# The Dice Game

• Nov 26th 2011, 01:06 PM
sothatwastaken
The Dice Game
Hello everyone! I was going through my sister's math practice sets and I found a probability question that is very challenging for me. I am trying to find a solution, but it is proving to be very difficult. Though this problem is from a university paper, I think it is much closer to High School Probability. Hopefully, someone can clear my concepts here. The question is as follows:

1. Consider a game with two players, Jim and Annie. Annie has a red die and Jim has a white die. They roll their dice and note the number on the upper face. Annie wins if her score is higher than Jim's (note that Jim wins if the scores are the same). If both players roll their dice once each what is the probability that Annie will win the game? (easy)

2. Now consider the same game where Annie can roll her die a second time and will note the higher score of the two rolls but Jim rolls only once. In this case, what is the probability that Annie will win? (easy, again )

3. In the case when both players can roll their dice twice, what is the Probability that Annie wins? (easy)

4. Investigate the game when they can roll the dice more than twice, but not necessarily the same number of times. (difficult )

I worked out the answers to 1. 2. and 3. as 15/36, 125/216 and 505/1296 respectively. However, part four has me stumped. Can anyone please point me in the right direction with this?
• Nov 26th 2011, 01:16 PM
Plato
Re: The Dice Game
Quote:

Originally Posted by sothatwastaken
3. In the case when both players can roll their dice twice, what is the Probability that Annie wins?

In case #3, how does Annie win?
Is that the max of her two outcomes is greater that either of his outcomes?
Or how?
• Nov 26th 2011, 01:24 PM
sothatwastaken
Re: The Dice Game
In case 2*, Annie can roll the dice twice and the higher value will be considered. Whereas Jim can roll the dice only once. If the higher of Annie's score is greater than Jim's score, then she win's.

I took three cases AAJ, AJA and JAA (order of them rolling the die), each time, probability of Annie winning is 125/216.

Example, for AAJ: Annie rolls 2, Annie rolls 5, Jim rolls 4 -> Annie wins.
But if Annie rolls 6, Annie rolls 2, Jim rolls 6 -> Annie Loses (original condition)

*EDIT: Oops, look like I explained the wrong thing :p
Anyways, here's the situation for part 3,
When they both roll their die twice, only the higher of the two values if considered for each. Example, if they throw in the order AJAJ and it is as: Annie rolls a 4, Jim rolls a 5, Annie rolls a 6, Jim rolls a 2 -> Annie wins because the higher of her two throws (6) is greater than the higher of Jim's two throws (5)

Using a very arduous process (trees), I found the value of this to be 505/1296
• Nov 26th 2011, 01:29 PM
Plato
Re: The Dice Game
Quote:

Originally Posted by sothatwastaken
In case 3, Annie can roll the dice twice and the higher value will be considered. Whereas Jim can roll the dice only once. If the higher of Annie's score is greater than Jim's score, then she win's.

That is not what is written. That is case #2.
In case #3, both roll twice.
• Nov 26th 2011, 01:30 PM
sothatwastaken
Re: The Dice Game
I am sorry :( Edited my previous post to reflect the correct statement.
• Nov 26th 2011, 03:16 PM
Plato
Re: The Dice Game
Quote:

Originally Posted by sothatwastaken
1. Consider a game with two players, Jim and Annie. Annie has a red die and Jim has a white die. They roll their dice and note the number on the upper face. Annie wins if her score is higher than Jim's (note that Jim wins if the scores are the same). If both players roll their dice once each what is the probability that Annie will win the game? (easy)

2. Now consider the same game where Annie can roll her die a second time and will note the higher score of the two rolls but Jim rolls only once. In this case, what is the probability that Annie will win? (easy, again )

3. In the case when both players can roll their dice twice, what is the Probability that Annie wins? (easy)

4. Investigate the game when they can roll the dice more than twice, but not necessarily the same number of times. (difficult )
I worked out the answers to 1. 2. and 3. as 15/36, 125/216 and 505/1296 respectively.

You are correct on the first three.
I think that #4 is far too vague the answer.
Place limits on the number of possible rolls.
Say up to four. That is enough to give a real challenge.
In the case of four, there are sixteen possible of scenarios to calculate.
• Nov 26th 2011, 11:58 PM
sothatwastaken
Re: The Dice Game
Thanks for checking my results on the first three Plato! Yes, when they roll the dice four times, total possibilities are 16 (Worried) - which is so much calculation.

I tried to find a pattern if they roll the dice the same number of times ...

If they both roll the dice 1 time, denominator = 6^2
If they both roll the dice 2 times, denominator = 6^4
If they both roll the dice 3 times, denominator = 6^6
.
.
.

I was hoping to find a pattern between 15, 125 and 505 (the numerator of the fractions when the dice is rolled 1, 2, 3 times respectively by each player) but I can't see anything concrete. Any ideas?