Rolling Kiriti's dice

We first find the probability for each roll 10,11,12. $P(10)=\dfrac{3}{36}$ $P(11)=\dfrac{2}{36}$ $P(12)=\dfrac{1}{36}$ Now, since this is given, to find the actual probabilities we must normalize. $P'(10)=\dfrac{P(10)}{P(10)+P(11)+P(12)}=\dfrac{1}{2}$ $P'(11)=\dfrac{P(11)}{P(10)+P(11)+P(12)}=\dfrac{1}{3}$ $P'(12)=\dfrac{P(12)}{P(10)+P(11)+P(12)}=\dfrac{1}{6}$ Now, we can simply calculate for each case. If the sum is 10, then we need a 5 or a 6 ( $\dfrac{1}{3}$ ). If the sum is 11, then we need a 4,5,6 ( $\dfrac{1}{2}$ ). If the sum is 12, then we need a 3,4,5,6 ( $\dfrac{2}{3}$ ).

Then, the probability is $\dfrac{1}{2}\dfrac{1}{3}+\dfrac{1}{3}\dfrac{1}{2}+\dfrac{1}{6}\dfrac{2}{3}=\dfrac{4}{9}$ The answer is $4+9=\boxed{13}$ .

First, we can list the ways to get a sum greater than or equal to 10:

1) Getting a sum of 10:

The dice are [4,6], [5,5] or [6,4]. Thus, there are 3 ways to get a sum of 10.

2) Getting a sum of 11:

The dice are [5,6] or [6,5]. Thus, there are 2 ways to get a sum of 11.

3) Getting a sum of 12:

The dice are [6,6]. There is only 1 way of getting a sum of 12.

In total, there are $1+2+3=6$ ways of getting a sum greater than or equal to 10.

Therefore there is a $\frac{3}{6}$ chance of the sum being 10. If the sum is 10, the number on the third dice must be 5 or 6 for the sum to be greater than or equal to 15. There is a $\frac{2}{6}$ chance of the number being 5 or 6. Thus, we take:

$\frac{3}{6} \times \frac{2}{6} = \frac{1}{6}$

Following similar logic for the sums of 11 and 12,

For the sum of 11: $\frac{2}{6} \times \frac{3}{6} = \frac{1}{6}$

For the sum of 12: $\frac{1}{6} \times \frac{4}{6} = \frac{1}{9}$

Then, $\frac{1}{6} + \frac{1}{6} + \frac{1}{9} = \frac{4}{9}$ . Adding $4+9$ , we arrive at the answer, 13.

The probabilities of the sum of a pair of dice being 10,11 and 12 are $\dfrac{3}{36}$ , $\dfrac{2}{36}$ , and $\dfrac{1}{36}$ respectively. Since we know that a 10 or higher was rolled, the probabilities of these events happening are $\dfrac{3}{1+2+3}=\dfrac{1}{2}$ , $\dfrac{2}{1+2+3}=\dfrac{1}{3}$ , and $\dfrac{1}{1+2+3}=\dfrac{1}{6}$ respectively. Now we do casework:

Rolls a 10: The probability of rolling a 5 or above next roll is $\dfrac{2}{6}=\dfrac{1}{3}$ . The total probability for this case is $\dfrac{1}{2}\cdot \dfrac{1}{3}=\dfrac{1}{6}$ .

Rolls an 11: The probability of rolling a 4 or above next is $\dfrac{3}{6}=\dfrac{1}{2}$ . The total probability for this case is $\dfrac{1}{3}\cdot \dfrac{1}{2}=\dfrac{1}{6}$ .

Rolls a 12: The probability of rolling a 3 or above next is $\dfrac{4}{6}=\dfrac{2}{3}$ . The total probability for this case is $\dfrac{2}{3}\cdot \dfrac{1}{6}=\dfrac{1}{9}$ .

The final probability is $\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{9}=\dfrac{4}{9}$ so our answer is $4+9=\boxed{13}$ .

