Strategic betting on the world series

Question

Your wealthy father-in-law gives you \$10000 to place a "double-or-nothing" bet on Boston Red Sox in the World Series [1]. In other words, your father-in-law expects to receive \$20000 if the Sox win, and \$0 if they lose.

To your horror, you then discover that the only kind of bets that are allowed at your neighborhood casino are double-or-nothing wagers on individual games. For simplicity, we assume that the casino accepts bets up until the start of each game and that the house cut is negligible. How much should you bet on the first game? I.e., what strategy would you follow so that your final outcome is certain to be equivalent to an overall double-or-nothing bet on the series?

To those who are not familiar with the World Series: the World Series is a "best-of-seven" event in which two teams play a series of games and the first team to win four games is the winner.

Answer 1

Let's call the other team the Los Angeles Dodgers. Let's solve this by

working backwards from our goal: we should have \$0 if the Dodgers win 4 games, and \$20000 if the Red Sox win 4 games.

Observe that at any point,

in order to reach one of two particular dollar amounts after the next game, we need to start with the average of the two amounts and then bet half their difference.

For instance,

after each team has won 3 games, we should have \$10000 so that we can bet \$10000 on the Red Sox and end up with the correct total.

Repeating this process,

Format: Red Sox wins, Dodgers wins: Total, Bet (comparison)

4, x: \$20000
x, 4: \$0

3, 3: \$10000, \$10000 (x, 4 vs 4, x)

3, 2: \$15000, \$5000 (3, 3 vs 4, x)
2, 3: \$5000, \$5000 (x, 4 vs 3, 3)

3, 1: \$17500, \$2500 (3, 2 vs 4, x)
2, 2: \$10000, \$5000 (2, 3 vs 3, 2)
1, 3: \$2500, \$2500 (x, 4 vs 2, 3)

3, 0: \$18750, \$1250 (3, 1 vs 4, x)
2, 1: \$13750, \$3750 (2, 2 vs 3, 1)
1, 2: \$6250, \$3750 (1, 3 vs 2, 2)
0, 3: \$1250, \$1250 (x, 4 vs 1, 3)

2, 0: \$16250, \$2500 (2, 1 vs 3, 0)
1, 1: \$10000, \$3750 (1, 2 vs 2, 1)
0, 2: \$3750, \$2500 (0, 3 vs 1, 2)

1, 0: \$13125, \$3125 (1, 1 vs 2, 0)
0, 1: \$6875, \$3125 (0, 2 vs 1, 1)

0, 0: \$10000, \$3125 (0, 1 vs 1, 0)

So the answer is

$3125.

Answer 2

First bet is \$3,125.

The second bet is also \$3,125 regardless of who wins.

The third bet is \$2,500 if Sox wins twice, or \$3,750 if tied so far.

The fourth bet is \$1,250 if Sox wins thrice, or \$3,750 if Sox wins twice so far.

The fifth bet is \$2,500 if Sox wins thrice, or \$5,000 if Sox wins twice so far.

The sixth bet is \$5,000.

The seventh bet is \$10,000.

If Sox loses more than they win, replace the number of wins in above with the number of loses, and bet against.

---

How does this work?

Start from the seventh bet. This means both teams have 3 wins. So this is equivalent to a single game. At this point out balance should be \$0, since otherwise there is no way for any bet to be equivalent to \$10,000 on a single game. Since our balance is \$0, then just bet \$10,000 as per normal.

Now, the sixth bet, if the score is 3-2 for Sox, then Sox winning must give us \$10,000 in total, and Sox losing must give us \$0 in total (as per previous point). So our balance must be \$5,000 at this point, and we should bet \$5,000.

We continue this process until the first bet.

This is easily visualized on the following graph:

$\require{AMScd}\require{enclose}$ $\def\¿{\enclose{roundedbox}}$ $\small \begin{CD} \¿{-10} @. \¿{-10} @. \¿{-10} @. \¿{-10}\\ @A-1.25AA @A-2.5AA @A-5AA @A-10AA\\ \¿{-8.75} @>1.25>> \¿{-7.5} @>2.5>> \¿{-5} @>5>> \¿{0} @>10>> \¿{10}\\ @A-2.5AA @A-3.75AA @A-5AA @A-5AA\\ \¿{-6.25} @>2.5>> \¿{-3.75} @>3.75>> \¿{0} @>5>> \¿{5} @>5>> \¿{10}\\ @A-3.125AA @A-3.75AA @A-3.75AA @A-2.5AA\\ \¿{-3.125} @>3.125>> \¿{0} @>3.75>> \¿{3.75} @>3.75>> \¿{7.5} @>2.5>> \¿{10}\\ @A-3.125AA @A-3.125AA @A-2.5AA @A-1.25AA\\ \¿{0} @>3.125>> \¿{3.125} @>3.125>> \¿{6.25} @>2.5>> \¿{8.75} @>1.25>> \¿{10}\\ \end{CD}$

The nodes represent our balance so far, and the edges represents the bet.

