How Many Dice Can You Roll the Same?

You’re rolling some dice. Whichever number you wind up having the most of (be it 1s, 2s, or whatever), how many of them can you expect to get on average?

Welcome to Fiddler on the Proof! The Fiddler is the spiritual successor to FiveThirtyEight’s The Riddler column, which ran for eight years under the stewardship of myself and Ollie Roeder.

Each week, I present mathematical puzzles intended to both challenge and delight you. Beyond these, I also hope to share occasional writings about the broader mathematical and puzzle communities.

Puzzles come out Friday mornings (8 a.m. Eastern time). Most can be solved with careful thought, pencil and paper, and the aid of a calculator. Many include “extra credit,” where the analysis gets particularly hairy or where you might turn to a computer for assistance.

I’ll also give a shoutout to 🎻 one lucky winner 🎻 of the previous week’s puzzle, chosen randomly from among those who submit their solution before 11:59 p.m. the Monday after that puzzle was released. I’ll do my best to read through all the submissions and give additional shoutouts to creative approaches or awesome visualizations, the latter of which could receive 🎬 Best Picture Awards 🎬.

This Week’s Fiddler

From Ryan Lafitte comes a delectable dilemma of dice:

When supervising the Board Games club at the high school where he teaches, Ryan saw a group break out TENZI. In this game, each player has 10 dice of the same color (in case the rolls get into each others’ areas, presumably).

To get started, you roll all 10 dice—whichever number comes up most frequently becomes your target number. In the event multiple numbers come up most frequently, you can choose your target number from among them. At this point, you put aside all the dice that came up with your target number.

From there, you continue rolling any remaining dice, putting aside any that come up with your target number. Once all 10 dice show the same number, you yell, “Tenzi!” If you’re the first to do so, you win.

Now, consider a simplified version of the game in which you begin with three total dice (call it “THREEZI”).

On average, how many dice will you put aside after first rolling all three?

This Week’s Extra Credit

From Ryan Lafitte also comes some Extra Credit:

Let’s return to the original game of TENZI, which has 10 dice.

On average, how many dice will you put aside after first rolling all 10? (What if, instead of 10 dice, you have N dice?)

Making the Rounds

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing another provocative puzzle from James Tanton, who’s fantastic at issuing sequences of related mathematical inquiries that build on each other. I’ll paraphrase this particular puzzle here, spoiler-free:

Suppose you have a sequence of numbers in which every term is the arithmetic mean (i.e., the average) of its two neighbors. Lo and behold, it’s an arithmetic sequence!

Now suppose you have a sequence of numbers in which every term is the geometric mean of its two neighbors. Wow, this time it’s a geometric sequence!

Okay, okay. What if every term is the quadratic mean of its two neighbors? That is, suppose every subsequence of three consecutive terms is of the form a, √((a²+b²)/2), b. What kind of sequence do you get?

And what if every term is the harmonic mean of its two neighbors? That is, suppose every subsequence of three consecutive terms is of the form a, 2/(1/a+1/b), b. What kind of sequence do you get?

In your investigations, be sure to check both increasing and decreasing sequences!

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Matt F 🎻 from Dayton, Ohio. I received 63 timely submissions, of which 53 were correct—good for an 84 percent solve rate.

In a game of “Rock, Paper, Scissors,” each element you could throw tied itself, beat one of the other elements, and lost to the remaining element. In particular, Rock beat Scissors beat Paper beat Rock.

Last week’s puzzle involved a similar game, “Rock, Paper, Scissors, Lizard, Spock” (popularized via The Big Bang Theory), which had five elements you could throw instead of the typical three. Each element tied itself, beat another two, and lost to the remaining two. More specifically, Scissors beat Paper beat Rock beat Lizard beat Spock beat Scissors beat Lizard beat Paper beat Spock beat Rock beat Scissors.

Three players were playing “Rock, Paper, Scissors, Lizard, Spock.” At the same time, they all put out their hands, revealing one of the five elements. If they each chose their element randomly and independently, what was the probability that one player was immediately victorious, having defeated the other two?

Let’s call the three players A, B, and C. Now, none of them was more or less likely to be immediately victorious than any of the others. So, by symmetry, each was equally likely to be the winner. Also, at most one of them could have actually been that lone winner—never two or three at the same time. That meant the probability of any one player winning (A, B, or C) was  precisely three times the probability of one specific player winning, such as player A.

Great! So then, what was the probability of player A winning?

Whatever A picked, there were two elements that B and C could have chosen that would have resulted in an immediate victory for A. It didn’t matter if B and C picked the same element, if B’s element defeated C’s, or if C’s element defeated B’s. Under any of these circumstances, A emerged victorious. For example, if A had picked Rock, then B and C could have picked either Scissors or Lizard.

Therefore, there was a 2-in-5 chance of A defeating B, and an independent 2-in-5 chance of A defeating C. That meant A’s chances of defeating both B and C in a single round was (2/5)·(2/5), or 4/25.

Again, the probability of any player immediately winning was triple this result, which was 12/25, or 48 percent—the answer to the puzzle. So about half the time a player won immediately.

What happened the rest of the time? While not asked for in the puzzle, a more complete analysis of this game was interesting in its own right.

Among the 5³, or 125, total ways the three players could have selected from among the 5 elements, solver Eric Widdison considered three broad cases:

• All three players selected the same element, resulting in a three-way tie. Among the 125 total scenarios, there were 5 ways this could happen (one way for each element), which meant the probability of this was 5/125, or 1/25.

• Two players selected the same element, while the third selected a different element. There were five ways to choose which element was selected by two players, four ways to choose the remaining element selected by the third player, and three ways to choose which players (among A, B, and C) made the same selection. Thus, the probability of this case was (5·4·3)/125, or 12/25. Half of this time (i.e., with probability 6/25), two players defeated the third, while the other half the time (again with probability 6/25), two players were defeated by the third.

• All three players made different selections. There were 5 ways to choose player A’s element, 4 ways to choose player B’s, and 3 ways to choose player C’s. Thus, the probability of this case was (5·4·3)/125, or 12/25. With some additional work, you could find that half the time (i.e., with probability 6/25), one player defeated the other two. But the other half the time (again with probability 6/25), the three players formed a loop (either A defeated B defeated C defeated A or A defeated C defeated B defeated A) with no clear winners or losers.

Having done this complete analysis, Eric arrived at the same answer. The probability of one player beating the other two outright could have happened when two players selected the same element (probability 6/25) or when all three players made different selections (again, probability 6/25). The total probability was 6/25 + 6/25, or 12/25.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 “Maths ForFun” 🎻 from the “Pacific NorthWest.” I received 31 timely submissions, of which 22 were correct—good for a 71 percent solve rate.

The rules for “Rock, Paper, Scissors” could concisely be written in one of the following three ways:

• Rock beats Scissors beats Paper beats Rock

• Scissors beats Paper beats Rock beats Scissors

• Paper beats Rock beats Scissors beats Paper

Each description of the rules included four mentions of elements and three “beats.”

Meanwhile, as previously mentioned, a similarly concise version of the rules for “Rock, Paper, Scissors, Lizard, Spock” (and adapted from the original site) was:

• Scissors beats Paper beats Rock beats Lizard beats Spock beats Scissors beats Lizard beats Paper beats Spock beats Rock beats Scissors

In this case, there were 11 mentions of elements and 10 “beats.” Including the one above, how many such ways were there to concisely describe the rules for “Rock, Paper, Scissors, Lizard, Spock”?