Gamer Janken, or How to Pick a First Player
dampfkraft.com
dampfkraft.com
Instead of trying to optimize the speed with which you figure out who goes first, it's a pre-game with its own rules.
You get knocked out if you lost to both of the individuals on your left and right. If you beat one and one beats you, you stay in for the next round.
For instance, if I throw ROCK and the two people to my side throw PAPER I'm out. The remaining players play another round (until you get down to two people, obviously).
If I throw ROCK and one to my left throws PAPER but the one to the right throws SCISSORS, I'm still in it for the next round.
Even with 6 or 8 people it still takes less than a minute, and the fact that you can so several rounds without getting knocked out builds up a kind of tension, along with the moment after a round where everyone around the table is looking around to evaluate who got knocked out.
I highly recommend you try it. Its pretty fun and really sets a light mood going into whatever game you're actually going to play.
However, I also have to plug this Android app our gaming group has been using for the last few years. It has one simple job, and does it very well.
https://play.google.com/store/apps/details?id=com.tendadigit...
Another free option that just requires you to select the player-count and then tells you who goes first (i.e. no fingers needed) is Who Goes First? [0]
But the most streamlined app is Fast First [1] for $1.99 which, every time you open it, generates images of players sitting around tables for player counts ranging from 2-7 (for Android and it looks like up to 17 with iOS, depending on the device) with a start-player selected for each player-count. That is, you get a screenful of images representing a table of 2 players, a table of 3 players, etc. each with its own random start-player selected. So, if you have, e.g., 5 players at your table, you just open the app and look at the image for 5 players and use the start-player shown in that image.
It does take away the fun interactivity of the finger-based apps, but, that makes it fast to use.
[0] https://play.google.com/store/apps/details?id=com.idunnolol....
(There doesn't appear to be an iOS version of this app by Daniel Lew -- there is a similarly named iOS app but it's different).
[1] Android: https://play.google.com/store/apps/details?id=com.pinnaclega...
In Ticket to Ride, i remember it as being the person who was most recently on a train, but on looking it up, it seems it was the player who had travelled the furthest to get to the game (either way, always me, as i used to play with friends who lived across town, at their place!).
So i tend to pick something relevant to the theme of the game, and pick the player who did that most recently. If nobody has done that thing, fall back to weaker versions of it.
Gardens of Alhambra? The player who most recently did some gardening. Fall back to watering a pot plant.
Hanabi? The player who most recently saw some fireworks. Fall back to anything that was on fire.
King of Tokyo? The player who most recently destroyed a skyscraper. Fall back to doing some DIY which involved knocking a wall down, then to breaking a cupboard door.
Settlers of Catan? The player who most recently really annoyed one of their friends.
I'm wondering though if you can do better with an even-odd variant; each person is assigned a number 1-n, and players secretly pick and simultaneously reveal a number 1-n. The winner is the person whos number is sum(player picks) mod n.
In the case of 3 players you get random numbers (reduced modulo 3): 1,4,7,10 => 1 40% of the time; 2,5,8 => 2 30%; 3,6,9 => 3 30%
Two random numbers produce: 1 @ 33%; 2 @ 30%; 3 @ 37%
And after adding the third number you get: 1 @ 30.9%; 2 @ 34.9%; 3 @ 34.2%
(These percentages are all exact BTW.)
So this is biased against player 1 (which is not actually the way I was expecting it to go). And if you want to cheat, you should assume the other players will add up to 3, and pick a number accordingly.
I assume 4,6,7,8,9 out of ten and anything out of 11 work similarly, but I haven't checked the numbers.
If "a number of fingers" is 1..N (or 0..^N) then it's unbiased, but just doesn't work at all for >10 (or >11 respectively).