This problem is not about geometry
You want the probability that Bret's last toss will exceed Shawn's by 11 or more.
There are 20x20 = 400 possible outcomes of the two tosses. Here are some of the ways Bret can win ion the last toss.
Bret can throw 12 while Shawn throws 1.
Bret can throw 13 while Shawn throws 1 or 2.
Bret can throw 14 while Shawn throws 1, 2, or 3.
.. (etc) Finally
Bret can throw 20 while Shawn throws 1, 2, 3, 4, 5, 6, 7, 8 or 9.
The total number of possibilities out of 200 that result in Bret winning is
1 + 2 + 3 + 4 + 5 + 6 + 7+ 8 +9 = 45.
The probability of Bret winning is 45/200 = 9/40 (using coprime numbers)
Therefore a = 9 and b = 40.
a + b = 49
Bret and Shawn are playing a game where each turn the players roll a 20-sided die fair and then add the number they get to their score. The winner is the player with the highest score at the end of the game (there may be a tie). Before the last turn, Shawn is ahead by 10 points. The probability that Bret can come back and win the game can be expressed as a/b where a and b are coprime numbers. What is a+b?
Details and assumptions:
~Both Shawn and Bret have 1 more roll on the last turn.
~If there is a tie, no one won.
4 answers
answer is 89
The total number of possibilities out of 400
The total number of possibilities out of 400 that result in Bret winning is
1 + 2 + 3 + 4 + 5 + 6 + 7+ 8 +9 = 45.
The probability of Bret winning is 45/400 = 9/80 (using coprime numbers)
Therefore a = 9 and b = 80.
a + b = 89
1 + 2 + 3 + 4 + 5 + 6 + 7+ 8 +9 = 45.
The probability of Bret winning is 45/400 = 9/80 (using coprime numbers)
Therefore a = 9 and b = 80.
a + b = 89
1 answer