# Math Help - Chance of winning

1. ## Chance of winning

Hi members,
Each of A,B and C throws with two dice for a prize.The highest throw wins ,but if equal highest throws occur ,the players with these throws continue . If A throws 10,find the chance of his winning.
My answer is $\frac{9}{16}$

2. ## Re: Chance of winning

With two dice, the probability of throwing
2 is 1/36
3 is 2/36
4 is 3/36
5 is 4/36
6 is 5/36
7 is 6/36
8 is 5/36
9 is 4/36
10 is 3/36
11 is 2/36
12 is 1/36.

If a player throws 10 he can only be beaten by 11 or 12 which has probability 2/36+ 1/36= 3/36= 1/12. The probability that either B or C or both roll 11 or 12 is 1/12+ 1/12- (1/12)^2= 12/144+ 12/144- 1/144= 23/144. If either of B or C but not both throw 10, which has probability 3/36+ 3/36- 2(3/36)^2= 12/144+ 12/144- 2/144= 22/144= 11/72 he will have probability 1/2 of winning on a second or later throw. If all 3 throw 10, which has probability (1/12)^2= 1/144, he will have probability 1/3 of winning.

3. ## Re: Chance of winning

P(A wins|A rolls 10) = P(B and C roll lower than 10) + P(B and C both roll 10)*P(A wins next round vs all three) + P(B or C rolls a 10, the other rolls lower)*P(A wins next round vs either B or C)

There are six ways to roll a 10 or above: (4,6), (5,5), (6,4), (5,6), (6,5), (6,6), so there is a $\dfrac{30}{36} = \dfrac{5}{6}$ chance of rolling less than 10.
P(B and C both roll lower than 10) = $\dfrac{5}{6}\cdot \dfrac{5}{6} = \dfrac{25}{36}$

P(B and C both roll 10) = $\dfrac{3}{36}\cdot \dfrac{3}{36} = \dfrac{1}{144}$
In general, all three players start with equal chances of winning, so $P\left(A\text{ wins next round vs. both }B\text{ and }C\right) = \dfrac{1}{3}$

P(B or C rolls a 10 and the other rolls lower) = P(B rolls 10, C rolls lower)+P(C rolls 10, B rolls lower) = $2\cdot \dfrac{1}{12}\cdot \dfrac{5}{6} = \dfrac{5}{36}$

In general, if there are only two players left, they have equal chances, as well, so $P\left(A\text{ wins next round vs. either }B\text{ or }C\right) = \dfrac{1}{2}$

Hence, $P\left(A\text{ wins}|A\text{ rolls }10\right) = \dfrac{25}{36} + \dfrac{1}{144}\cdot \dfrac{1}{3} + \dfrac{5}{36}\cdot \dfrac{1}{2} = \dfrac{331}{432}$

4. ## Re: Chance of winning

Originally Posted by SlipEternal
P(A wins|A rolls 10) = P(B and C roll lower than 10) + P(B and C both roll 10)*P(A wins next round vs all three) + P(B or C rolls a 10, the other rolls lower)*P(A wins next round vs either B or C)

There are six ways to roll a 10 or above: (4,6), (5,5), (6,4), (5,6), (6,5), (6,6), so there is a $\dfrac{30}{36} = \dfrac{5}{6}$ chance of rolling less than 10.
P(B and C both roll lower than 10) = $\dfrac{5}{6}\cdot \dfrac{5}{6} = \dfrac{25}{36}$

P(B and C both roll 10) = $\dfrac{3}{36}\cdot \dfrac{3}{36} = \dfrac{1}{144}$
In general, all three players start with equal chances of winning, so $P\left(A\text{ wins next round vs. both }B\text{ and }C\right) = \dfrac{1}{3}$

P(B or C rolls a 10 and the other rolls lower) = P(B rolls 10, C rolls lower)+P(C rolls 10, B rolls lower) = $2\cdot \dfrac{1}{12}\cdot \dfrac{5}{6} = \dfrac{5}{36}$

In general, if there are only two players left, they have equal chances, as well, so $P\left(A\text{ wins next round vs. either }B\text{ or }C\right) = \dfrac{1}{2}$

Hence, $P\left(A\text{ wins}|A\text{ rolls }10\right) = \dfrac{25}{36} + \dfrac{1}{144}\cdot \dfrac{1}{3} + \dfrac{5}{36}\cdot \dfrac{1}{2} = \dfrac{331}{432}$
Hi SlipEternal,
You are perfectly correct.I cannot describe my pleasure in words after reading your answer.Thank you very much.