We're all out of straw
October 19, 2011 9:37 AM   Subscribe

How do you (mostly) randomly choose one person out of four (or more?) with no props?

For two people, there's rock-paper-scissors, and for three people, there's odd man out. But the only way I've ever seen people do more than that is the drawing of straws. Is there some way to choose from four people so that they basically have an even chance of winning?

For 2^n people you could obviously do RPS tournaments, but what about 5? And I was thinking of only one-round things. Is this even possible?
posted by thewumpusisdead to Sports, Hobbies, & Recreation (21 answers total) 3 users marked this as a favorite
Persons A and B do Rock Paper Scissors. Persons C and D do Rock Paper Scissors. The winners do Rock Paper Scissors against each other. Each has an equal chance of coming out as the winner.
posted by Proginoskes at 9:40 AM on October 19, 2011 [1 favorite]

Nevermind. I didn't see the one-round only restriction.
posted by Proginoskes at 9:41 AM on October 19, 2011

Best answer: mod(n) odds & evens. Do you know the game odds & evens? For two people, one person chooses odd and the other person chooses even, then each person puts up one or two fingers -- if the sum is odd the odd person wins, if the sum is even then the even person wins. You can increase this for groups of 3 or more people by dividing and using the remainder -- so for 3 people, one person gets 0, one person gets 1, one person gets 2. Each person puts out 1-3 fingers, and the sum is then divided by 3 and the remainder is determined, and the person who had that number wins (so, say, if 2 people put out 2 fingers and one person puts out 3, the sum is 7, so the remainder after dividing by 3 is 1, so the person who had 1 wins). For e.g. 6 people, each person gets a remainder from 0-5, each person puts out between 1 and 6 fingers, and the person who has the right remainder wins. Works easily up to 10, then you have to start writing things down.
posted by brainmouse at 9:42 AM on October 19, 2011 [8 favorites]

Every person picks a number. Everyone who picked an odd (or even) number is disqualified. Repeat as many times as necessary.
posted by griphus at 9:42 AM on October 19, 2011

If the people are from a similar population (i.e. all males), you can do the shortest/tallest person. It's like the human version of drawing straws.
posted by yodangson at 9:42 AM on October 19, 2011

Wiat, do you mean "one round" as in it filters down to a single person from X people in one round?
posted by griphus at 9:43 AM on October 19, 2011

Even rock-paper-scissors for 2 people (let alone 4) isn't just 1 round, since there can be ties, requiring another round to break the tie. Same with choosing a number.
posted by John Cohen at 9:44 AM on October 19, 2011

Oh, sorry, you said no props. Apologies.
posted by Cool Papa Bell at 9:44 AM on October 19, 2011

Best answer: You play fingers.

Everyone stands in a circle. You go "one, two, three, go!" just like RPS. But instead of rock paper scissors, each person puts in 0, 1 or 2 fingers. Add up the total number of fingers thrown, and starting from the person running the game count around the circle up to that number. The person you land on is the winner. Done.
posted by auto-correct at 9:45 AM on October 19, 2011 [10 favorites]

posted by Jehan at 9:46 AM on October 19, 2011

Okay, everyone is assigned a number 1-X where X is the number of people. Then they pick a number 1-X.

So let's say you've got five people and they pick five numbers.

Easy case: they pick a set of numbers where the digits eventually add up to less than or equal to X. So with five people picking 1, 5, 5, 6 and 2 you get a sum of 19. Add the digits and you get 10. Add those digits and you get 1. Person 1 is your choice.

Harder case: they pick a set where it adds up to greater than X. For instance: 5, 5, 5, 1, 1. That adds up to 17, the digits of which add up to 8. In that case just subtract X from the number. With five people you subtract 5 from 8 and get 3. Person 2 is your choice.
posted by griphus at 9:53 AM on October 19, 2011

...person 3 is your choice, rather.
posted by griphus at 9:54 AM on October 19, 2011

Best answer: A similar method to Griphus is to just figure mod{number of people} from the sum of random numbers given by the people to be chosen from. In other words, everybody picks a number, you add them up, divide by the number of people, and the remainder+1 is which person is picked. Let's say, people pick 4, 7, 2, and 2. That's 15, divided by 4, has a remainder of 3 -- add 1, and person #4 is picked. if it was 4,7,2, and 3, that's 16, or remainder 0 -- plus one, person #1 gets picked.

As a silly option, you know that way in baseball of chosing who goes first, by going hand over hand up the length of a baseball bat until somebody is on top? Do that with your arm or leg. The person whose hand is at the end gets chosen.
posted by AzraelBrown at 10:01 AM on October 19, 2011

With auto-correct's fingers method, the result of the count will tend towards the total number of people in the group (and therefore towards the person running the game being chosen), assuming that each of {0, 1, 2} each have an equal chance of being chosen, which is a reasonable assumption. That probably isn't random enough for your purposes.
posted by Kwine at 10:06 AM on October 19, 2011 [1 favorite]

Best answer: The 'fingers' concept isn't terribly random. The average number of fingers-per-person will probably be about 1, so you'll usually count through the group about once, picking either the person at the beginning or the end. The ones in the middle will be picked less often.

With few people it's not such a big deal, but in a group of eight, the ones in the very middle are only half as likely to be chosen as the people at the ends. (Yes, I ran a quick simulation.)

I much prefer the modular-arithmetic methods.
posted by vasi at 10:09 AM on October 19, 2011 [4 favorites]

Write down a random date. The person whose birthday is closest to that date wins.

You could also use last two digits of SSN, or the number of first letter of their mom's maiden name (1=A, 26-Z)
posted by stupidsexyFlanders at 10:21 AM on October 19, 2011

Is there a reason Eenie Meenie Miney Moe wouldn't work? Obviously for the one-round restriction you'd use the "and you are it" version, rather than the "you are not it" method.
posted by katemonster at 10:26 AM on October 19, 2011

Guyyyyyys, this is exactly what Eeny, meeny, miny, moe is forrrrrrrrrrrrr.

End Childish huffing, sighing, and whining.

Commence cute kid video
posted by bilabial at 1:27 PM on October 19 [+] [!]

Except, I put this comment accidentally in the how do you choose a dog thread. not so funny there, is it? So I asked the mods to delete it there.

posted by bilabial at 10:31 AM on October 19, 2011 [1 favorite]

Response by poster: These are some good ones! I like the mod(n) ones and since it's mostly programmers I don't think it's a big deal to geek out over something like this.
Eeny-meeny is definitely not random enough for my incredibly sophisticated purposes (picking the person who gets to choose where to go to lunch). And I guess fingers isn't quite good enough either.
posted by thewumpusisdead at 11:15 AM on October 19, 2011

The winner could be the person with the birth date closest to the current date, counting the year or not.
posted by maurreen at 2:11 PM on October 19, 2011

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