0

BTW: I posted this on r/askmath but got no response - noting in case someone has seen this before. The content is the same.

Consider the following problem, from the Heard on the Street book (Q4.11 in the 2021 edition):

They call this the “World Series” problem in the U.S. Sports teams “A” and “B” are to play each other until one has four wins and is declared the series winner. You have \$100 to bet on Team A to win the series. You are, however, only allowed to bet on individual games, not the final outcome directly, and, you must bet a positive amount on each game. So, if Team A wins the series, you must walk away with \$200, but if Team A loses the series, you must walk away with zero, and you must do so having placed a non-zero bet on every game. Your best assessment is that Team A has a 70% chance of winning any game and Team B has a 30% chance. How do you place your bets?

I understood the main idea: that is, start backwards. Let (a, b) be a pair where a represents the number of games won by team A and b represent the number of games won by team B.

For a given node in a tree, we take the mean of the child nodes - so the value for (3, 3) is the mean of the value for (4, 3) (which is \$200) and (3,4) (which is \$0), and hence we bet \$100 for node (3,3).

What I don't quite understand is - why are we taking the mean? Because I would have thought that's kind of like finding the expectation, which by definition involves probabilities (which is useless in this case). Why can't it be, say 0.9 of the node (4,3) and 0.1 of (3,4)?

I have probably missed something crucial, and would appreciate some help. Thanks in advance.

P.S: the below image may be useful:

An image showing the replicating tree as given in the book, and some annotation of mine. I don't have the reputation to embed an image.

Leaderboard
  • 335
  • That tree appears to contemplate equal odds for each team. That is, it looks like every bet on team $a$ is double or nothing. – lulu May 12 '22 at 20:05
  • @lulu But it shouldn't matter what odds we give to each team, should it? Or am I misunderstanding something? – Leaderboard May 12 '22 at 20:12
  • Double or nothing bets mean equal probabilities for both, hence all those arrows occur with probability $.5$, so the straight mean is in fact the expected value. – lulu May 12 '22 at 20:14
  • I think lulu's comments require a clarification. If the gambler bets (say) ten dollars on a game, do they stand to win a full ten dollars; or is it less because their team has a higher than $50%$ chance of winning? – paw88789 May 12 '22 at 20:19
  • So if I understand correctly, the reason we are assuming 50% probability for each state is because in the given betting design (where supposing you bet $$k$, if you win, you get $$2k$, and if you lose, you get 0), and to ensure equality, the expected return must be the same. And indeed, that is the case, as $E(X) = 0.5 \times 2k + 0.5 \times 0 = k$. Am I right there? – Leaderboard May 12 '22 at 20:23
  • So the win probability for each team is irrelevant to the calculation at hand (although it does affect how likely you are to end up with $$200$ or with nothing). – paw88789 May 12 '22 at 20:43
  • Think of it this way: the probability is implicit in the offered odds. Working from backwards induction, you know how much you need to have at each stage which tells you how much you needed to have bet on the parent node. This point of view is purely outcome based. I don't know or care how likely various outcome are, so long as I guarantee having the right amount of wealth in each state. – lulu May 12 '22 at 21:18
  • @lulu Re my answer, while I suspect that the win factor is irrelevant, I regard this as unclear. For example, suppose that you bet $~\displaystyle \frac{87.5}{8}$ on the 1st game, win, let it ride, and lose the 2nd game. It is clear to me that your game 3 bet should not exceed the amount inherent in the analagous strategy of achieving $175$$ by the end of game 4, if games 3 and 4 win. However, this does not necessarily imply that there might not be a superior strategy that involves betting less on game 3. It is also unclear, at this point what the relevance is of the win factor. – user2661923 May 13 '22 at 04:49
  • @user2661923 It's not a question of a strategy, this is a replication. It is meant to be exact. That is, for each possible state $(a,b)$, indexed by the number of games each team has won, you must have a certain amount of stored wealth and a certain amount you must bet. Those two numbers can be computed by backwards induction, as in the link the OP provides. – lulu May 13 '22 at 10:26
  • @user2661923 The case of even money bets is misleadingly simple, which is why I called attention to it. That case (and only that case) has the property that the global bet (the bet on the series, not on a single game) has the same odds as the game bet. If, say, each game bet on $A$ pays $150$ on a $100$ bet then $A$ will be heavily favored for the series (I get an $A$ win probability of $.8267$ in this case. So, that would be the bet you'd seek to replicate through individual game bets. – lulu May 13 '22 at 10:30

1 Answers1

1

Without loss of generality, assume that you are making each bet at even odds. This means that if you wager $(x)$, the result for the game will be either $(+x)$, or $(-x).$

If, the moment that team A has 3 wins, that your betting bankroll at that point is less than $100\$$, then the stated objective can not be guaranteed. That is, if you have less than $100\$$, and team A wins the next game, then regardless of how much you bet, you failed to reach $200\$$.

Let $~\displaystyle f(n) = \frac{200}{2^{4-n}} ~: ~n \in \{0,1,2,3\}.$

When team $A$ has $n$ wins $~: ~n \in \{0,1,2,3\}$, if your betting bankroll is below $f(n)$, then you can't reach your stated objective.

However, you want to maximize your probability of achieving your objective. This means that assuming that the series goes at least $6$ games, if (for example), team A has $3$ wins, after the first $5$ games are played, then you want to have $150\$$ exactly (if feasible), at that point.

This is because, attempting to have more after game 5, increases your risk, with no gain in the probability of reaching your objective. That is, you can bet $50\$$ on game 6, and then, if necessary, bet $100\$$ on game 7.$

Suppose, however that team A has 3 wins after game 4. Then, you want to have $175\$$, so that you can (if needed), bet $25\$$, then bet $50\$$, then bet $100\$$.

This implies that if team A has 3 wins after game 3, you want to have $187.50\$$ at that point.

This is where the analysis gets tricky. You want to bet the minimum amounts in all games that precede team A having 3 wins, so as to be ready in case the first moment when team A has 3 wins is after 3 games, 4 games, or 5 games.

Note that you may therefore assume that you will have $3$ double-up opportunities, since you are assuming that team A will get to $3$ wins, and each win is an opportunity for you to double your money.

I therefore advocate the following strategy:

  • Your initial goal is to get to $87.50\$$ by the end of game 3, through $3$ double ups. Therefore, in game 1, you should bet $\displaystyle \frac{87.50\$}{8}$, and let the winnings ride, unless and until team A loses before winning $3$ straight games.

  • Assume that team A has achieved it's very first loss, somewhere in the first $3$ games. Now, consider how many games are left, before game 4 is reached, and how many wins (below $3$ wins) that team A is lacking. You then alter your strategy, by pursuing the adjusted objective of reaching $175\$$ by the end of game 4, based on the idea that team A might achieve its 3rd win in game 4.

  • Then, if team A loses its second game, before game 5, you similarly alter your strategy so as to achieve $150\$$, in case team A wins its 3rd game, in game 5.

  • Then, if team A loses its third game, before game 6, you alter your strategy so as to achieve $100\$$, in case team A wins its 3rd game, in game 6.

Now, this is where things get really complicated. How do you know whether the strategy that I have outlined above is optimal? Although my instinct suggests that it is, that is obviously inconclusive.

Further, to what extent should the optimal strategy be based on A's win factor in each game (i.e. it is given in this problem that A has a win factor in each game of $70\%$)?

Personally, I regard the entire problem as very complicated. I think that if an elegant analytical derivation of the optimal strategy is feasible, it will require someone with a far greater knowledge of Probability theory than I have.

user2661923
  • 35,619
  • 3
  • 17
  • 39