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From: Eric Cavalcanti <eric.domain.name.hidden>

Date: Mon, 11 Oct 2004 18:08:10 +1000

Norman,

I haven't (once again) read your email very carefully.

I think you have made a mistake in the calculations.

You always lose on average in that version of the game.

To see that, imagine you are the last one to flip your

coin (this does not affect the analysis, but makes it easier

to visualize). The other players have either come to an

even or an uneven result. In the case N=6, it is either

3/3 or one of {4/2, 5/1, 6/0, 0/6, 1/5, 2/4}

In the first case, whichever result you get in your flip,

you are gonna be in the Losing Flip. In the second case,

your result won't affect the Losing Flip (because since

there is an even numbers of other players, the Losing Flip,

whichever it is, has at least 2 players more than the Winning

Flip.

So it means that the amount that your outcome affects the

result depends on the probability of a draw prior to your

flip. It is easy to see that this probability decreases

for large N, being zero for infinite players. It means

that with infinite players, the losing flip is independent

of your outcome, so that you are equally likely to be in

either.

Is that right or have I made another mistake?

Eric.

On Mon, 2004-10-11 at 16:31, Norman Samish wrote:

*> Eric, you say you get out even with a large number of players. My
*

*> calculation disagrees with that. Notice that Kory's tables (assuming I
*

*> figured them properly) show that with 7 or more players, they WIN money each
*

*> game - with the amount they win increasing with the number of players.
*

*>
*

*> What's the limit when there are infinite players?
*

*>
*

*> Norman
*

*> ----- Original Message -----
*

*> From: "Eric Cavalcanti" <eric.domain.name.hidden>
*

*> To: "Norman Samish" <ncsamish.domain.name.hidden>
*

*> Sent: Sunday, October 10, 2004 11:21 PM
*

*> Subject: Re: The FLip Flop Game
*

*>
*

*>
*

*> On Mon, 2004-10-11 at 14:20, Norman Samish wrote:
*

*> > In one version of Flip Flop, each of an odd number of players simply flips
*

*> > a
*

*> > coin. The majority result, heads or tails, pays the casino $1 each while
*

*> > the minority result gets paid $2 each. Based on these rules, I worked out
*

*> > Kory's tables for 3, 5, 7 and 9 players.
*

*> >
*

*> > The results show that the player's expectation changes according to how
*

*> > many
*

*> > players there are.
*

*> >
*

*> > For example, if there are 3 players then the long-term odds are that each
*

*> > game costs each player 25 cents. If there are 5 players, the average cost
*

*> > goes down to 6.3 cents per game. If there are 7 players, they make on the
*

*> > average 3.1 cents per game. If there are 9 players they make about 9
*

*> > cents
*

*> > per game.
*

*> >
*

*> > It isn't clear to me why this should be so.
*

*> > Norman
*

*>
*

*> Because the larger the number of players, the less
*

*> you are to be in the Losing Flip. In the limit when
*

*> N is very large, you are equally likely to be in any,
*

*> and you get out even from the game on average.
*

*>
*

*> Of course, you can just work out the maths for all
*

*> the possibilities in each case and see why you are more
*

*> likely to be in the majority. But to gain an intuition,
*

*> the thing that makes it be more likely for you
*

*> to be in the majority is the fact that your presence
*

*> raises the chance that whatever side you are is the
*

*> majority. In the limit of large N, your presence does not
*

*> make much difference.
*

*>
*

*> Eric.
*

*>
*

*>
*

*>
*

*>
*

Received on Mon Oct 11 2004 - 04:08:51 PDT

Date: Mon, 11 Oct 2004 18:08:10 +1000

Norman,

I haven't (once again) read your email very carefully.

I think you have made a mistake in the calculations.

You always lose on average in that version of the game.

To see that, imagine you are the last one to flip your

coin (this does not affect the analysis, but makes it easier

to visualize). The other players have either come to an

even or an uneven result. In the case N=6, it is either

3/3 or one of {4/2, 5/1, 6/0, 0/6, 1/5, 2/4}

In the first case, whichever result you get in your flip,

you are gonna be in the Losing Flip. In the second case,

your result won't affect the Losing Flip (because since

there is an even numbers of other players, the Losing Flip,

whichever it is, has at least 2 players more than the Winning

Flip.

So it means that the amount that your outcome affects the

result depends on the probability of a draw prior to your

flip. It is easy to see that this probability decreases

for large N, being zero for infinite players. It means

that with infinite players, the losing flip is independent

of your outcome, so that you are equally likely to be in

either.

Is that right or have I made another mistake?

Eric.

On Mon, 2004-10-11 at 16:31, Norman Samish wrote:

Received on Mon Oct 11 2004 - 04:08:51 PDT

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