United States
Member #202,920
December 14, 2019
900 Posts
Offline
Alice has a pair of standard six-sided dice. She says to her little brother Brad, "I'm gonna keep rolling these dice until I get either a sum of 7 or 2. If the sum is 7, I win. If the sum is 2, you win. Whaddya say Brad?"
Brad says, "No way, cheater. Getting sum of 7 is six times more likely than getting a sum of 2! How about you keep rolling until you get either a sum of 7 TWICE IN A ROW, or a sum of 2. If you roll a sum of 2, I win. If you roll a sum of 7 twice in a row, you win."
Alice accepts the new terms of the game. Is this a fair game? If not, what is the probability Alice wins?
United States
Member #197,030
March 28, 2019
1,640 Posts
Offline
Quote: Originally posted by db101 on Feb 3, 2023
Alice has a pair of standard six-sided dice. She says to her little brother Brad, "I'm gonna keep rolling these dice until I get either a sum of 7 or 2. If the sum is 7, I win. If the sum is 2, you win. Whaddya say Brad?"
Brad says, "No way, cheater. Getting sum of 7 is six times more likely than getting a sum of 2! How about you keep rolling until you get either a sum of 7 TWICE IN A ROW, or a sum of 2. If you roll a sum of 2, I win. If you roll a sum of 7 twice in a row, you win."
Alice accepts the new terms of the game. Is this a fair game? If not, what is the probability Alice wins?
The game is still not fair, because now Alice has only a 6/13 (∼46.15%) chance of winning. Brad's likelihood of winning is 7/13 (∼53.85%).
Here's how I solved it. Suppose Brad's probability of winning is Z. On any given roll there are three outcomes:
Sum = 2, with probability 1/36
Sum = 7, with probability 6/36
Sum = something other than 2 or 7, with probability 29/36
If the first roll is a sum of 2 (first case), he wins automatically.
If the first roll is a sum of 7 (second case), he wins if the next roll is a sum of 2. If the next roll is a sum of 7, he loses, and if the next roll is something other than a sum of 2 or 7, the game continues as if the first two rolls never happened.
If the first roll is a sum of neither 2 nor 7 (third case), the game continues as if the first roll never happened.
So you get
Z = [1/36] + [(6/36)(1/36) + (6/36)(29/36)*Z] + [(29/36)*Z]
which works out to Z = 7/13. And so Alice's probability of winning is 1 - Z = 6/13.
(I also confirmed it running simulations of the game seeing how often Alice wins.)
United States
Member #202,920
December 14, 2019
900 Posts
Offline
Quote: Originally posted by cottoneyedjoe on Feb 5, 2023
The game is still not fair, because now Alice has only a 6/13 (∼46.15%) chance of winning. Brad's likelihood of winning is 7/13 (∼53.85%).
Here's how I solved it. Suppose Brad's probability of winning is Z. On any given roll there are three outcomes:
Sum = 2, with probability 1/36
Sum = 7, with probability 6/36
Sum = something other than 2 or 7, with probability 29/36
If the first roll is a sum of 2 (first case), he wins automatically.
If the first roll is a sum of 7 (second case), he wins if the next roll is a sum of 2. If the next roll is a sum of 7, he loses, and if the next roll is something other than a sum of 2 or 7, the game continues as if the first two rolls never happened.
If the first roll is a sum of neither 2 nor 7 (third case), the game continues as if the first roll never happened.
So you get
Z = [1/36] + [(6/36)(1/36) + (6/36)(29/36)*Z] + [(29/36)*Z]
which works out to Z = 7/13. And so Alice's probability of winning is 1 - Z = 6/13.
(I also confirmed it running simulations of the game seeing how often Alice wins.)
I like your algebra solution. What made this one hard and not intuitive, for me at least, is that you know the prob of rolling snake eyes in 1 roll is 1/36, and the prob of rolling 7s back to back in 2 rolls is also 1/36, so it feels like it should be fair. But you know it can't be because having to do something twice in a row is harder
us
Member #226,814
November 8, 2022
646 Posts
Offline
This type of Bayesean conditional reasoning can be used to solve the probability of a winning game of tennis given p=probability of winning a point in tennis. You will find a large slope at p=50% of the P(Win Game) vs P(Win Point) plot. ie, small point win advantage turns into a large probability of game win. You can compute the exact value of this slope.
Also, you can derive why a higher speed 1st serves make sense to maximize p, despite greater fault rates of higher speed serves.
