This week the #puzzle is: Will the Odds Be Ever in Your Favor? #odds

With the regular season over, there are two clear favorites for baseball’s American League Most Valuable Player (MVP) award according to ESPN:

– Aaron Judge of the New York Yankees, whose odds are -150. – Cal Raleigh of the Seattle Mariners, whose odds are +110.

The “-150” odds mean that for every $150 you bet on Judge to win MVP, you’ll earn $100 if he actually wins. The “+110” odds mean that for every $100 you bet on Raleigh to win MVP, you’ll win $110 if he actually wins. Note the differing results (winning $100 vs. betting $100) that comes with the odds being negative vs. positive.

While these betting lines may be informed by an assessment of Judge’s and Raleigh’s real chances, they may also be informed by how much money people are betting on each player.

Suppose all bettors have wagered on either Judge or Raleigh with the odds above. Some fraction f of dollars wagered have been in favor of Judge, while 1−f has been wagered on Raleigh. For what fraction f will the oddsmaker earn the same amount of money, regardless of which player earns the MVP award?

And for extra credit:

Suppose there are two leading candidates, A and B, for MVP in the Fiddler Baseball League. There are two parts to this Extra Credit, so please read carefully!

Part 1:

The odds for A winning the award have been set to +100x, where x > 1.

Let f represent the fraction of dollars wagered in favor of A. For many values of f, the oddsmaker can set the odds for B so that they’ll make the same amount of money regardless of whether A or B wins the award. However, below a certain value of f, it’s impossible for the oddsmaker to do this.

What is this critical value of f? (Your answer should be in terms of x.)

Part 2:

Now, the odds for A winning the award have been set to -100y, where y >1. Again, for many values of f, the oddsmaker can set the odds for B so they’ll make the same amount whether A or B wins the award.

What is the critical value of f below which this isn’t possible? (Your answer should be in terms of y.)

Will the Odds Be Ever in Your Favor?

Intermission

So, I didn’t get the extra credit point last week. I am still not quite sure what I should have done. (Apart from knowing more about geometry and trigonometry.) I think my methods must have contained some kernel of correctness, but I can’t figure out how to reach the actual correct value. Sigh. Any help appreciated.

Highlight to reveal (possibly incorrect) solution:

Spreadsheet

Calculations:

• If AJ wins, the oddsmaker will, for every $1 spent on AJ, pay out $(100+150)/150 or $5/3. They will lose $2/3.
• If CR wins, the oddsmaker will, for every $1 spent on CR, pay out $(110+100)/100 or $2.1. They will lose $1.1.
• A fraction f was bet on AJ, the rest, 1-f, on CR.
• If AJ wins, the oddsmaker will, in all, keep, for every $1 spent, f * (-$2/3) + 1-f.
• If CR wins, the oddsmaker will, in all, keep, for every $1 spent, f + (1-f)*(-$1.1).
• The winnings will be the same for them when:
• f * (-2/3) + 1-f = f + (1-f) * (-1.1)
• -2/3 * f + 1 – f = f – 1.1 + 1.1 * f
• -5/3 * f + 1 = 2.1 * f – 1.1
• 2.1 = (2.1 + 5/3) * f
• 2.1 = 113/30 * f
• 21/10 * 30/113 = f
• 63/113 = f, appr. 0.55752

And for extra credit:

Desmos

Calculations:

• If A wins, the oddsmaker will, for every $1 spent on A, pay out $(1+x). They will lose $x. (Note that I don’t require x > 1.)
• If B wins, the oddsmaker will, for every $1 spent on B, pay out $(1+z). They will lose $z.
• A fraction f was bet on A, the rest, 1-f, on B.
• If A wins, the oddsmaker will, in all, keep, for every $1 spent, f * $(-x) + 1-f.
• If B wins, the oddsmaker will, in all, keep, for every $1 spent, f + (1-f) * $(-z).
• The winnings will be the same for them when:
• f * (-x) + 1-f = f + (1-f) * (-z)
• f * (-x) + 1-f = f + 1-f + (1-f) * (-z-1)
• f * (-x) – f + 1-f = 1-f + (1-f) * (-z-1)
• f * (-x-1) + 1-f = 1-f + (1-f) * (-z-1)
• f * (-x-1) = (1-f) * (-z-1)
• f / (1-f) * (-x-1) = -z-1
• f / (1-f) * (x+1) = z+1

Test cases:

• We already know the relationship between 2/3, 1.1 and f = 63/113. 1 – f = 50/113
• x = 2/3, f = 63/113, z+1 = 63/113 / 50/113 * (1+2/3) = 63/113 * 113/50 * 5/3 = 63/50 * 5/3 = 63/3 * 5/50 = 21 * 1/10 = 2.1 <=> z = 1.1. Correct.
• x = 1.1, f = 50/113, z+1 = 50/113 / 63/113 * (1+1.1) = 50/113 * 113/63 * 2.1 = 50/63 * 21/10 = 50/10 * 21/63 = 5 * 1/3 = 5/3 <=> z = 2/3. Correct.

x must be greater than 0. (Why bet on something, if you can’t win more than you bet?) Likewise z must be greater than 0.

Case 1: x > 1.

• f / (1-f) * (x+1) = z+1 & z > 0
• f / (1-f) * (x+1) > 1
• f * (x+1) > 1-f
• fx + f > 1-f
• fx + 2f > 1
• f * (x + 2) > 1
• f > 1 / (x + 2)

Case 2: The odds are -100y, y > 1. This can be restated as odds +100x, 0 < x < 1.

• Bet 100y, potentially the oddsmaker will lose 100/y
• Bet 100, potentially the oddsmaker will lose 100x
• Bet 100yx, potentially the oddsmaker will lose 100x = 100x * x * y
• 100x = 100xxy
• 100 = 100xy
• 100 / 100y = x
• 1/y = x

Putting these 2 results together:

• f > 1 / (1/y + 2)
• f > y / (1 + 2y)

Phew! This was hard. I keep getting confused by the odds.