The board game is about to begin. And suddenly we realize that we need to decide the starting player. The dice procedure is tiresome. And we can't find our dice right now. Or any other artifact that might be useful.


What procedure can we use?


  1. You can use no objects apart from the players around the table (pick the player whose shape is closest to a cube shape?)
  2. It must be fair, i.e. a flat distribution among the players (count the number of atoms each player has in their body and the player whose number first appears as a sequence in pi gets it?)
  3. It must work for any number of players more than or equal to 2 (what if I'm the only player?!)
  4. It must be quick and easy to do (we're ready to play here!)
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    $\begingroup$ You have trouble deciding who goes first in a single-player game? I think maybe there are other more pressing issues to address... $\endgroup$ Jul 12 '18 at 20:06
  • $\begingroup$ This seems to be the definition of "too broad." $\endgroup$
    – Chowzen
    Jul 12 '18 at 20:38
  • 1
    $\begingroup$ @Chowzen how do you figure that? The definition of "too broad" from the VTC popup reads, paraphrased: "limit to a specific problem, enough detail to identify an adequate answer, and avoid asking multiple questions at once". There's only one problem, one question, and the method to identify an adequate answer is given as an itemised list of requirements. What am I missing here? $\endgroup$
    – Bass
    Jul 12 '18 at 21:00
  • $\begingroup$ It is not necessarily too broad, but somehow it doesn't feel like a puzzle to me. Maybe it's suitable for Board & Card Games or Lifehacks? $\endgroup$
    – Glorfindel
    Jul 12 '18 at 21:04
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    $\begingroup$ @Bass Maybe I'm missing misinterpreting the intent of the definition, certainly possible! I understood "too broad" to be a question that invites a large number of technically "correct" answers. Your answer seems to fit the question perfectly, but the OP seems to now be saying the equivalent of "guess what answer I'm thinking of." $\endgroup$
    – Chowzen
    Jul 12 '18 at 21:08

Number the players 0 to n-1. Then

Each player secretly chooses a number from 0 to n-1, and puts up that many fingers under the table. Use finger binary if you are short of fingers. Reveal on count of three. Count fingers (mod n),

the result is the number of the player that gets to start.

Why this is fair (flat distribution, unpredictable result even if repeated, safe from collusion):

no matter what the other players have chosen, each player's choice could change the result to any other player exactly as likely.

Why this is easy:

people sitting around a table is the ideal setup for modular addition: numbering the players in clockwise order makes the process as simple as always pointing at the next person while counting up to each person's revealed number in turn.

Why this is quick:

Always picks the starting player on the first go.

Example with 2 players:

Each player chooses either 0 or 1.

player 1 starts, if the players chose different numbers.
player 0 starts if they chose the same number.

(Also works if you are alone.)

  • $\begingroup$ Great system, way more efficient than mine ! I'm definitely stealing this. But what's with all the spoiler tags? What are you trying to hide? $\endgroup$ Jul 12 '18 at 20:10
  • $\begingroup$ @PierreCathé Thanks! It's customary to hide any clever bits from answers, so that when someone views the question at a later date, and wants to try and solve the puzzle for him/herself, they won't accidentally get spoiled by the answers given earlier. $\endgroup$
    – Bass
    Jul 12 '18 at 20:17
  • $\begingroup$ Right, I didn't think this was a puzzle, just a genuine question, but it makes sense if you look at it this way. $\endgroup$ Jul 12 '18 at 20:25
  • $\begingroup$ So my point was that you needn't have all the players participate. If one person picks two other people and asks for a number from 1 to n then counts from themselves, that will do. I didn't mean to imply that I was looking for a totally different answer. Just that your answer could be improved (imho). $\endgroup$
    – Dr Xorile
    Jul 13 '18 at 3:44
  • $\begingroup$ Not sure if this is an Asian/ Singaporean thing, but here we frequently use a similar method to pick an "odd man" out for children games (e.g. hide and seek). We call it "Open Numbers". $\endgroup$
    – Xenocacia
    Jul 13 '18 at 5:47

similar to bass but kinda different. the amount of players - 1 (n - 1) is the amount of numbers available, and you show your number by holding up your fingers. (eleven is one finger up on each hand, twenty is two fingers on one hand, and a fist in the other, etc. get creative if there are alot of people) everyone shows numbers at same time. if your number matches someone elses then you are out. do it again if everyone chose a different number. this will get you down to two players. the last two play "chopsticks" with each other. https://en.wikipedia.org/wiki/Chopsticks_(hand_game)

  • $\begingroup$ This was my first thought, but if people all chose the same # or if in an even #'d group each person chose a number that someone else did, you would have to redo it a few times. $\endgroup$
    – Sensoray
    Jul 12 '18 at 20:43

A simple solution is

Iterated Cumulative Rock-Paper-Scissors

It goes like this:

1. Everyone simultaneously picks rock, paper or scissors
2. Each player counts how many other players they beat
3. If there is a single player with the highest score, they are the starting player
4. Otherwise, everyone who isn't tied for first place is eliminated, everyone who is tied for first place goes again.


Just gotta put this out there:

Whoever "calls dibs on going first" or "took a dump last"


How about

Putting your fingers on the side of your nose.
Last one to do so counts the contestants off clockwise from himself using the number of letters in his first name, last letter goes first.

  • 1
    $\begingroup$ That system can easily be manipulated : If we are five players and my name has five letters, then I just wait to put my finger on my nose last and BOOM, I go first. $\endgroup$ Jul 12 '18 at 21:16

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