Conference rusure::math

    Tell Susan that the first person has the advantage.

    Suppose B is the bill.  The first guesser should choose X very
    close to the expectation's midpoint.  (Below, I assume that the
    pay-off is independent of the bill's amount).  That is, choose X
    so that:
    
	P( B > X ) <= 1/2,  and  P( B < X ) <= 1/2.

    Surely there is such an X.  (why?)
    
    Now suppose the second guesser chooses X+1;  his probability of being
    right is P(B>X), which is <= 1/2.  Suppose the second guesser chooses
    X-1; his probability of being right is P(B<X), which is also <= 1/2.

>	P( B > X ) <= 1/2,  and  P( B < X ) <= 1/2.
>
>    Surely there is such an X.  (why?)
    
    	Right.  X=B.  Anything larger has 1/1 probability of being >X
    	and anything smaller has 1/1 probability of being <X.
    
    So you're saying first guesser should choose B exactly.  But how's
    Susan supposed to do that ?
    
    Anyway, how do you refute the .0 argument of saying second person
    has larger range covered.
    
    /Eric

> thinks bill is between $4 and $5 so she guesses $4.50.  Her son thought
> bill was between $3.50 and $4.50 so when she says $4.50, he says $4.49.

    I hope her son isn't named Bill, or this could get mighty confusing.

    To me, it looks like Eric is right for the wrong reason.  When Susan
    says $4.50, Sonny should answer with the guess that maximizes the
    amount in HIS range.  It works out to be the same.
    
    One could devise a computer game to simulate this.  Have a random
    "bill generator" select a number N, and two intervals (A, A+100)
    and (B, B+100) where N is contained in both intervals.
    Your opponent guesses N, then you get to guess N after seeing his guess.
    I estimate the person who goes second will win 2/3 of the time.
    Shouldn't be that tough to work thru the math if someone wanted to.

      John

    Let's look at a different game.
    
    We have a source of random real numbers, following a "well-behaved"
    distribution with a well defined but unknown mean.  You can picture the
    distribution as normal if you wish.  The game is to guess the mean
    as accurately as possible.  Player 1, is given a single output,
    while player 2 is given two outputs from the source.  Who has the
    advantage?  Clearly player 2 does.  By averaging the two numbers
    player 2 can, on the average, do better in the game by a factor
    of about sqrt(2) (exactly that if normal).  Player 1's best strategy
    is to guess the number given to him/her.
    
    What if one of the numbers given to 2 is the same as the one given
    to 1?  Does that change anything?  Answer: no, not if the goal is
    to guess the mean as accurately as possible.  But it *does* change
    things if the goal is to simply guess the mean more accurately than
    the opponent and if player 2 knows which number is the same as that
    received by player 1.
    
    Let's assume that the second sample, and thus the average of the
    two samples, is larger than the first sample.  If player 2 guesses
    the average of the two samples, (s)he wins if the actual result
    is greater than that average, or if the actual result is greater
    than the half-way point between his/her guess and player 1's guess.
    By moving his/her guess down toward player 1's guess, the size
    of that latter region is increased without losing any other area.
    So the best strategy for player 2 is to make a guess as close as
    possible to player 1's guess on the same side as the average of
    the two samples, which is the same (since one sample is the same
    as player 1's best guess) as the side of player 1's guess that
    player 2's second sample is on.
    
    Except for the fact that the game involves a discrete source of
    random variables, rather than continuous, this latter game is
    equivalent to the game in .0 assuming that the two players are
    approximately evenly matched, and they are both "unbiased" guessers
    at the price.  (The former assumption doesn't really make any
    difference if violated, but it complicates the analysis;  the
    latter adds an extra factor to the ideal strategy if the bias
    is known).  We can view the two players as sources of random numbers
    with some distribution around the mean of the actual price.  The
    first player has their own a priori guess as a sample from that
    source, while the second player has the first player's guess plus
    their own guess made before they heard the first player's.
    
    The second player has a clear advantage, because they have more
    information.
    
    					Topher

    I should add that there are circumstances, unlikely to apply in
    the actual game, where the first player does have an advantage.
    
    If the accuracy of the guesses is sufficiently high (variance of
    the sampling sufficiently small) then there is a good chance that
    the first player can guess the right amount exactly (this is possible
    because the distribution is discrete;  for a continuous distribution
    the probability of getting *exactly* the right answer is zero).
    
    The first player therefore has the advantage of being able to block
    the second player from guessing the exact answer.  To take advantage
    of this the first player needs to get "right on the money" fairly
    often.  The exact accuracy needed depends (I think sensitively)
    to assumptions about the distribution of guesses around the actual
    value.
    
    						Topher

    I agree with the original noter that the second guesser has the
    advantage.  The reason is when it is time for the son to make a
    guess, he has the extra informtion of what the mother's estimate
    of the bill is.  The reason there is a controvercy is because
    the problem is not really well formulated in the original note;
    hence, is open to different intepretations.
    My assumptions are:
    
    1.  The mother believes that the distribution function for the bill is
        a bell curve with maximum at $4.50.
    
    2.  The son believes that the distribution function for the bill
        is a bell curve (same shape as his mother's) with maximum at $4.00.
    
    3.  The distribution of the actual bill is a bell curve with maximum
         at $4.25.
    Eugene

re .2:

Hmmm.  I gues I wasn't clear enough.  In the following,

>	P( B > X ) <= 1/2,  and  P( B < X ) <= 1/2.
>
>    Surely there is such an X.  (why?)

