Suppose you are one of two rational players tied with $5000 in two-player Final Jeopardy!

I wanted to call your attention to an article in the October issue of
Mathematics Magazine (Vol. 67, No. 4), "Wagering in Final Jeopardy!"
by George T. Gilbert and Rhonda L. Hatcher at Texas Christian
University.  The authors look for optimal betting strategies for all
possible distributions of scores at the start of Final Jeopardy!,
under the assumption that the goal is to win or tie for first without
regard to the amount of money won.  They consider strategies for
playing against actual contestants (whose strategies are assumed to
follow empirical distributions) and strategies for playing against
rational opponents.

-- David A. Karr <karr@cs.cornell.edu>, October 1994

In article <384b73$eno@netnews.upenn.edu> harris99@futures.wharton.upenn.edu
(David Harrison) writes:
>Well, it seems to me that you are either going to wager nothing or
>everything.  So if you have better than a 50% shot of getting it, you'd
>wager it all.

Bearing in mind that it makes very little difference to you whether the
other person loses or manages to tie your score, i.e. the primary goal
is not to lose, is the above really a complete analysis for the case of
two rational players?

Suppose both players have 60% chance of getting the question right.
Suppose they each bet everything.  Then the outcomes are

P(A right, B right) = .36       tie
P(A right, B wrong) = .24       A wins
P(A wrong, B right) = .24       A loses
P(A wrong, B wrong) = .16       tie

So A wins with probability .24 and ties with probability .52.  If A
were to bet less than everything, the first outcome would change to a
loss and the last one to a win, so A wins with probability .40 and ties
with probability 0.  Assuming a tie is very nearly as good as a win,
this should dissuade A from changing his bet, and the same goes for B.
So this is an equilibrium point.

Now suppose they each bet nothing.  Then the outcomes are

P(A right, B right) = .36       tie
P(A right, B wrong) = .24       tie
P(A wrong, B right) = .24       tie
P(A wrong, B wrong) = .16       tie

So A ties with probability 1.  If A bets any positive amount, A wins
with probability .6 but ties only with probability .16.  So again,
assuming ties are very nearly as good as wins, this is an equilibrium
point.  What's more, this equilibrium is much better for each player
than the other equilibrium.  So which ones will rational players
choose?

P.S. According to the article I cited earlier, the top two contestants
in a three-way Final Jeopardy! each get the question right just over
50% of the time.  I would guess this would still be true if the third
contestant was eliminated by a negative score.

-- David A. Karr <karr@cs.cornell.edu>

[ Editor: Matthew T. Russotto <russotto@bigdog.fred.net> pointed out a
flaw in this analysis; Karr followed up: ]

To recap and clarify: The assumptions of the Jeopardy! analysis are
that whoever is in first place, or ties for first place with a positive
score, receives the money and can play again.  Anyone else receives a
consolation prize and is sent home.  Gilbert and Hatcher assume that
the payoff function for each player is 1 for a win (regardless of the
final score), 1 - epsilon (where epsilon is very small) for a tie with
positive score, and 0 for a loss or a tie at 0.  I used these
assumptions too, though I think that totally ignoring the actual
winning score is somewhat unrealistic; it doesn't explain why when the
scores are A, B, and C with A > 2B > 2C, the leading player should
usually wager A - 2B - 1 (depending on the category).

Here's what I should have written for the payoff matrix when both
players bet everything with 60% chance each of getting the question
right:

P(A right, B right) = .36       tie
P(A right, B wrong) = .24       A wins
P(A wrong, B right) = .24       A loses
P(A wrong, B wrong) = .16       BOTH PLAYERS LOSE

Now A's expected payoff for betting everything, assuming that B bets
everything, is

        0.24*1 + 0.36*(1 - epsilon) = 0.6 - 0.36*epsilon

while the payoff for betting less than everything (assuming B
continues to bet everything) is 0.4.

So my conclusion was correct, i.e. this is an equilibrium point.

But suppose the probabilities of a correct answer are very different,
in particular, a category has been chosen in which it is known that A
will get the question right 70% of the time and B only 20% of the time.
Now the payoff matrix for "both bet it all" is

P(A right, B right) = .14       tie
P(A right, B wrong) = .56       A wins
P(A wrong, B right) = .06       A loses
P(A wrong, B wrong) = .24       BOTH PLAYERS LOSE

Now it turns out this is not an equilibrium point, because B can
improve B's expected payoff by at least 0.1 simply by betting less than
everything while A continues to bet everything.  Paradoxically, A has a
similar motivation.  In fact, I claim the "both bet everything"
strategy is an equilibrium point if and only if the players'
probabilities of getting the question right sum to at least 1, that is,
the important thing is their total strength rather than their overall
strength.

But in any case, unless one of the players has at least 1 - epsilon
chance of getting the question right, the "both bet nothing" strategy
is an equilibrium and it pays each player better than the "bet
everything" equilibrium.

-- David A. Karr <karr@cs.cornell.edu>, October 1994

[ Editor: see also competition/games/jeopardy. ]