Two gamblers, A and B, are about to play a game.
decision/two.gamblers New
New entry, edited from material collected for the archive and not yet in a released version.
Problem
Two gamblers, A and B, are about to play a game. Each will privately write three amounts adding up to $1000 on a slip of paper. Then they will compare their amounts, the largest against the largest, the smallest against the smallest, and the median against the median. The gambler with the larger amount in two of these three comparisons will win the game and take the $1000. What should gambler A write on the slip of paper? -- MathSoft puzzle #11, December 1995
Show solution
[ Editor: from the MathSoft, Inc. puzzle web page, updated February
1996. Most of the formulas and the diagrams on that page were images,
which are not in the collected material; only the text is given here. ]
Consider the following diagram, where x denotes A's choice of the
smallest amount and y denotes A's choice of the median amount. If z
denotes A's choice of the largest amount, then clearly z = 1000 - x - y,
so we immediately have two constraints: y >= x and 1000 - x - y >= y.
The shaded region gives all the feasible points (X,Y) on which B will win
[relative to A's choice of (x,y)]. Our aim, therefore, is to minimize the
area of the shaded region. The region clearly can be partitioned into
four triangular subregions, and we apply a well-known determinant formula
to compute each of these areas.
[ Editor: the four determinants, their sum, and its simplification and
derivatives were images. ]
Solving this 2-by-2 linear system ... Let's not be *too* hasty in
accepting this as the correct solution. Look at a contour plot:
The contour plot shows that our candidate solution is, in fact, a
*saddle point*. This is confirmed by the second derivative test, which
is positive. The minimum point we desire is along the upper boundary
y = 500 - x/2.
[ Editor: the substitution and the one-variable minimization were
images. ]
Take x = 231, y = 384, z = 385 (rounded to the nearest dollar). This is
the optimal choice of dollar amounts for A.
Note: our treatment has been rather incomplete since, in the above
diagram, we've *assumed* that y > 1000/3 and 1000 - x - y < 500 ! So
there are other cases which must be examined, e.g.,
y > 1000/3 and 1000 - x - y > 500
y < 1000/3 and 1000 - x - y < 500
y < 1000/3 and 1000 - x - y > 500
each of which gives rise to a slightly different diagram and hence a
slightly different triangular partition of the feasible region. The case
discussed earlier gives A's best option for winning the game.
Also note: we've assumed here that B makes a completely random choice.
If there is additional information about B's game plan, then this alters
the above solution.