19 Replies to “Probability Goes to the Dogs”

  1. Was there an option to bet the field on this one?

    I would have thought some algorithm in the background would check the sum of the probabilities – or is that asking too much?

  2. …from an even more basic perspective: does it not immediately seem wrong that 6 of the runners have prices of $6 or better while at the same time there is no “short-priced” favourite?

    1. Well, not “wrong”. It’s a remarkably balanced field, which can happen. But the six (not 6) runners ≤ $6 is a proof that the odds are definitely against the punter. So are the eight runners being ≤ $8.

      1. OK, wrong was a poor choice of words. The book-maker can choose whatever “odds” they want.

        Whether this is in their best interest or not is another matter.

        1. If the dogs are even, they are even. The point of the example is the simplicity of the demonstration of rigged odds.

  3. As I said before, I am not an expert in betting – but I have been learning over the summer, not through experience mind you! Let me offer a few observations.

    * In this context, the reciprocal of the odds is not a probability. Indeed, it is not intended to be a probability.

    * The aim of the book-maker is to make a profit and the odds will reflect this.

    * I do not even know the meaning of a statement like “The probability that MyDog will win is 0.2.”

    * I wonder about the meaning of a “fair coin”. When defining probability, Aust Curr M v9 in its glossary writes “the probability that a fair coin toss will come up ‘heads’ is 0.5”. Is this example defining a fair coin in terms of probability (- which the example is trying to define)? Can you show me a fair coin?

    Confusing as all this has been for me, I am slowly getting to the bottom of it. And Yes, I am keeping a diary for my PD!

        1. An easy way to see it. Suppose the “odds” (i.e. payouts) on your dogs are $A and $B and $C. Now bet $BC on the A dog and $AC on the B dog and $AB on the C dog. So, the total you wager is

          W = $(BC + AC + AB).

          And, no matter which dog wins, your payout is

          P = $ABC.

          Then W/P gives you the ratio of wager to return, and

          R = W/P = $(1/A + 1/B + 1/C).

          Thus R = 1 is a fair market, and the extent to which R > 1 indicates the extent to which the odds (overall) are rigged against you.

        2. And there’s always a film clip. (They cite odds in the form e.g. 5-1, which corresponds to Australian odds of $6).

    1. Hi, Terry. A bit more detail:

      *) I think “fair coin” is an abstraction the same way a point or a number is an abstraction.

      *) I agree that the “probability” of a MyDog winning cannot mean the same as the probability of 13 coming up in roulette. In the second case second, we can abstract to a “fair table”, and give the odds as 1/37 (or 1/38, depending upon the table type). In the former, we/gamblers must do something else.

      Both scenarios are more complicated than they might seem, and in both cases mathematics can play a role. Burkard talks some about the first type of scenario in the video below. He has thoughts about doing a Mathologer on sports/animal gambling, but I have to do the drafting …

  4. Hi,
    If the sum of the reciprocals is 1 then the game is fair otherwise arbitrage may be possible if different bookmakers offer different odds
    Steve R

  5. Do the odds associated with the outcome of a future event change over time as we get closer to the event?

  6. I am referring to the odds published by an on-line betting agency. Suppose that there is an upcoming sporting event between two teams A and B, and the agency publishes odds on A winning, a draw, and B winning. Do these odds change over time?

    1. Yes, absolutely. They adjust the odds as bettors bet, and for weather and such things. And, possibly as other betting agencies adjust their odds. (I’m not sure of the mechanics of the latter.)

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