Conference rusure::math

    Calling 3n/2 is not the best strategy, since it makes no use of the
    information you already have. If you are calling first you might like
    to adjust 3n/2 to allow for what you are hiding. If you are playing
    later in the round you can adjust 3n/2 according to what you deduce
    about the other players likely plays from their calls.
    
    In general there is a big difference between computer simulations of
    games like this and real life games like this where real beers are at
    stake. A little arithmetic helps (I hesitate to says maths) and a lot of
    psychology and knowledge of the other players previous behaviour can
    help a lot, since when it really counts n is small (ie. 2 or 3) and
    very few players will behave randomly.
    
    Brian

Something not defined in this description is what happens if noone gets the 
right total. Is that round scratched, or does the closest 'win' the round?
If the latter, what about ties?

Let's assume that the round gets scratched. Now let's look at the game for 
a small number of players and see how it works. It's very difficult to 
spell out exact strategies for everyone, but some dominant strategies can 
be identified.

When there are only two or three players, the last player has a slight
advantage: s/he knows not only what s/he holds but what the earlier players
have guessed. To overcome this advantage, the earlier players should play
so as to maximize the later players' errors: i.e. hold 0 or 3 coins and
guess high or low, resp., i.e. bluff a lot. This maximizes the chance that
the round will be scratched. 

On the other hand, with a large number of players the last player may find
that all the reasonable guesses have been taken. In this case s/he should
play to skew the total to one extreme or the other, i.e. hold 0 and take 
the largest low number available, or hold 3 and guess the smallest high 
number available. This maximizes the chance that the actual total is still 
guessable. In general, the last player should always hold 0 or 3 50% of the 
time and guess according to the best information available.

In either case the intermediate players have little to say about the 
outcome. They do well to play randomly, guess as accurately as possible 
(e.g. hold 2 and guess (3n+1)/2) and hope the swingers on either end cancel 
each other out.

re .1

You are absolutely correct in saying that a call of 3n/2 ignores available
information. Indeed, when the play is down to 2 players, the 2nd player would
be very foolish to call 3 after a call of 0 when holding 1 coin.

On the other hand, the first player has no information to go on and might
reason as follows:

	"If I hold 	0: the possible outcomes are:	{0, 1, 2, 3}
			1: 				{1, 2, 3, 4}
			2: 				{2, 3, 4, 5}
			3: 				{3, 4, 5, 6}
	The one value common to all possible outcomes is 3; I'll call that"	

re .2

If no player guesses the correct outcome, nobody retires and the game proceeds 
to the next round. This - to my mind - somewhat reduces the value of bluffing 
as, unlike in Poker, it rarely leads to a win but merely maximises the chances 
for a standoff. And practical (read: bitter :-) experience shows that the longer
you stay in, the greater your chances of losing. Tell me if I am wrong.

I am fascinated by your strategy for the last player.

Players at each end of the order have advantages: The early players have
the maximum of choice of plays, while the later ones have more information.
Over the long haul I think I'd rather be last, since I think I can make
better use of the information than of the choice. If you tend to lose this
game consistently it's mostly dumb luck, but bluffing occasionally ought to
help your percentages. In any event, dragging the game on should help when
you are in a bad position, e.g. in the middle.

Yes, this looks like a game which falls into the "interesting" class: there is
an opportunity to do some thinking, but there is no clear optimal strategy.

.3>And practical (read: bitter :-) experience shows that the longer
>you stay in, the greater your chances of losing. Tell me if I am wrong.

The word 'longer' in the sentence above is ambiguous.  Are you talking about
rounds of guesses or rounds of beer? ;-)

In terms of rounds of guesses: unless you have a winning strategy for small n,
you are best off trying to get out when n is large, just to avoid danger.

When n is large and you are an early guesser, I suspect your best strategy is 
to hold a random number of coins and guess as near as possible to 3n/2.  This
is a good strategy if everybody is counting on dumb luck.  It has the 
additional advantage of being unrevealing: no later guesser can use the
information provided by your guess.  If you can guess the strategies of the 
later guessers, you might improve on this.  For example, if you expect them to 
draw obvious conclusions from your guess, you might hold 0 coins and guess
close to 2+(3n/2).  Then they will also guess hign and if you have guessed 
wrong, at least they are likely to be wrong as well.

When n is large and you are a late guesser, you have all the previous guesses
to reason from.  If you suspect that early guessers are playing the simpler
strategy in the paragraph above, you can do no better than guess the 
remaining number closest to 3n/2.  If you suspect them of a more complex 
strategy, you have to guess what it is and play accordingly, for example:

	their guess	their strategy		your best guess

	high		hold high		high
	high		hold low		low

This will get pretty complex since early guessers may use different strategies,
and the late guessers must consider all early guessers and their guesses.  I
think I would play simple in the early rounds and save my brainpower for my 
beer and the rounds where n is small.

I'll enter this note now, before the network eats it.  I may talk about small
n in a later note.

.3>On the other hand, the first player has no information to go on and might
>reason as follows:
>
>	"If I hold 	0: the possible outcomes are:	{0, 1, 2, 3}
>			1: 				{1, 2, 3, 4}
>			2: 				{2, 3, 4, 5}
>			3: 				{3, 4, 5, 6}
>	The one value common to all possible outcomes is 3; I'll call that"	

Assume first that this game is being played exactly once among total
strangers.  Then you are correct in thinking that the first player has no
more information than the coins in his/her hand.  The guess 3 actually gives
the first player no better chance of guessing right than any other number
in the appropriate row above, but it does deny the second player any new
information.  Since 3 = 3n/2, it also obscures whether the first player is 
using a strategy more complex than 3n/2.

Now consider the second player, who knows that the first player has guessed
3.  The second player has already picked the number of coins to hold, but 
under the assumption above, the second player might reasonably expect a guess
of 3, and hold stones accordingly.  By a reasoning similar to .2, I believe 
that second player should hold 0 or 3 stones randomly.  Let's assume the 
portable random number generator said to hold 0.  

Assume further that second player is either a Bayesian (or a frequentist who
knows how to use Bayesian reasoning) and reasons as follows:

	My opponent must be holding {0,1,2,3} coins.  I have no information
	which would affect the relative probabilities of these outcomes, so
	I must assign them the same probability.  So the expected number of
	coins s/he holds is 1.5, so I can guess 1 or 2 with equal confidence.
	I consult my random number generator again and come up with 2 as
	my guess.

If first player held 3 coins, s/he wins, if 2, second player wins, if 0 or 1 
then they play again.

Presumably second player now goes first.  But now the situation is different.
Both players have heard their opponents guesses and seen their holdings.  So 
their optimal strategies will depend on what they have seen.  If neither
wins this round, they will play again, with even more information.

This is getting too complex for me, and I haven't even had any beer.