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robert A · Jul 30, 2007, 05:25 PM

Hi all!

In my class we have a system of "lots" that the teacher uses to choose a student to do any number of things. Its works like this. Those interested join a circle and each extend one or 2 fingers. The teacher announces a number, usually between 5 and ten times the amount of students present in the circle, and begins to count the fingers round and around, until the final number is the winner.

say for example seven students joined in, each would put out one or two fingers, the number could be 59, and the count would begin around in a circle, starting from a random student. One number on each finger, so if a person put out 2 fingers they would be counted twice.

There has for a long time been speculation that those a enter with 2 fingers have a better chance of winning, they are simply doubling the number of tickets they have in the raffle!

Is our hunch verifiable?

Of course there are many variables, such as how many participate, what number is chosen, and where the count begins, but I'm guessing there would be some general rules as to whether one or 2 fingers would give an edge.

I had thought about making a spreddsheet where I could enter the number to find the best probability, but that's for another days (when I learn to make a spreadsheet! :) )

Thanks for any ideas!

Capuchin · Jul 30, 2007, 11:37 PM

I think your hunch is correct, 2 fingers will give the best probability.

lets say there are 5 students, and the teacher's number is random, and the teacher starts counting from a random position.

If they all put 1 finger in, they all have the same chance of being picked, if they all put 2 fingers in, they all have the same chance of being picked.

If there's say, 4 1-finger studetns and one 2-finger student, then the 1 finger students each have 1/6 chance of getting picked, whereas the 2 finger student has a 2/6 chance.

Through a bit of logical thinking, you can see that it is never a disadvantage to have 2 fingers in, you will always get picked more if you have 2 fingers in, irrespective of the number of fingers everyone else puts in.

Now, if they teacher's numbers aren't totally random, and she always starts from the same position, then there are better places to sit, and more complicated reasons for choosing how many fingers to put out. But that's for someone with more time than I have :)

mikezapwnzor · Jul 30, 2007, 11:38 PM

Sounds like a hard way to pick someone to do something... ha ha

robert A · Jul 31, 2007, 08:38 PM

It actully works quite quickly. :)

What bother me is that I can seem to prove that having 2 fingers is better than one. For any given number I choose, it seems that, depending on the amount of fingers used by the other students, there are for me (just an example) three winning possible combinations. In the first I win regardless of wehter I use one or 2 finger. In the second I win with 1 finger and lose with 2. and in the third I lose with one and win with 2. in all the remaining combinations I lose.

What gives? Why don't I fair better with 2 than one?

Thanks

ebaines · Aug 2, 2007, 10:22 AM

Your odds are always better throwing 2 fingers rather than 1. Let's call the number of fingers thrown by the student f (f can be either 1 or 2), and the total of all fingers thrown by all the other students F. The total number of fingers thrown by all is therefore f+F. As the teacher counts around the circle he "uses up" f+F numbers. The student's chance of winning is therefore equal to f/(f+F). For f=1, this is: 1/(1+F), and for f=2 this is 2/(2+F). Since 2/(2+F) is greater than 1/(1+F), your odds are always better throwing 2 fingers than 1.

Here's the proof that 2/(2+F) is greater than 1/(1+F), for all positive values of F. Start with:

$ 2+2F\ >\ 2+F $

Divide both sides (1+F)*(2+F):

$ \frac {2+2F} {(1+F)*(2+F)}\ >\ \frac {2+F} {(1+F)*(2+F)} $

Simplify:

$ \frac {2*(1+F)} {(1+F)*(2+F)}\ >\ \frac {2+F} {(1+F)*(2+F)} $

$ \frac 2 {2+F}\ >\ \frac 1 {1+F} $

robert A · Aug 3, 2007, 12:19 PM

thanks, that formula does it.

is there a way to determaine the actual probability based on the number of perople, or the number of fingers?

thanks

Capuchin · Aug 3, 2007, 12:37 PM

number of your fingers/total number of fingers

that's the probability, assuming the teacher is totally random.

robert A · Aug 3, 2007, 12:45 PM

What I mean is the probabilty per Situation.

For example, based of the number crunching I've done, when there are 6 people and the number is 62, the number 1 will win 18 times (of the 64 possible combinations) and the number 2, 46. But when the number is 63 one finger will win 26 times (again out of the possible 64) and two fingers 38 times.

Is there a forulma that describes these results?