Can This Be The Answer??!!

[Pokerlogist]

Here's a different type of solution that has the virtue of being 100% finite and determinant. It needs only 12 coin tosses.

Since there are 9 people each person needs to have a 1/9 chance of being chosen. To do this you will toss a coin 12 times in a row. Since you are a math whiz and poker expert you know that there are 2^12=4,096 possible sequences of heads and tails. Before tossing the coin, you assign each of the eight persons 455 of the possible different sequences. That gives each person a 455/4096=.111=~1/9 chance of being chosen. You assign 456 sequences to yourself which still gives you a 456/4096=.111=~1/9 chance. This gives everybody a fair chance and limits coin tossing to 12. Whoever has been assigned the observed sequence of heads and tails from the 12 tosses gets to go on the boat trip.

Will Google dump me for this idea or give me some of their $467/share stock?

[jay_shark]

You do know that 455/4096 is not exactly 1/9. In fact , it cannot terminate in 12 flips of a coin for the simple reason that 4096 is not divisible by 9 .

[Pokerlogist]

Actually, it has to terminate because the above solution covers all possible sequences. (455*8)+456=4096. That part works.

[PairTheBoard]

[ QUOTE ]
Here's a different type of solution that has the virtue of being 100% finite and determinant. It needs only 12 coin tosses.

Since there are 9 people each person needs to have a 1/9 chance of being chosen. To do this you will toss a coin 12 times in a row. Since you are a math whiz and poker expert you know that there are 2^12=4,096 possible sequences of heads and tails. Before tossing the coin, you assign each of the eight persons 455 of the possible different sequences. That gives each person a 455/4096=.111=~1/9 chance of being chosen. You assign 456 sequences to yourself which still gives you a 456/4096=.111=~1/9 chance. This gives everybody a fair chance and limits coin tossing to 12. Whoever has been assigned the observed sequence of heads and tails from the 12 tosses gets to go on the boat trip.

Will Google dump me for this idea or give me some of their $467/share stock?

[/ QUOTE ]

Actually, if you're going to get the other 8 people on the boat to consider it fair I think you're going to have to take a probabilty for yourself that's less than 1/9 rather than greater than 1/9.

If by "deterministic" you mean that you can guarantee fair resolution in a predetermined finite number of flips, say N, then no matter how you do it, what it amounts to is finding 9 Mutually Exclusive Exhastive Events with equal liklihood on a Sample space consisting of 2^N equally likely outcomes. You are asking that 2^N be divisible by 9. Time would be better spent refiguring how to make the water last for all 9 people.

PairTheBoard

[Alex-db]

Can't you just let everyone pick a different sequence of 4 flips from the 16 possible, and discard and repeat if sequences 10-16 occur?

[KrewMaster]

Why are you guys making it so complicated?
This is what I think is a great SOLUTION!

Procedure 1:
Everyone stands on a line, 1, 2....., 9.
Then the first person takes the coin and flips it. If it's either heads or tails that number is assigned to him, then he passes the coin to the next person in line and he does the same thing. When the coin is at the last person, he flips it and does the same thing. (Assigns the value to himself.)

Procedure 2: (for super-ultra fair choice)
Then it starts over again. If you think it should be really really superulrafair, person number 9 goes in front of the line and starts the next flip, and then number 8 is in the back of the line and the last person to flip the coin.

Procedure 1 and 2 is repeated again and again until every person has a different line then all the others. Procedure 2 is really unnecessary in my opinion since it's stil 50/50 to get the coin in either way.

The least possible is getting different numbers in 4 flips. That is: (example)

Player:
1. Heads, Heads, Heads, Heads
2. Tails, Tails, Tails, Tails
3. Heads, Tails, Tails, Tails
4. Tails, Heads, Heads, Heads
5. Heads, Heads, Tails, Tails
6. Tails, Tails, Heads, Heads
7. Heads, Tails, Heads, Tails
8. Tails, Heads, Tails, Heads
9. Heads, Tails, Tails, Heads

When they are all different like this, if the person with the lowest or highest number gets to go on the lifeboat, it'd be fair since I can't see how any coinflip would make any 2 of the 9 persons get the same low or high number.

Hope it's correct

EDIT:

OF COURSE: When the values are assign, someone tosses the coin the number of times that was required to get everyone on different numbers. Then you give one point for every "hit" when the coinflip is equal to your assigned number.

[PairTheBoard]

[ QUOTE ]
Can't you just let everyone pick a different sequence of 4 flips from the 16 possible, and discard and repeat if sequences 10-16 occur?

[/ QUOTE ]

[ QUOTE ]
If by "deterministic" you mean that you can guarantee fair resolution in a predetermined finite number of flips, say N,

[/ QUOTE ]

Your solution and none of the other ones here satisfy this, as has aready been pointed out by several authors of such solutions. You can't give me a predetermined or predesignated number of coin flips which will guarantee a resolution by your method. Certainly by your method the 4 flips do not guarantee resolution. Neither does 16 flips. Or 32. Or any number you wish to name.

PairTheBoard

[PairTheBoard]

What if everybody keeps flipping nothing but heads until they run out of water and all die?
PairTheBoard

[Pokerlogist]

One last defense of the 12 toss solution:
The stipulation of the original problem was that the procedure be FAIR. It specifically did not use the word EQUAL. Fairness is a subjective term with varying degrees. The 12 toss solution gives each and and every person a .111 chance when rounded to 3 significant digits. It is true that one person with 456 sequences would have an .0004 extra chance. That tiny amount should be "fair" enough for this situation. If not, more tosses could be added until the discrepancy is small enough to be acceptable.

I would like to see a solution that is 100% determinative AND 100% equal. It looks impossible right now.
[BTW the 12 toss solution is really a specific case of the general idea presented in bigpooch's first post in this thread.]

[PairTheBoard]

[ QUOTE ]

I would like to see a solution that is 100% determinative AND 100% equal. It looks impossible right now.


[/ QUOTE ]

It is impossible as others have stated both with and without proofs. I believe what I said constitutes a simple proof as well.

[ QUOTE ]
If by "deterministic" you mean that you can guarantee fair (equal) resolution in a predetermined finite number of flips, say N, then no matter how you do it, what it amounts to is finding 9 Mutually Exclusive Exhastive Events with equal liklihood on a Sample space consisting of 2^N equally likely outcomes. You are asking that 2^N be divisible by 9.

[/ QUOTE ]

PairTheBoard