Winning At A Simple Game of Dice

The simplest way to solve this problem is by noting that the probability that David's roll is larger than Calvin's is the same as the probability that Calvin's roll is larger than David's. Therefore, the desired probability is $1$ minus the probability of the rolls being equal all divided by $2$ .

The probability of the rolls being equal is $\frac{6}{6\cdot6}=\frac{1}{6}$ . The desired probability is then $\frac{1-\frac{1}{6}}{2}$ = $\frac{5}{12}$ . Adding $5$ and $12$ gives us $\boxed{17}$ .

Total sample space is $6\times 6=36$ . Among these $36$ cases in $6$ cases both the numbers are same.

Total favorable cases are $15=\frac{36-6}{2}$ ( when David's dice is bigger than the number on Calvin's dice).

So the probability is $\frac{15}{36}=\frac{5}{12}$ .

Answer is $5+12=\boxed{17}$

In how many ways can the number on David's die be larger than the number on Calvin's?

1. If David's die shows $6$ , there are 5 ways (Calvin's die can show $1, 2, 3, 4, 5$ )

2. If David's die shows $5$ , there are 4 ways (Calvin's die can show $1, 2, 3, 4$ )

3. If David's die shows $4$ , there are 3 ways (Calvin's die can show $1, 2, 3$ )

4. If David's die shows $3$ , there are 2 ways (Calvin's die can show $1, 2$ )

5. If David's die shows $2$ , there is 1 way (Calvin's die can show $1$ )

Hence, this can happen in $5+4+3+2+1 =15$ ways.

The total number of possibilities is given by $6^{2} = 36$ .

$P_{D>C} = \dfrac{15}{36} = \dfrac{5}{12} \Longrightarrow 5+12 = \boxed{17}$

There are three types of rolls here, $(a,b) , (a,a)$ and $(b,a)$ where $b>a$ . There are six $(a,a)$ pairs - (1,1), (2,2), (3,3) ... (6,6). That leaves 30 $(a,b)$ and $(b,a)$ pairs. Since they are symmetric, there are 15 of each. Thus we have 15 pairs out of the total 36 pairs - $\frac{15}{36}$ . Thus sum is $5 + 12 = \boxed{17}$ .

There are 6 possible outcomes in this situation, wherein Calvin can get any of the 6 sides. If Calvin gets a 1, there is a $\frac{5}{6}$ chance that David will get a higher number. If Calvin gets 2, the chances are $\frac{4}{6}=\frac{2}{3}$ , and so on. Basically, the chances for David to get a higher number than Calvin is $D=\frac{6-C}{6}$ , where $C$ & $D$ are the numbers Calvin and David got, respectively. In the end, the six possible fractions are $\frac{5}{6},\frac{4}{6},\frac{3}{6},\frac{2}{6},\frac{1}{6}, and \frac{0}{6}$ . So there are $5+4+3+2+1+0=15$ possible C-D combinations where $D>C$ . There are $6^2=36$ possible combinations all in all, so the probability of David getting a higher number is $\frac{15}{36}=\frac{5}{12}$ . $5+12=\boxed{17}$ .

There are $6\cdot6 = 36$ total possible outcomes. Once the dice have been thrown, there are three possibilities.

1. Calvin and David rolled the same number, and there are 6 ways for this to happen.
2. Calvin rolls a bigger number than David.
3. David rolls a bigger number than Calvin.

Clearly however many ways there are for Calvin to get a bigger number than David is equal to the amount of ways for Calvin to get a smaller number than David. Thus, the number of ways for David to roll a bigger number than Calvin is $(36 - 6)/2 = 15$ .

The probability is then, $\frac{15}{36} = \frac{5}{12}$ , and so the answer is $5 + 12 = \boxed{17}$

total number of ways= 36 no of possibilties =15 15/36 5/12 5+12=17

(5+4+3+2+1)/36 =5/12 5+12=17

to solve this one must calculate the individual probabilities of the numbers appearing on david's dice FOR A PARTICULAR NUMBER appearing on calvin's dice. for eg, if calvin got a 1 on his dice, the problem conditions are only fulfilled for 2,3,4,5,6 appearing on david's dice. and so on for all numbers. the answer is (1/6)(5/6 + 4/6 +3/6 +2/6 +1/6 +0/6), equal to 15/36 or 5/12. hence the answer is 17.

For 1 showing up on Calvin's dice, there are 5 favourable cases for David(2,3,4,5,6) similarly for C's 2 , D has 4 favourable cases ....therefore total number of favourable cases=5+4+3+2+1=15 total number of cases=6*6=36 Thus a=5 and b=12

Total number of possible outcomes = 36 , thus n(S) = 36.

Now let the number shown on calvin's dice be 1 , so possible outcomes on David's dice = 2,3,4,5,6 (5 outcomes) Similarly for 2 , possible outcomes = 3,4,5,6 (4outcomes) For 3 = 4,5,6 For 4= 5,6 For 5 = only 6 Hence total number of favourable outcomes n(E) = 5+4+3+2+1 = 15

Probability = n(E)/n(s) = 15/36 = 5/12. Hence 5+12=17