Can This Be The Answer??!!

[fnord_too]

Oh, here is another way that I just thought of, I think this works!:
One person is designated the flipper.
The other 8 divide into two groups, one gets heads and one gets tails.
Winning group continues, and the flipper then flips to see which group he is in, so then you have either a group of 4 or 5 to choose from with everyone having a 50/50 shot of being in the remaining group, no chance of freak flipping stopping you.
If 4 - easy.
If 5, 1 flipper, 2 groups of 2 leaving 2 or 3.
If 2 - easy
If 3 1 flipper, 2 groups of 1 leaving 1 or 2, both of which are easy.

I think that works, there is no advantage or disadvantage to being the flipper.

EDIT - bah, that does not work either
EDIT 2 - but it is very sophistic and you could probably convince someone not too horribly estute that it does, at least for long enough to get the on the spot job offer.


Before I came up with that just now, here is what I was thinking:
I had a thought about this. I don't think this works, but I may play with it some.

You do the 4 bit thing I suggested earlier. To make it end, you record the number generated each time. When one of the non 0-8 numbers gets to n, you sum all the other totals and mod it by 9. That person loses. So, for example, say n is 50, and you eventually get to:
Number - Times occured
0 - 0
1 - 0
2 - 0
3 - 0
4 - 0
5 - 0
6 - 0
7 - 0
8 - 0
9 - 13
10 - 22
11 - 12
12 - 50
13 - 44
14 - 2
15 - 9

So the sum of non 12's is 102 = 3 mod 9. So 3 loses.

Is there a n for which this has equal probability for 0-8 mod 9? This has a definite maximum number of flips. Alternately, what if we capped the flips at say n = 10,000, so if no 0-8 came in that time, we mod the number of 9's by 9. Does that work for any value of n?

Both these questions are I think answerable. I will think about it some. Actually, I think this may be the same as saying flip the coin n times and mod the number of heads by 9. Did anyone try this approach? I did not see it, but it may have been implied that this cannot work in one of the things above.