# 2 player game rolling a die

• Aug 31st 2009, 05:29 PM
Robb
2 player game rolling a die
Homer and Marge are about to play a game with the following rules. They take turns rolling a standard fair die, starting with Homer. The game ends when someone rolls a 5 or 6. That person wins the game.
(b)
Suppose that the game ends in a draw if ever two 1's come up in a row. With this change to the rules, find the probability that Homer will win the game.

• Sep 5th 2009, 03:56 AM
Robb
Hmm, anyone able to help with how to treat the draw?

I know that the probability of Homer winning (ignoring the new rule) is;
define W as winning, and N as not winning. Then $\displaystyle P(w)=1/3$ and [tex]P(N)=2/3[/math.
So for homer to win;
$\displaystyle P(w)=1/3$
$\displaystyle P(NNW)=\frac{2}{3}\cdot \frac{2}{3}\cdot \frac{1}{3}=\frac{4}{9}\cdot \frac{1}{3}$
$\displaystyle P(NNNNW)=\frac{2}{3}\cdot \frac{2}{3}\frac{2}{3}\cdot \frac{2}{3}\frac{1}{3}=(\frac{4}{9})^2\cdot \frac{1}{3}$
etc.
So the probability for homer to win is just the sum of these;
$\displaystyle P(H)=P(w)+P(NNW)+P(NNNNW)+....$
which is the sum of a geometric process, so $\displaystyle P(H)=\frac{1}{3} \cdot \frac{1}{1-\frac{4}{9} } =\frac{1}{3}\cdot \frac{9}{5}=\frac{3}{5}$

So now I just have no idea how to incorporate the draw into the game. I know that at the end of each players turn, there will be 3 states (1) they won, (2) they rolled a 1 (3) they rolled a 2, 3, 4. In state (1) the game ends, in state (2) The other player will roll, and if a 1 the game ends. and (3) the game continues if rolls a 1, and ends if roll a 6.
• Sep 5th 2009, 04:39 AM
pedrosorio
If you consider a Markov-Chain with the following states:

Homer won,
Marge won,
Homer rolled 1,
Marge rolled 1,
Homer rolled 2,3 or 4,
Marge rolled 2,3 or 4,
The game ended in a draw,

You can represent the transition probabilities as a matrix A and may find the answer to your question, by finding the limit (A^n) and multiplying that by the initial vector (which would be 0's on all positions and a 1 in the "Marge rolled 2,3 or 4").

Using Matlab and finding A^10000 (it converges), and then multiplying by the initial state vector I found that:

P(Homer winning) = 0.5647
P(Marge winning) = 0.3686
P(Draw) = 0.0667
• Sep 5th 2009, 05:02 AM
Robb
Hmm, I think I understand this now.. we still want to know the probability of homer winning, so we dont want to include the possibility of a draw occuring;

So for any roll;
$\displaystyle W \to \mbox{win (5,6)}; P(W)=\frac{1}{3}$
$\displaystyle N \to \mbox{not win (1,2,3,4)}; P(N)=\frac{2}{3}$
$\displaystyle D \to \mbox{draw(1)}; P(D)=\frac{1}{6}$
$\displaystyle ND \to \mbox{not win,but not roll a 1(2,3,4)}; P(ND)=\frac{1}{2}$

So the probability of homer winning is:
$\displaystyle P(H)=P(W)+P((N \cdot ND) \cdot W) + P((N \cdot ND) ( N \cdot ND) \cdot W)+...$
$\displaystyle P(H)=\frac{1}{3}+\left(\left(\frac{2}{3}\cdot \frac{1}{2}\right)\cdot \frac{1}{3}\right)+\left(\left(\frac{2}{3}\cdot \frac{1}{2}\right)^{2}\cdot \frac{1}{3}\right)+...=\frac{1}{3} \cdot \frac{1}{1-\frac{2}{6} }=\frac{1}{2}$

Thanks pedrosorio, but no matlab for this question.. :P
• Sep 5th 2009, 05:15 AM
pedrosorio
I think you're missing something in your calculation because the second term for example, should be:

P((ND * N) + (N * ND)) * P(W) , no?
• Sep 5th 2009, 05:39 AM
Robb
Hmm, yeah, i think you are on the right track.. I think what I have done is incorrect... as what i am saying is for the 2nd probablilty,
Homer rolls a 1,2,3,4 and marge rolls a 2,3,4...

The probability has to be less then what it was before the new rule was introduced.... How about;
P(H)=P(w)+P(D*ND+ND*D)*P(w)+P(D*ND+ND*D)^2*P(w)+.. .
P(H)=1/3+(1/6*1/2+1/2*1/6)*1/3+....=1/3*[1/(1-1/6)]=2/5

ie. this is saying homer rolls a 1, marge rolls a 2,3,4 or homer rolls a 2,3,4 and marge rolls a 1. This way for every roll a 1,2,3, or 4 is covered so homer/marge wont win on that roll and the game cannot end in a draw.
sorry for the lack of latex
• Sep 5th 2009, 06:09 AM
pedrosorio
P(H)=P(w)+P(D*ND+ND*D)*P(w) ...

This can't be, it must be

P(w)+P(N*ND + ND*N)*P(w)...

I agree that the probability of Homer winning should be under 3/5 for sure (because of the draw rule).

The probability of homer winning on the 3rd try would be:

(P(N)*P(ND)*P(ND)*P(N) + P(N)*P(ND)*P(N)*P(ND) + P(ND)*P(N)*P(ND)*P(N)) * P(w) = 3 * (P(N)*P(ND))^2 * P(w)

In general I think you'll have probability of winning on nth try:

n*(P(N)*P(ND))^(n-1) * P(w)
• Sep 6th 2009, 01:36 AM
Robb
argh, still have no idea how to solve this;
redifine X to be ND (ie. 2,3,4 is rolled), and D to be a 1 is rolled. Then homer can win if the following rolls happen:

Homer wins on his first roll;
W
Homer wins on his second roll;
XXW
XDW
DXW
Homer wins on his third roll;
XXXXW
XXXDW
XXDXW
XDXXW
XDXDW
DXXXW
DXXDW
DXDXW
Homer wins on his fourth roll;
XXXXXXW
XXXXXDW
XXXXDXW
XXDXXXW
XXDXXDW
XXDXDXW
XXXDXXW
XXXDXDW
XDXXXXW
XDXXXDW
XDXXDXW
XDXDXXW
XDXDXDW
DXXXXXW
DXXXXDW
DXXXDXW
DXDXXXW
DXDXXDW
DXDXDXW
DXXDXXW
DXXDXDW

etc. :( I cant work out how to calculate a recursive formula for this, or see if its a geometric series. I tried working out the probability of a draw but still cant work it out...
• Sep 9th 2009, 05:04 AM
centuryletters
Hi Robb, I'm doing the same problem as you, are you by any chance at ANU?