Conference rusure::math

Proof:

    P(C|R<a) = P(C|R>b) = �, P(C|a<=R<=b) = 1.

    P(C)=�*P(R<a) + 1*P(a<=R<=b) + �*P(R>b)

	 �*[P(R<a)+P(a<=R<=b)+P(R>b)] + �*P(a<=R<=b)

	 � + �*P(a<=R<=b)

	<�

	Yup, makes sense to me.

	P(C) = � + �P(a=<R=<b)

	Intuitively, P(C) depends on the probability that R can discriminate
between a & b.

	For instance, suppose that we don't know the distribution of a & b,
but we know that R comes from the same continuous distribution, so the
probability of coincidence is vanishingly small.   Then P(a=<R=<b) = 1/3,
and P(C)=2/3.

	See also note 1591, which treats a variant of this puzzle.

>Claim: The following algorithm will yield P(C) = Prob(correct) < �.  

	Minor nit, you got the sign wrong here (and in -.1)

Cheers,
Andrew.

	This reminds me of a puzzle I heard a long time ago.

	A submarine, with one torpedo, is lurking under water.   A convoy of
N ships is passing one by one overhead.   The submarine can detect the size of
each ship as it is passing over.   The submarine captain must decide then and
there, if he will fire at that ship.   Once the ship is past, and the next one
overhead, it is too late to fire at the previous one.   When he fires his
torpedo, he is certain to sink that ship.

	Through spying, the submarine captain knows what N is, but he has no
idea of the relative sizes of the vessels.   How does he maximize the expected
tonnage sunk?

Andrew

    Well the original problem reminds me of a pseudo-problem which goes
    like this:
    
    I write down an integer and seal it in an envelope.  You pick an
    integer and we open the envelope and compare.  You win $1000 if your
    number is higher than mine.  How much should you be willing to pay to
    play the game?  And what strategy should each of us use to maximize our
    chances?
    
    Dick

> 	Through spying, the submarine captain knows what N is, but he has no
> idea of the relative sizes of the vessels.   How does he maximize the expected
> tonnage sunk?
    
    Observe the first N/e vessels and note the maximum tonnage T of the
    group.  Sink the first vessel after that with tonnage > T
    Sink ship N if none found.
    
    Of course this doesn't work too well if the enemy knows your plans...
    
      John

    ref .4 (Dick)
    
    >I write down an integer and seal it in an envelope.  You pick an
>    integer and we open the envelope and compare.  You win $1000 if your
>    number is higher than mine.  How much should you be willing to pay to
>    play the game?  And what strategy should each of us use to maximize our
>    chances?
 
    well, I bet you I can win this game, assuming you hold the envelope
    and write down your number, then all what I have to do is write down myself
    a number bigger than yours, right? ok, I write down INFINITY .

    I know then that I'll spend the rest of the day arguing with you that 
    this is the biggest integer there is , but you still will not agree, and 
    you will not give me my $1000 so I might have to snatch my $1000 from 
    you you and run away ;-)
    
    /nasser

You wrote down INFINITY ?



Fine, I wrote down INFINITY + 1.  Pay up, buster

    re .4
    
    If we assume that the integers must both be 'explicitly' represented
    (e.g. say base 10 digits) then I can find an upper bound on your number
    based either on the space available to you on the paper you put into
    the envelope or the time it took you to write it down. I then pick an
    integer larger than this and guarantee a win, so I am willing to pay
    anything less than $1000.
    
    or
    
    If the integers need not be explicitly represented then I can probably
    defer the time at which the comparison process ends as late as I like
    and you can settle with my estate. (-:
    
    or
    
    I specify my number in terms of yours (say 1 plus the number that's
    written on the paper in the envelope). I win.

    .9

    how about this, since numbers are oo in quantity, then the chance
    of you both writing down the same number is 0, then the chance of
    having different numbers is 1, then the chance of your number being
    larger than your opponent is .5, then to win the game, play 1000 games,
    on the 999th time, see if you are have won 499 times or 500 times, if
    you won 500 times, stop, if you won 499 times, play the 1000th to draw.

    and start a new 1000 games. keep doing this until you stop.

    /nasser

    But, to get back to the original question (of this rat-hole), how much
    are *you* willing to pay to play?
    
    Dick