Another Aspect to Halloween Candy

[Ghost Post]

Note that the Fairy Princess strategy of (n-3,0,1,0,2) is no good
when there are only 3 pieces of candy! In this situation, assuming
that an indifferent party will vote Aye or Nay with equal odds, what
is her strategy? And what is the expected take of each greedy
brat?

[mithrandir]

5th person: 0,0,0,0,3
4th person: doesn't matter

now, the next person's strategy depends on the risk hes willing to take:

3rd person: 0,0,3,0,0(with 50% chance of success) or 0,0,2,1,0

i would go with 0,0,2,1,0, as the expected amount is 2 vs. 3/2

2nd person: 0,2,0,1,0 , 0,1,0,1,1 or 0,1,0,2,0

expected amount is: 2/4 vs. 1/2 vs. 1/2
so, the second person could choose any of these.

so fairy: 2,1,0,0,0 or 1,1,1,0,0

#2 will except since 1 is better than his expected take of 1/2. The expected for either is 1 for fairy.

[Ghost Post]

P.S. A few other assumptions are needed. First, where the proposer
is indifferent between a number of possible proposals, s/he chooses
between them with equal odds. Second, the per piece value of candy
to the kids is constant. This means that a 1/3 chance of getting 3
pieces is equal to a certainty of getting 1 piece, etc.

[mithrandir]

expected take with my method is:
fairy: 1
2nd: 7/8
3rd: 19/24
4th: 7/24
5th: 1/24

[Ghost Post]

I think you're partly right, mith. Here's what I just came up with:

#4 and #5 are trivial, so let's start with #3.

As you noted, #3's single best proposal is 0,0,2,1,0.
This will always be accepted by #4, so the expected take of the five
children is 0,0,2,1,0.

On to #2.

As you noted, #2 is indifferent between three possible proposals he
might make. If he proposes 0,2,0,1,0, #4 and #5 are both indifferent
voters, so there's a 1/4 chance that the proposal is accepted, and a 3/4
chance that it's rejected (i.e. #3 votes as above, and the children get 0,0,2,1,0.)
Do the math, and we get expected takes of 0,1/2,3/2,1,0.

If #2 proposes 0,1,0,2,0, only #5 is indifferent, so there's a 1/2 chance that it's
accepted and a 1/2 chance that we move on to #3 and the children get 0,0,2,1,0.
Again, we factor in the odds. The expected take is 0,1/2,1,3/2,0.

Same thing for 0,1,0,1,1, and we get 0,1/2,1,1,1/2.

#2's three proposals' average yield, respectively:

0, 1/2, 3/2, 1, 0
0, 1/2, 1, 3/2, 0
0, 1/2, 1, 1, 1/2

Since #2 chooses between the proposals with equal odds, we average and get:

0, 1/2, 7/6, 7/6, 1/6.

Now, on to #1.

The only way to swing two votes and have any candy herself is to propose
1, 1, 0, 0, 1. And this will be accepted.

Now, how about 4 candies...

[mithrandir]

Yeah, you're right, I forgot to use the expected values to see what plan #1 needed. I'll think about 4.

[mithrandir]

Alright for 4 I get:

5,4 trivial
3:00310
2:02020 or 02011 expected is 0 1 1.5 1.25 .25
1:21001 or 20101 or 20011 or 12001
expected is: 1 1 11/16 19/32 23/32

[Ghost Post]

You forgot that 0,1,0,2,1 has the same expected value (1 piece) for #2. It would be accepted 100% of the time, whereas the two proposals giving #2 two pieces would only be accepted 50% of the time.

So, since all three are equal, the average at #2's turn is 0,1,1,3/2,1/2.

Now #1 has five equal choices:

proposal – 2,1,0,0,1
expected take – 1,1,1/2,3/4,3/4

proposal – 2,0,1,0,1
expected take – 1,1/2,1,3/4,3/4

proposal – 1,2,0,0,1
expected take – 1,2,0,0,1

proposal – 1,0,2,0,1
expected take – 1,0,2,0,1

proposal – 1,0,0,2,1
expected take – 1,0,0,2,1

And the average is 1,7/10,7/10,7/10,9/10

[mithrandir]

Oops again, I need to pay more attention to what I'm doing I think. I believe for 5 it reverts to n-3,0,1,0,2.