rolls a higher dice then B then A wins and vice versa. if they roll the same number its a draw.

What is the probability that A wins if they play a single game?
it wouldnt be 1/2 i dont think because they can draw, but both A and B's chance of winning are the same?

Also if they play 100 games what is the probability A wins 50 or more games?
would i just use the probability from the question above and do like the
P(50<X<100) binomial distribution thing.

thanks

2. Hmmm... I'd be inclined to say that the probability player A wins in a single game is [tex]\dfrac{1}{3}[tex].

My reasoning, consider it like a game of "Rock, Paper, Scissors". There's only three possible outcomes: Winning, Losing and Drawing and each of these has an equal chances of occuring. Say A rolls a 2, B has to roll a 3 or greater to win but adversely, B could end up in the same situation.

But this is only speculation. I'm gonna look into a bit more.

As for the second part, treat the game as if it only had two options: Winning and lossing, where winning has $\dfrac{1}{3}$ chance of occuring and losing has $\dfrac{2}{3}$ chance.

It's been a long time since I've done statistics, I'm afraid so sorry I can't be of more use.

3. Originally Posted by b0mb3rz
hey guys,
just having trouble with this question.

if 2 people play a dice game and each rolls a dice, if A rolls a higher dice then B then A wins and vice versa. if they roll the same number its a draw.

What is the probability that A wins if they play a single game?
it wouldnt be 1/2 i dont think because they can draw, but both A and B's chance of winning are the same?

Also if they play 100 games what is the probability A wins 50 or more games?
would i just use the probability from the question above and do like the
P(50<X<100) binomial distribution thing.

thanks

First one: Conditonal probabilites
Let W be the event that A wins and the notation that $B_1,B_2...B_6$ is the event that player B rolls a 1,2,3..6

Then
$P(W)=P(W|B_1)P(B_1)+P(W|B_2)P(B_2)+...P(W|B_6)P(B_ 6)$

Now we need to compute the above probabilities.

first $P(B_i)=\frac{1}{6}$ for $1 \le i \le 6$

Now $P(W|B_1)$ is the event that A wins given B rolled a 1. This event has prob $=\frac{5}{6}$.

$P(W|B_2)$ is the event that A wins given B rolled a 2. This event has prob $=\frac{4}{6}$. The pattern continues this gives

$P(W)=\frac{5}{6}\frac{1}{6} +\frac{4}{6}\frac{1}{6}+\frac{3}{6}\frac{1}{6}+\fr ac{2}{6}\frac{1}{6}+\frac{1}{6}\frac{1}{6}+\frac{0 }{6}\frac{1}{6}=\frac{5+4+3+2+1}{36}=\frac{15}{36}$

The 2nd way to get this result is to make a table of all possible outcomes.
Let the ordered pair (A,B) be the number A and B rolled respectivly then

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Now we can just count the out comes there are 36 possible events and
A > B only on the bottom triangle below the diagonal for a total of 15
giving the same as the probability above
$\frac{15}{36}$