binary math and random chance

[jasong]

I realize most people don't care about esoteric math concepts, but an argument was started over whether 3 coin flips gave 8 even chances.

Before I start let me say that this is an argument about MATH, not an argument about whether or not my decision in the contest should be honored. I have totally accepted Lauren's decision in the matter and am simply disturbed with the accidental abuse statistical analysis is getting.

Now to begin:

We have 8 choices: 0-7, with the zero being the "eighth" choice since we're doing this in binary and switching.

The first choice is basically 0 or 4, the second choice is 0 or 2, and the third choice is 0 or 1. Lauren commented that there's a greater chance of getting 2 of one and 1 of another(heads or tails), which is true, but I must point out we're not counting heads and tails, we're figuring binary values(like a computer).


For the sake of understanding things, I'm going to add 1 to each choice. This is how it works:

The first choice decided whether the number will be 1-4 or 5-8. 50-50 split in choice and randomness

the second choice is to add or not to add 2, either way 2 pots of 2 numbers to choose from.

the third choice uses the last 2 numbers and randomly picks between the two.

Every time 50% of the numbers make it through and 50% don't.

Also, might I add that since this is binary, if the total number of heads and tails mattered, wouldn't the total number of 0s and 1s matter in a computer? And yet nobody ever asks that question since it has no play on the vast majority of processes, including random numbers.

[Helix_Von_Smelix]

Hey, and not one question mark:Pokes: :Pokes: :crazy:

[bwkaz]

Originally posted by jasong
Lauren commented that there's a greater chance of getting 2 of one and 1 of another (heads or tails), which is true, but I must point out we're not counting heads and tails, we're figuring binary values(like a computer). Right, order matters. (Permutations, not combinations.)

I'll continue with the "just add one when you're done" version (which is exactly the same probability-wise as "turn 0 into 8", it's just slightly less confusing. It'd be easier if we assigned people numbers starting at 0, but most people don't start counting at 0, so I won't).

Since 0-0-0 gives a different winner (person #1) than 1-1-1 (#8), you can't lump them into the same "pot" and say "well, that's only two chances out of eight". What Lauren says is true -- you are less likely to get 3 of a kind -- but it doesn't affect the fairness of the method. The same thing applies to 0-1-1 (winner is #4), 1-0-1 (winner is #6), and 1-1-0 (winner is #7) -- they can't be lumped together, because a different person would win with each permutation. Same with 0-0-1 (winner is #2), 0-1-0 (winner is #3), and 1-0-0 (winner is #5).

There are eight different outcomes, because we assign a "value" to each individual flip.

(I don't disagree with what happened with the drawing, either -- it's not like I was in it or anything. But I don't like to see probability / stats / math getting accidental abuse either. ;))


since this is binary, if the total number of heads and tails mattered, wouldn't the total number of 0s and 1s matter in a computer? Well, actually...

http://www.catb.org/~esr/jargon/html/B/bit-bucket.html


Another variant of this legend has it that, as a consequence of the "parity preservation law", the number of 1 bits that go to the bit bucket must equal the number of 0 bits. Any imbalance results in bits filling up the bit bucket. A qualified computer technician can empty a full bit bucket as part of scheduled maintenance. ;)