stuck on this problem

[student222]

I spend 2 and half hour trying to solve this problem but I got nothing, I try to look for example on probability books and I did not get any thing
so can any one help me by solving this complex problem.
I have a test next week and I really need to solve it.
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In a game RISK, two players compete in a game of dice rolling for conquest of the world. One player in on “offense” while the other is on “defense”. For this problem, the player on offense is allowed to roll multiple dice while the player on defense rolls a single die. Whoever rolls the higher number wins (i.e., the highest number rolled by the offense is compared with the number rolled by the defense). In case of a tie, the defense is declared the winner. The loser must remove one army from the board. Find the probability of the offense winning and the probability of the defense winning in each of the following scenarios:
(a) Both players roll only single die
(b) Offense rolls two dice while defense rolls one die
(c) Offense rolls three dice while defense rolls one die

 

[HallsofIvy]

I spend 2 and half hour trying to solve this problem but I got nothing, I try to look for example on probability books and I did not get any thing
so can any one help me by solving this complex problem.
I have a test next week and I really need to solve it.
------------------------------------------------------------------------
In a game RISK, two players compete in a game of dice rolling for conquest of the world. One player in on “offense” while the other is on “defense”. For this problem, the player on offense is allowed to roll multiple dice while the player on defense rolls a single die. Whoever rolls the higher number wins (i.e., the highest number rolled by the offense is compared with the number rolled by the defense). In case of a tie, the defense is declared the winner. The loser must remove one army from the board. Find the probability of the offense winning and the probability of the defense winning in each of the following scenarios:
(a) Both players roll only single die

This one, at least, is fairly straight forward. There are a total of 6x6= 36 ways two dice can come up. 6 of them are equal, 15 with offense larger than defense, 15 with defense larger than offense. The probability defense wins is (15+ 6)/36= 21/36= 7/12.

(b) Offense rolls two dice while defense rolls one die

Now there are 6x6x6= 216 possible outcomes. Six of them will be ties- all three die the same- leaving 210. Two thirds of them, 140, will have one of the first two dice higher than the third. One third of them, 70 will have the third die highest. The defender will win in 70+ 6= 76 out of the 216 rolls so the probability is \(\displaystyle \frac{76/216}= \frac{4(19)}{4(54)}= \frac{19}{54}\).

(c) Offense rolls three dice while defense rolls one die

I'll leave this one for you- there are 6x6x6x6= 1296 possible outcomes. Six of them will be ties, leaving 1296. How do you divide those among the four dice?

 

[student222]

This one, at least, is fairly straight forward. There are a total of 6x6= 36 ways two dice can come up. 6 of them are equal, 15 with offense larger than defense, 15 with defense larger than offense. The probability defense wins is (15+ 6)/36= 21/36= 7/12.


Now there are 6x6x6= 216 possible outcomes. Six of them will be ties- all three die the same- leaving 210. Two thirds of them, 140, will have one of the first two dice higher than the third. One third of them, 70 will have the third die highest. The defender will win in 70+ 6= 76 out of the 216 rolls so the probability is \(\displaystyle \frac{76/216}= \frac{4(19)}{4(54)}= \frac{19}{54}\).


I'll leave this one for you- there are 6x6x6x6= 1296 possible outcomes. Six of them will be ties, leaving 1296. How do you divide those among the four dice?

Thank you ... This really help me