Nonlinear odds-to-probs conversion

#1
  • The New England Patriots are playing the Buffalo Bills.
  • The fractional odds are Patriots 1/2 and the Bills 2/1, where a bet on the Patriots risks $2 to win $1, and a bet on the Bills risks $1 to win $2.
  • A rational bettor has $1 to bet.
  • A $1 wager on the favorite Patriots yields a $0.50 profit, if they win.
  • Or, a $1 wager on the underdog Bills yields a $2.00 profit, if they win.
  • Therefore, the relative payout on the Bills is 4X that of the Patriots (2.00/0.50; this simple ratio mathematically ties the teams’ odds together, making the odds-to-probability relationship nonlinear).
  • As odds-and-probs have an inverse relationship, the probability that the Bills win is 1/4th that of the Patriots.
  • Therefore, the Bills have a 20% chance of winning, which is 1/4th that of the Patriots 80% chance of winning.
Can you show that this is NOT the case?

The conventional conversion of odds-to-probability is inverse-linear (see link below), where the probabilities are derived independently for each outcome from its odds. This overstates the probability of the underdog, while understating the probability of the favorite ... hence, the well-known 'longshot bias' (see link below).

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Linear odds-to-probs conversion: https://www.google.com/search?q=odd...9j0l2j69i59.4738j0j4&sourceid=chrome&ie=UTF-8

Longshot bias: https://www.google.com/search?ei=Pb...-wiz.......0i71j0i67j0i131j0j0i10.lPjHNNGo270
 
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#2
This chart shows the odds-to-probs 'misunderstanding' -- assuming the above is correct -- of the implied probability from the underdog's odds (the difference between linear-inverse and nonlinear-inverse conversion).

As noted earlier, this could account for 'longshot bias' in betting.



Note: The absolute misunderstanding -- the difference between linear and nonlinear conversions -- appears to be shrinking as the underdog's odds increase, but the relative error 'misunderstanding' goes thru the roof.
 
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#3
Assuming that 'odds' is the fractional odds for an outcome (or the payout on a $1 bet), the odds and its implied probability have a direct inverse-proportional relationship: Prob.x = f(1/Odds.x) and Odds.x = f(1/Prob.x).

Therefore, with any zero-sum event with n competitive outcomes:

Odds.1 x Prob.1 = Odds.2 x Prob.2 = ... = Odds.n x Prob.n

In the example of the hypothetical football game:

Patriots: 1/2 x 0.8 = 0.4
Bills: 2/1 x 0.2 = 0.4

Using the conventional inverse-linear odds-to-probs conversion -- where 1) Prob.x = f[1/(Odds.x+1)], and then 2) normalizing the overround -- distorts this natural relationship between odds and probability.

As a test of functionality of odds-to-probs conversion, take listed odds in a zero-sum event and convert them to implied probabilities in each of the outcomes. Then, take those calculated probabilities and convert them back into their implied odds (no house take). How do those implied odds compare to the listed odds?
 
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#4
On the conversion of probabilities into their implied odds, the following relationship exists for a game between Team 1 and Team 2:

2-outcome.generic,9-11.jpg

Therefore, using the above NFL example:

2-outcome.Pats-Bills,9-11.jpg

QED ... patents pending, for two-or-more event outcomes.
 
#6
There was no response to my odds-to-probs-to-odds challenge (above, 3rd post)...

"As a test of functionality of odds-to-probs conversion, take listed odds in a zero-sum event and convert them to implied probabilities in each of the outcomes. Then, take those calculated probabilities and convert them back into their implied odds (no house take). How do those implied odds compare to the listed odds?"

...so I took the liberty to answer my own question, using the earlier NFL example.

That's wtf ... and thanks for asking, I wouldn't want to be unclear.
 
#7
For an event with two outcomes (fractional odds):

Prob-to-odds,9-11.jpg
…you may ask yourself, where did this nifty equation come from? * [see below]

DERIVATION

For a two-outcome event (outcomes #1 or #2) with no house take, the fractional odds of an outcome is the square-root of the relative probability of the opposite outcome … for instance, a 50/50 chance, such as a coin flip, would have a relative probability of 1, and therefore odds = 1, for both heads and tails. The derivation of this relationship is provided below.

In a settled wager, the amount risked, or bet, on an outcome is the same as the profit from the bet on the opposite outcome. The bettors just compensate each other after the event, based on whose choice won. In the earlier NFL example, a $2 bet on the Patriots results in a $1 profit if the Patriots win, while a $1 bet on the Bills results in a $2 profit if the Bills win. There’s $3 in the pot with combining both bets ($2 + $1) … who wins the pot depends on the game result.

The inherent bet symmetry can be summarized for a two-outcome event, with no house take:

Bet.1 = Profit.2

Bet.2 = Profit.1

In addition, the definition of fractional odds on the outcomes is:

Odds.1 = Profit.1/Bet.1

Odds.2 = Profit.2/Bet.2

So, substituting for Profit.x in each fractional-odds equations…

Odds.1 = Bet.2/Bet.1

Odds.2 = Bet.1/Bet.2 = 1/Odds.1

And we already know (see post #3)…

Odds.1 x Prob.1 = Odds.2 x Prob.2

Therefore,

Odds.1 x Prob.1 = (1/Odds.1) x Prob.2

So, solving for Odds.1…

1573506792649.png

…am I right? … am I wrong? *

*
 
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