What is the Probability of Winning a Dice Game?

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In summary, the student got 1/9 correct on the first roll, 2/3 correct on the second roll, and 13/1458 correct on the third roll, for a total of 43/180.
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pretzel1998
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Hi There!

This is my first time posting on this forums, so I hope I am following the right etiquette.

Im having some severe difficulties with the problems under Question 3 in this past exam paper. http://www.nzqa.govt.nz/nqfdocs/ncea-resource/exams/2014/91267-exm-2014.pdf
Can someone please check my answers and explain the questions where I faceplanted? Thanks

3. (a), I got 1/9 which I am pretty sure is good.
3. (b), I got 2/3 which I am pretty sure is good.
3. (c,i). I got 1/81, I am not very confident about this answer because the wording of the question is incredibly bad. I did 4/36 X 4/36 as that's the probability he gets 5 on the first roll, and then 5 on the second roll to win?
3. (c,ii). I got 13/1458, I don't think I got this right because it seems like such a crazy probability. I did the probability he got the sum of 5 on the first roll (4/36), multiplied by the probability that he did not get a sum of 5 on the second roll (13/18 I think? I might be wrong), the multiplied again by the probability that he will get a sum of 5. so in total (4/36 X 13/18 X 4/36 = 13/1458)? Really confused about this question.
3. (d) This question I had absolutely no idea what to do. I some how got 7/18, but I think I am wrong.

Thanks so much!
 
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  • #2
pretzel1998 said:
Hi There!

This is my first time posting on this forums, so I hope I am following the right etiquette.

Im having some severe difficulties with the problems under Question 3 in this past exam paper. http://www.nzqa.govt.nz/nqfdocs/ncea-resource/exams/2014/91267-exm-2014.pdf
Can someone please check my answers and explain the questions where I faceplanted? Thanks

3. (a), I got 1/9 which I am pretty sure is good. I agree.
3. (b), I got 2/3 which I am pretty sure is good. I agree.
3. (c,i). I got 1/81, I am not very confident about this answer because the wording of the question is incredibly bad. I did 4/36 X 4/36 as that's the probability he gets 5 on the first roll, and then 5 on the second roll to win? I think that for part (c) you are meant to assume that he got a 5 on the first roll, so you only need to find the probability of scoring 5 on the second roll. If so, then the answer would be 1/9 rather than 1/81.
3. (c,ii). I got 13/1458, I don't think I got this right because it seems like such a crazy probability. I did the probability he got the sum of 5 on the first roll (4/36), multiplied by the probability that he did not get a sum of 5 on the second roll (13/18 I think? I might be wrong), the multiplied again by the probability that he will get a sum of 5. so in total (4/36 X 13/18 X 4/36 = 13/1458)? Really confused about this question. Here again, if you ignore the 1/9 that came from the first roll, you would get 13/162 rather than 13/1458, and I think that would be the answer they are looking for.
3. (d) This question I had absolutely no idea what to do. I some how got 7/18, but I think I am wrong.
On the first roll, M has a 2/9 probability of winning and W has a 1/9 probability. To make the game fair, the second roll must be arranged so that M's and W's overall probability of winning are both 1/2. The probability of a second roll being necessary is 2/3. If the second roll is arranged in such a way that M's probability of winning it is 15/36 (and W's probability of winning it is 21/36), then M's chances of winning the game are 2/9 (on the first roll) plus (2/3)x(15/36) (on the second roll). That comes to $\color{red}{\frac29 + \frac5{18} = \frac12}$, as required. So I think the answer should be 15/36.

Thanks so much!
...
 

FAQ: What is the Probability of Winning a Dice Game?

What is the probability of rolling a specific number on a single die?

The probability of rolling a specific number on a single die is 1/6 or approximately 16.67%. This is because there are six possible outcomes (numbers 1-6) and only one of those outcomes will result in the desired number.

What is the probability of rolling a specific number or higher on two dice?

The probability of rolling a specific number or higher on two dice can vary depending on the specific numbers and whether the roll is with replacement (the same die can be rolled more than once) or without replacement (each die can only be rolled once). Generally, for a roll with replacement, the probability would be (7 - desired number)/6, and for a roll without replacement, the probability would be (13 - desired number)/36.

What is the probability of rolling a specific combination on two dice?

The probability of rolling a specific combination on two dice can also vary depending on the specific combination and whether the roll is with or without replacement. For example, the probability of rolling a pair of 6s on two dice with replacement is 1/6, while the probability of rolling two 6s without replacement is 1/36.

How many different outcomes are possible when rolling two dice?

When rolling two dice, there are 36 possible outcomes (6 possible outcomes for the first die multiplied by 6 possible outcomes for the second die).

How does the probability change when rolling multiple dice?

The probability changes when rolling multiple dice as the number of possible outcomes increases. For example, when rolling three dice, the number of possible outcomes is 216 (6 x 6 x 6), and the probability of rolling a specific number or higher on all three dice would be (19 - desired number)/216. The more dice that are rolled, the more the probability approaches 1 (100%).

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