It’s official, the Boston Celtics and the Dallas Mavericks will face off in the NBA Finals. The best-of-seven series tips off this Thursday and will run through June, until one team achieves the four wins required to be crowned champion.
Home-court advantage plays a serious role in basketball. Since the finals comprise a maximum of seven games, and seven is an odd number, one team may get an extra home game. The NBA knows this confers an advantage, so they award the extra home game to the team with the better win/loss record in the regular season (this year that goes to the Celtics). They also stagger the schedule, so that the privileged team plays at home in games one, two, and, if necessary, five and seven, while the other team hosts games three, four, and, if necessary, six.
This week’s puzzle investigates the role of scheduling in the outcome of the championship. Could we eliminate home-court advantage if we front-loaded the Mavericks’ home games?
Did you miss last week’s puzzle? Check it out here, and find its solution at the bottom of today’s article. Be careful not to read too far ahead if you haven’t solved last week’s yet!
Puzzle #45: There’s No Place Like Home
The Celtics and Mavericks will face off in a best-of-seven series where the first to four wins takes the trophy. Suppose that both teams have a 55% chance of winning their home games and 45% chance of winning their away games (there are no ties). If the Mavericks hosted the first three games and the Celtics hosted game four and, if necessary, games five, six, and seven, then who would be more likely to win? What if the series were best-of-101 and the Mavericks hosted the first 50 games?
Try to solve this without resorting to messy probability calculations.
I’ll be back Monday with the answer and a new puzzle. Do you know a cool puzzle that you think should be featured here? Message me on X @JackPMurtagh or email me at gizmodopuzzle@gmail.com
Solution to Puzzle #44: Blank Dice
Last week I gave you a trio of puzzles about labeling blank dice, one of which was an Amazon interview question. I want to give a shout-out to all of you. The comments section had a lively discussion about the value (or lack of value) in puzzle-y interview questions, and many of you commented or emailed alternative solutions to the puzzles that I hadn’t considered.
Suppose you have one normal die and one blank die. Label the blank die with some subset of the numbers 0, 1, 2, 3, 4, 5, 6 so that when you roll both dice, all sums from 1 to 12 are equally likely.
Answer: Label the blank die with 0, 0, 0, 6, 6, 6. Half of the time you’ll roll a zero, in which case the sum of both dice will be a number from 1 to 6, each with equal frequency. The other half of the time you’ll roll a 6, in which case the sum will be from 7 to 12, again with equal frequency. This solution is unique.
Given two blank dice A and B, label them with the digits 1 through 12 once each (no repeats) so that when you roll them, there is a 50% chance that A rolls higher than B and a 50% chance that B rolls higher than A.
The answer that makes the most intuitive sense to me is to label A = [1, 2, 3, 10, 11, 12] and B = [4, 5, 6, 7, 8, 9]. Half of A’s rolls (1, 2, and 3) will be smaller no matter what B rolls, while the other half of A’s rolls (10, 11, and 12) will be larger no matter what B rolls.
Label three blank dice using the digits 1 through 18 once each (no repeats) so that when you roll them, each die has an equal chance of being the highest.
Inspired by the solution to the previous problem, let’s label A so that one-third of its rolls are guaranteed to be the highest no matter what the other dice roll, while the other two-thirds of A’s rolls are guaranteed to be the lowest: A = [1, 2, 3, 4, 17, 18]. Now we have the numbers 5 through 16 remaining and two more dice to label. Notice that A satisfies the puzzle condition regardless of how we label dice B and C. So in effect, we’ve reduced things back to the two-dice case, just with slightly shifted numbers. Following our strategy from the two-dice case, we’ll label:
B = [5, 6, 7, 14, 15, 16] and C = [8, 9, 10, 11, 12, 13]
Again, we assign all of the remaining extreme digits to B, so that it rolls higher than C exactly half the time, regardless of what C rolls.
Enfy posed the question of whether this can be extended to four dice. I’m not sure and welcome any ideas in the comments!