Two players each draw a card from a uniform deck of numbers between 0 and 1 and may simultaneously choose...

decision/draw.or.stand

Problem

Two players each draw a card from a uniform deck of numbers between 0
and 1 and may simultaneously choose to keep it or replace it with a new
draw; higher card wins.  What is the best strategy?

You and a friend play a game using a large deck of random numbers from
0 to 1.  You each draw one card face down and look at its value.
Neither player can see the other's card.

You each decide whether or not to draw another card to replace the
first card.  Simultaneously, using hand signals (ala
rock-paper-scissors), each player indicates if he wishes to draw or
stand.  Thus, neither player knows what the other will do before
indicating his intentions.  If a player draws, he discards the original
card.  The player with the higher card then wins.

(1) After a few games, you note that your friend stands if his first
card is greater than .5 in value.  Your first card in a new game is
.56.  Should you draw or stand?

(2) The next time you play, you note that your friend has a new
strategy regarding when to draw or stand.  He tells you that he got
this strategy from a math student.  He thinks it will optimize his
results against the most intelligent strategy you can employ.  If the
math student is correct, what is the strategy?

-- Bill Ryan [Editor: signed "Bill Ryam"; posted as BillRyan22]

Show solution

[ Editor: a reply to part (1). ]

(a) If you stand on .56, then your chances of winning are (.5)(.56) +
(.06)(1) + (.44)(0) = .34.  If you draw again, then you will get some
number x.  If x is less than .5, then your chances of winning with it
are (.5)x = x/2.  If it is greater than .5, then your chances of
winning are (.5)x + (x-.5)(1) = (3x-1)/2.  So, if you draw again, then
your chances of winning are (Int(x=0;1/2) x/2 dx) + (Int(x=1/2;1)
(3x-1)/2 dx) = .375 (easier than integrating is drawing the graph and
calculating the areas by geometry).  So your chances increase by about
10% if you draw again.

(b) eep.  Another day.

-- Matthew Daly <mwdaly@pobox.com>

(1) Daly is right: draw.  Standing on .56 wins with probability .34,
drawing with probability .375.  More generally, against a friend who
stands above .5 you should stand only on cards above 7/12 = .5833.

(2) The math student's strategy: stand if the card is greater than
(sqrt(5) - 1)/2 = 0.618..., the golden ratio less 1; draw otherwise.
It guarantees a 50% chance of winning whatever you do, and nothing
guarantees more.

Proof.  Suppose your friend stands exactly when his card exceeds t.
His final card is then a fresh uniform card with probability t, or a
card uniform on (t,1) with probability 1-t, so the chance that his
final card is below y is

  G(y) = t y               for y <= t,
  G(y) = (1+t) y - t       for y >= t.

If you stand on x you win with probability G(x); if you draw you win
with probability equal to the average of G over (0,1),

  E = t^3/2 + (1+t)(1-t^2)/2 - t(1-t).

G is increasing, so your best reply is also a threshold: stand when
G(x) > E.  The friend's threshold is unexploitable exactly when your
best reply is the same threshold, i.e. G(t) = t^2 = E, which
simplifies to

  1 - t - t^2 = 0,   t = (sqrt(5) - 1)/2 = 0.6180...

Then the best you can do against him is to use the same rule, which
wins half the time by symmetry; so every strategy of yours wins at
most 1/2 against it (ties have probability 0, so the game is
zero-sum).  Conversely, against any other threshold t you can do
better than 1/2 (your best reply has threshold different from t), so
the golden threshold is the only optimal one.  For example, the best
reply to t = .5 is the threshold 7/12 found above, and the best reply
to t = .65 is about .594.

Whittington's reply (stand or draw at random, without looking) leaves
his final card uniform, so it is easily beaten: stand on cards above
.5 and draw otherwise, and you win 5/8 of the time.

Check: a C++ computation of the best-reply threshold (found by
bisection, with E integrated numerically) returns t itself exactly at
t = 0.618034, the best reply to it wins 0.500000, and 4,000,000
simulated games of golden threshold against itself give 0.5002.
-- Claude (rec.puzzles Oracle, 2026-11-02)

[ Editor: the material collected earlier follows. ]

The only reply collected is this one, to part (2):

Without even looking at your card, at some random time after it is
dealt, glance quickly at the seconds display of your digital watch.
If even, stand.  If odd, draw.

-- Fritz Whittington <fritz@ti.com>