A math problem that might help you win one million dollars

Louis-Charles just won $1 million dollars…for making a few sports picks. The details are the 31-year old package handler correctly made 25 picks in a row before anyone else in ESPN’s Streak for the Cash game.

The most important part of the game was that it emphasized streaks, also known as runs or consecutive wins. Creating a streak is harder than winning a lot of games, but it is a skill worth developing… especially if ESPN repeats the contest in the future.

The thing is streaks can be strange, as today’s probability problem will illustrate.

The problem of consecutive wins

I came across an interesting problem called “successive wins” in Frederick Mosteller’s Fifty Challenging Probability Problems with Solutions.

Here is the setup:

“To encourage Elmer’s promising tennis career, his father offers him a prize if he wins (at least) two tennis sets in a row in a three-set series to be played with his father and the club champion alternately: father-champion-father or champion-father-champion, according to Elmer’s choice. The champion is a better player than Elmer’s father. Which series should Elmer choose?”

A game theory approach…

You can reach the answer even without doing any math. Just think like the problem maker and reason backwards.

Since the father is the worse player, it would make sense that Elmer play him more than the champion. The choice of father-champion-father seems obvious… perhaps it’s a little bit too obvious. A common test-taking skill is avoiding such answers, as the website SparkNotes suggests:

[A]void choosing the answer that looks “right” at first glance. On easy items, this choice may well be the correct answer. For items numbered 12 or higher, an answer that screams “Ooh! Ooh! Pick me and hurry on!” should also be handled like a live snake. SAT distractors are designed to catch students in a hurry. More often than not, an answer choice for a hard item that looks too good to be true is exactly that-too good to be true.

[source]

This puzzle appears in a book of challenging probability problems, so the obvious answer should be held with suspicion. After ruling that choice, one can conclude the correct choice must be champion-father-champion.

(I know this sounds like that Princess Bride scene, but thankfully this reasoning has an ending. And in fact, champion-father-champion is the right answer.)

While we have reasoned to the correct answer, we have not solved a more interesting question: why should playing the tougher opponent more make Elmer’s winning streak more likely?

The explanation

The reason has to do with the order of play. While champion-father-champion will have fewer expected wins, Elmer doesn’t care about wins per se. He cares about creating a streak.

In this puzzle, Elmer needs to win at least two sets in a row. How many ways is that possible? Begin by writing out the possible outcomes of the sets and identifying the win streaks. We’ll abbreviate for win (W) and loss (L):

Notice that Elmer has precisely three ways to make a streak: by winning all the sets (WWW), winning the first two (WWL), and winning the last two (LWW).

The key feature here is that Elmer needs to win the second set in order to make a streak.

As the second set is critical, Elmer would be wise to play the worse player–his father–in that slot. And that is the reason why champion-father-champion is the correct answer.

The size of the advantage

If we make a few more assumptions we can also calculate the size of the advantage.

Suppose Elmer’s odds of winning a set depend purely on whom he plays, and suppose each set can be considered an independent event. Let f and c denote the probability Elmer can beat his father and the champion in any given set, respectively.

We can then add up the probabilities of the winning events and compare their size:

Taking the example of f=0.6 and c=0.4, the chances are 38.4 percent for champion-father-champion and only 33.6 percent for father-champion-father. Choosing the right order increases the chances dramatically–nearly 5 percentage points, or more than a 14 percent increase.

It’s not how you start but how you finish…

If you need to make a streak, do not consider only expected wins. Consider the path of opponents in the middle, and if you can, schedule the tough picks at the beginning or the end.

In fact, that’s more or less what happened to the million dollar prize winner. He picked from just a handful of sports that he was most comfortable with, despite a rocky start:

Louis-Charles’ 24-pick streak began with a Tampa Bay Rays victory over the Philadelphia Phillies in Game 2 of the World Series. Since then, [his] wins have come exclusively on college football, NFL and NBA action.

Ironically, he started the Streak For The Cash game with five straight losses.

[source]

It’s strange he lost several picks at the start and went on to win it all. I’m sure there are others who had a better set of 30 games, but none had as many consecutive wins. The whole experience puts a twist on an old saying: to the streakers go the spoils of war…

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  1. 8 Responses to “A math problem that might help you win one million dollars”

  2. Now, I did the math and it works out, but it seems counterintuitive.

    In my mind I don’t see why the probability for a C-F or F-C winning streak changes depending on the outcome of a third (or first) match.

    Couldn’t we rewrite the probability table as:

    CFC:
    Champion (Win) – Father (Win) – Champion (Any)
    Champion (Any) – Father (Win) – Champion (Win)

    FCF:
    Father (Win) – Champion (Win) – Father (Any)
    Father (Any) – Champion (Win) – Father (Win)

    Which would make the probabilities:

    CF*1, 1*FC, FC*1, 1*CF (respectively), or CF for all?

    Not that I doubted the math but I decided to simulate this and the results matched up pretty well. I did a thousand trials of each and got the following results:

    FCF = 319 streaks
    CFC = 366 streaks

    For a difference of (drumoll) 14.7% (using .6 for F and .4 for C).

    But it still seems a bit counterintuitive to me.

    By Scott on Dec 9, 2008

  3. Upon rethinking I’ve figured out what I was missing: the importance of the second match.

    As you state, in order to have a streak you MUST win the second match. Thus the maximum number of streaks is limited by how many second matches you win which is greater when that second match is the father.

    By Scott on Dec 9, 2008

  4. I applaud your approach of simulating the outcomes. I think that’s the best way to see how the trends turn out…

    And you got it. It is precisely the importance of the second set that drives the counter intuitive result. This problem illustrates why we must resist casual guesses when money is on the line!

    By Presh Talwalkar on Dec 9, 2008

  5. Amazing article, Presh.
    Scott: Your analysis was superb.

    By Reddy on Dec 9, 2008

  6. That was fun. But I think the point is made easier -once you have drawn out the state space, it then becomes “obvious” that to succeed you have to win the second match.

    Who would you like to play then in the second match? The weaker opponent of course.

    By michael webster on Dec 10, 2008

  7. It’s NOT that winning the SECOND set is important, it’s that winning TWO sets in a row is important (winning the FIRST set is just as important as winning the second to make a streak). In my view, the explanation is even simpler: In an alternating series, to make two in a row, you have to win from both players once. Would you rather try twice against the better player (C) or try twice against the easier player (F)? If for abstraction sake we turn playing ‘Father’ into a walkover, it immediately shows why the CFC is the better choice.

    By Rick on Dec 12, 2008

  8. As shown the second match is necessary while neither the first or last are (you can lose either the first or last and still get a streak by winning the other one). This fact alone shows why the second match is more important. The reason why I initially found this counterintuitive had to do with thinking about it the way most people would: winning the most matches. Under this frame of mind you’d want to go against the easier player more. However, for streaks, the middle matches have higher importance so you want the easier matches focused there.

    By Scott on Dec 12, 2008

  9. Thanks for the additional explanations guys. I think it is easy to get caught up in the details of explaining it in words…which can be hard since our intuition focuses on expected wins, as Scott points out. Once the focus is about the necessary condition for a streak, namely winning the middle game, the rest follows somewhat trivially…

    By Presh Talwalkar on Dec 17, 2008

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