X is the value that's *chosen*, and B is the random variable.

I've started on a formal proof.  Doesn't anybody else see why this works?
Moral support requested.

    I thought about the problem again, and I retract .6 .  I need to
    think about it more (perhaps not :-).
    Eugene

I would restate the competition to be as follows:

	Player 1 guesses the bill
	Player 2 guesses whether the bill is high or low

An algorithmic solution for player 2 is almost identical to that for player 1,
but in my opinion an intuitive solution is much easier for player 2.

.7> Hmmm.  I gues I wasn't clear enough.  In the following,
.7>
.7>>	P( B > X ) <= 1/2,  and  P( B < X ) <= 1/2.
.7>>
.7>>    Surely there is such an X.  (why?)

    There is such an X since CDFs are continuous monotonics defined everywhere.
    But the first guesser doesn't know P().

      John

    Re .7:
    
    Okay, you have moral support.
    
    However, I think we need some more discussion about the knowledge each
    of the players has.  If the first player's believed distribution for
    the price does not correspond to the actual distribution, they cannot
    find the optimal guess that your proof requires.  Can the second player
    take advantage of that even if they do not have better information
    about the distribution?
    
    
    				-- edp

    I stick by my belief that player 2 has the advantage.  Let me try
    to describe why again.

    Player 1 is attempting to guess the price.  She makes some kind
    of internal estimate.  This may be a point estimate (e.g., somewhere
    around $4.25) or it may be a distribution (e.g., 95% sure between
    $2 and $3, equally likely for all values).  On the basis of this
    internal estimate -- it doesn't matter how -- she chooses a single
    best choice.  If we observe the game played with the same player 1,
    and for some particular actual price M, many many times (all other
    factors beyond our observation), we will observe a distribution
    of player 1's guesses.

    Call the random variable representing player 1's guesses G1, with
    distribution g1.  For simplicity let us assume that g1 is symmetric
    about the actual price M (e.g., player 1 is equally likely to guess
    10� high as 10� low -- this does not effect the outcome of the analysis).
    Of course, since the price cannot be a fraction of a penny, the
    distribution is discrete.

    Now if player 1's guess is exact then she wins whatever player 2 does.
    This is her advantage.  Let us call the probability that player 1
    guesses the exact price P1E (for Probability of player 1 guessing
    Exact price).

    To show that player 2 has an advantage we have to show that there
    is *a* strategy which give player 2 an advantage.  We don't have
    to worry about whether that strategy is optimal, i.e., whether
    player 2 could do even better.

    Strategy: player 2 goes through exactly the same process as if
    she were player 1 ignoring player 1's guess, coming up with a single
    guess.  This guess can be represented by a random variable which
    we will call G2.  If G2>G1, player 2 guesses G1+1, if G2<G1, player
    2 guesses G1-1, if G2=G1 player 2 chooses between the two randomly.

    In the last case (G2=G1), and assuming that G1~=M, player 2 has
    a 50% chance of winning.  Since neither player has an advantage
    in this case, it does not effect our judgment of who has the
    advantage.  So we can simply exclude this case from our further
    analysis.

    Let the distribution of G2 be g2.  For simplicity assume g1=g2
    (roughly correct if they are evenly matched; the analysis is complicated
    if it is not true but, I don't think it changes the results much,
    as long as they are evenly matched (or player 2 is better than player
    1)).  We further assume that G1 and G2 are independent (if they
    are not, then, once again the analysis is made more complex but
    it shouldn't effect the results;  we must then divide g1 and g2
    into dependent and independent parts and repeat do the same
    analysis on the independent parts.  This analysis breaks down if
    if G1 and G2 are completely dependent, i.e. if player 2's best guess
    is always the same as player 1's.  This seems to be the assumption made
    in .1).

    Now player 2 will win if player 1 fails to guess the exact price
    and if G2 is not beyond G1 relative to M, so the probability of player
    2 winning is Prob(G1~=M)*Prob((G1<M & G2>G1) or (G1>M & G2<G1)).
    Now Prob(G1~=M) = (1-P1E).  Because we assumed that G1 is unbiased
    it is equally likely that G1<M and G1>M, and because we assumed
    that it was completely symmetric it doesn't matter which we choose,
    so let's assume that G1<M (i.e., player 1 happens to choose low).

    If player 1 doesn't guess the price exactly, player 2 will win
    either if she guesses on the opposite  side of the true price
    (probability .5) or if she guesses between player 1's guess
    and the true price (inclusive of the latter).  The latter probability
    is player 2's advantage.  Player 2 has the total advantage if
    the probability of doing that is greater than P1E.

    How can we estimate that?  To get a ballpark figure, let's assume
    that G1 and G2 are distributed normally.  We could then perform
    an integration of the normal distribution function (the integral of
    z(x)*Z(x)dx from 0 to infinity where z(x) is the standard normal pdf
    and Z(x) is the standard normal cpf), but that's rather difficult.
    Alternately we could perform a simulation, but there is an easier
    way to get a rough estimate.  The average distance of a random
    sample from the mean of a standard normal distribution is SQRT(2/pi)
    (according to several statistics handbooks I checked).  The probability
    of player 2 choosing a value greater than -SQRT(2/pi) is approximately
    0.79.  That means that player 1 must be able to guess the exact price,
    to the penny mind you, roughly 1 time out of 3 in order to break even.
    Since I would say a player would be doing very well to guess within 10�
    of the price, this would seem to be unrealistic.

    I therefore conclude that player 2 has a strong advantage.

				    Topher