Since the first one can be greater or equal to 10, you have more than 1 value to contend with. First, find all the combinations of dice that form a number bigger or equal to 10. You can have

• 5 + 5 = 10
• 4 + 6 = 10
• 6 + 4 = 10
• 6 + 5 = 11
• 5 + 6 = 11

• 6 + 6 = 12

  This shows that 1/2 the time the answer is 10, 1/3 of the time it is 11, and 1/6 of the time it is 12. After that, you have to find what adds up to 15 or more. For 10, there is 5 or 6, or 1/3 of the time. Multiply that by 1/2 and you get 1/6. Put this aside for later.

  Next, we have 11. For 11, there are 4, 5, and 6, or 1/2 of the time. Multiply this by 1/3 to get 1/6. Put this also aside for later

  Lastly, we have 12. For 12, there are 3, 4, 5, or 6. This means 2/3 of the time. Multiply this by 1/6 and simplify to get 1/9.

  Now, add everything up. 1/6 + 1/6 + 1/9 = 4/9. 4+9 = 13

The answer is 13.

Sorry for mistakes in formatting. This is my first solution.

The total number of ways we can have the sum of two dice be greater than or equal to 36 is in one of the following 6 ways: $(6+6,6+5,5+6,6+4,4+6,5+5)$ . All these 6 cases can occur with equal probability.

Case 1: 6+6

In this case, the third dice can take values 3,4,5,6 so that the sum of the three dice result is $\ge 15$ . The probability of one of these 4 occurring is $4 \over 6$ .

Case 2: 6+5, 5+6

In this case, the third dice can have values 4,5,6. The probability of one of this occurring is $3 \over 6$ .

Case 3: 6+4, 4+6, 5+5

In this case, the third dice can take values of 5 or 6 to make the sum of the three dice $\ge 15$ . The probability of this occurring is $\frac{2}{6}$ .

So the total probability is $\frac{1}{6}\frac{4}{6} + \frac{2}{6}\frac{3}{6}+\frac{3}{6}\frac{2}{6} = \frac{4}{9}$ .

baye's theorem - P(A\B) = P(A intersection B) / P(B)

Event A : sum of all the three throws is greater than or equal to 15

Event B : sum of first 2 throws is greater than or equal to 10 . therefore ,P(B) = 6/36 and P(A intersection B) = 16/216

Let us consider the first two die. To roll a 10 or higher, we can roll a 10, 11, or a 12. Since there are 3 ways to roll a 10, 2 ways to roll an 11, and only one way to roll a 12, the probability of rolling a 10 (given that it is greater than or equal to 10) is equal to $\frac{3}{1+2+3}=\frac{1}{2}$ , the probability of rolling an 11 is $\frac{1}{3}$ , and the probability of rolling a 12 is equal to $\frac{1}{6}$ . Now, we can look at the third die, dividing into 3 cases.

Case 1, if the first two die sum up to 10:

In order for us to obtain 15 or higher, the third die must be 5 or higher, a probability of $\frac{1}{3}$ . The total probability for this case is equal to the probability that the first two sum up to 10 times the probability that the third die rolls 5 or higher. This is equal to $\frac{1}{2}\times\frac{1}{3}$ , or $\frac{1}{6}$ .

Case 2, if the first two die sum to 11:

The probability of this is equal to $\frac{1}{3}$ , as stated in the first part. Now we can roll a 4 or higher, a probability of $\frac{1}{2}$ . Multiplying, we again have $\frac{1}{3}\times\frac{1}{2}=\frac{1}{6}$ .

Case 3, if the first two die sum to 12:

There is a $\frac{1}{6}$ chance of this happening. Now, we can have the third die be a 3, 4, 5, or 6, a probabilty of $\frac{2}{3}$ . Multiplying, $\frac{1}{6}\times\frac{2}{3}=\frac{1}{9}$ .

Adding up our three probabilities, we obtain $\frac{1}{6}+\frac{1}{6}+\frac{1}{9}=\frac{8}{18}=\frac{4}{9}$ . Since we want the sum of $a+b$ , we have $4+9=\boxed{13}$ .