"Fore!"
In the game of golf, scoring is as follows on a hole: a birdie beats a par, and a par beats a bogey.
A "four ball" or "best ball" match is when two teams, each with two players, compete: the best score by either player is that team's score for the hole (so if Player A gets a birdie and Player B gets a bogey, that team's score for the hole is a birdie.)
Al and Ben play a 36-hole best ball match against Chuck and Dan.
Al and Ben are incredibly consistent players: both of them get a par on 100% of the holes they play.
Chuck and Dan are more erratic. On any given hole, Chuck has 1 chance in 6 of making a birdie, 1 chance in 3 of making a par, and 1 chance in 2 of making a bogey. This happens at random (imagine rolling a normal 6-sided die for Chuck's result every hole: a 1 is a birdie, a 2 or 3 is a par, and a 4, 5, or 6 is a bogey.) Dan has the same erratic proclivities as Chuck.
The team that wins more holes is the match winner.
If things go "according to Hoyle", which team should win the match?
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1 solution
We know for a fact that Al and Ben's score on every hole will be par.
Al and Ben win a hole when Chuck and Dan's score is bogey. That happens when both Chuck and Dan make bogey on a hole, which has a chance of (1/2 * 1/2) = 1/4 on any given hole, * 36 holes, so Al and Ben should win 9 holes.
Chuck and Dan win a hole when either one of them gets a birdie. Chuck will get on average (1/6 * 36) = 6 birdies, and Dan will also get 6 birdies on average, but (1/6 * 1/6) = 1/36 of the time they both get a birdie, and that only counts as winning 1 hole, so Chuck and Dan should win (6 + 6 - 1) = 11 holes.
Therefore, Chuck and Dan should win the match by 2 holes.