You start filling 10 at the right edges (and -10 at the top), and 0 in the diagonal. The anytime we see a node where its right node and top node are filled, we fill that node with the average. The edge is the difference between the two adjacent nodes, which would be the amount of bet we would need to place. If positive bet for, if negative, bet against.

Answer 3

Let's say Boston have had 3 wins so far, winning or losing a certain amount. Let the remaining amount be $n$:

Bet A: In case Boston win right after that, the next bet $b_1$ must satisfy $n+b_1 = 20000$.

Bet B: If they lose instead with neither team guaranteed to win the WS yet, you have $n-b_1$ dollars left. In case Boston win, $n-b_1+b_2 = 20000$ must be satisfied.

Bet C: If they lose again with neither team guaranteed to win the WS yet, you have $n-b_1-b_2$ dollars left. In case Boston win, $n-b_1-b_2+b_3 = 20000$ must be satisfied.

Bet D: If they lose again with neither team guaranteed to win the WS yet, you have $n-b_1-b_2-b_3$ dollars left. In case Boston win, $n-b_1-b_2-b_3+b_4 = 20000$ must be satisfied.

Bet E: If they lose yet again, then Boston have lost the series, so you should have $n-b_1-b_2-b_3-b_4 = 0$ dollars left. Then $b_1 = k, b_2 = 2k, b_3 = 4k, b_4 = 8k, n=15k, k=20000/16=1250$.

If Boston have lost $l$ times so far while having won thrice, $n+b_1 = 20000$, $n = b_1+...+b_{4-l}$, and $b_1 = k,...,b_{4-l}=2^{3-l}k$. This means $n=(2^{4-l}-1)(20000/2^{4-l})=20000-(20000/2^{4-l})$

If Boston have won $w$ times so far while having lost thrice with the remaining amount being $n$:

Bet A: In case Boston lose right after that, satisfy $b_1=n$.

Bet B: If they win instead with neither team guaranteed to win the WS yet, you have $2n$ dollars. In case Boston lose, $b_2 = 2n$.

Bet C: $b_3 = 4n$

Bet D: $b_4 = 8n$

Bet E: If they win yet again, then Boston have won the series, so $k=1250$ if $w=0$.

In general, $n+b_1+...+b_{4-w}=2^{4-w}n=20000$.

Reconciling the two:

Imagine that both teams have won twice so far. Since the next bet could lead to $n=20000-(20000/4)=15000$ if it's Boston, or $n=20000/4=5000$ if it's their opponent, the current remaining amount in a 2-2 tie must be 10000.

Possibly leading to a 2-2 tie, let's say Boston are 2-1 up. Either $n=10000$ if it's to become 2-2, or $n=20000-20000/8 = 17500$ if it's to become 3-1, so the current remaining amount in a 2-1 scoreline must be 13750.

Possibly leading to a 2-2 tie, let's say Boston are 2-1 down. Either $n=10000$ if it's to become 2-2, or $n=20000/8 = 2500$ if it's to become 1-3, so the current remaining amount in a 1-2 scoreline must be 6250.

A 2-0 scoreline becomes either 3-0 or 2-1, so the current remaining amount must be (18750 + 13750)/2 = 16250.

A 0-2 scoreline becomes either 0-3 or 1-2, so the current remaining amount must be (1250 + 6250)/2 = 3750.

In a 1-1 tie, the result is one of them leading 2-1 in the next game, so the current remaining amount must be (13750 + 6250)/2 = 10000.

A 1-0 scoreline becomes 2-0 or 1-1, so the current remaining amount must be (16250 + 10000)/2 = 13125.

A 0-1 scoreline becomes 0-2 or 1-1, so the current remaining amount must be (3750 + 10000)/2 = 6875.

Just make sure you get these values until one wins thrice.