# Re: The World Serious Problem. [SPOILER]

Dr. Ernest N. Prabhakar (ernest@pundit)
Wed, 26 Feb 97 20:58:18 -0800

Hey, you never told us the answer to the last puzzle, did you?

Okay, let's do this by induction.

If there is only one game, you simply bet 1000. Anytime a team is
one game away from winning and your balance is x, you must bet 1000-x.
By symmetry, any time the series is tied, you must also have zero
balance.

For the two game case, we can obviously bet 500 on each of the first
two games. If the same team wins both, we are +-1000, as desired. If
we end up tied, we are at 0 balance, and we can just bet 1000 as in
the 1 game case.

Now let's move to the three game case. We need to bet the same
amount, which I call X/2, on the first two games. After two games, we
either have:
a) 2-0: +-X
b) 1-1: 0 (which reduces to the two game case)

For (a), we bet 1000-X. If the leader wins, game over. If they
lose, we have Y = 2X-1000, and a 2-1 game. As before, we have to bet
1000-Y. If the leader loses again, we have a 2-2 game with a balance
of 2Y-1000, which is 2(2X-1000)-1000 = 4X - 3000. By symmetry this
has to be zero, giving X = 3000/4 = 750, and X/2 = 375.

For the four game case, we have the same initial condition, where
symmetry requires we bet the same amount of the first two games,
giving us balance W (2-0) or zero (1-1). The latter is the same as
the three-game case, of course.

Let us focus on the cases where the teams split the next three games
to end up 3-2 and have a balance V. At that point, you must bet 1000
-V in case you win, but if you lose 2V-1000 must be zero, so V = 500.
The two paths of interest are:

1) 2-0:W -T> 3-0:W+T -[1000-(W+T)]> 3-1:2(W+T)-1000 -> 3-2:4(W+T)-3000

2) 2-0:W -T> 2-1:W-T -[ W-T]> 3-1:2(W-T) -> 3-2:4(W-T) - 1000

The other possibilities merely terminate if they branch away from
these, and act as constraints.

>From 2-0 we place a bet T. For case (1), we have W+T [3-0], so we
must bet 1000-(W+T). If we lose, we end up with S = 2(W+T)-1000[3-1].
In case we win, we have to bet 1000-S = 2000-2(W+T), which if we
lose gives us 2S-1000 = 4(W+T)-3000 balance, which should be 500. So,
W+T = 3500/4, T = 875-W.

The question is whether (2) gives a compatible answer. A bet of T
gives us a balance of W-T if we lose. To maintain symmetry in case we
lose again, we must bet W-T, which gives us R = 2(W-T) if we
win[3-1]. We must bet 1000-R in case we win, which we if we lose
gives us 2R-1000 = 4(W-T) - 1000 = 500, or W-T = 1500/4, T = W - 375.

Combining gives us 875-W = W-375, or 2W = 1250, so:
W = 625
T = 250
W-T = 375
W+T = 875
S = 750
R = 750

So S=R at [3-1], which should be true (but need not be).

To summarize, we bet 312.50 for the first two games. If they are
split, we go to the three game case. Otherwise, (assuming we've won,
else invert) we have 625 [2-0] and bet 250, giving us either 875 [3-0]
pr 375 [2-1]. In the first case we bet 125, in the latter 375. The
options are then:
4-0: 1000 (end)
3-1: 750
2-2: 0 (2-game case)

If the series is at 3-1, we bet 250 again, giving either 1000[4-1] or
500[3-2]. We then bet 500, giving us zero if we lose [3-3], at which
point we bet 1000 as in the one game case.

So, the answer to the question is you must bet 312.5 on the first
match (unless I messed up the math somewhere).

You wrote:
> The question is, what bet can you place on Game 1 of the series so
> that no matter what the eventual outcome of the series (4-0, 4-1,
> 4-2, 4-3, 3-4, 2-4, 1-4, 0-4), you will win exactly \$1000 if your
> team wins the series, and you will lose exactly \$1000 if your team
> loses the series. Your answer should be either in the form of an
> amount, a range of amounts, or the statement that there is no way to
> do it.
>
> ----