Compute the probability implied by a fair bet for each team

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Betting on sporting events is big business both in the US and abroad. Consider, for instance, next winter's American football tournament known as the Superbowl. Billions of dollars will be bet. The vast majority of bets are placed on individual teams, though more complex bets are available. The attached table shows the odds posted on July 15, 2017, by one of the largest on-line betting shops in Great Britain. (Online gambling is illegal in California.) Odds are offered on the 32 potential participants in the tournament. The table is interpreted as follows: the Los Angeles Chargers, for example, have odds of 40. If you bet on the Chargers, the betting shop pays you 40 times your bet and you get your bet back if the Chargers win the Superbowl, and you lose your bet if the Chargers do not win. We say the odds in favor of the LA Chargers winning are 1 to 40. The odds posted are for a team winning: you may not bet on a team losing using its posted odds (though, you could theoretically create a bet on a team losing by using bets for the other teams). There are no ties.

If a bet on the Los Angeles Chargers at these odds were a fair bet, then on average you would neither win nor lose by making this bet. Suppose that you bet £1 on the Chargers. Let p denote the probability of winning implied by a fair bet; hence, (1-p) is the probability of losing. Then, £40p is the anticipated amount to be won, £1(1-p) is the anticipated amount to be lost, and, if the bet is fair,
40p = 1(1-p).

Solving for p yields p = 1/41. More generally, a fair bet with odds of 1 to b implies that the fair bet probability p of winning is p=1/(b+1).

1. Using the odds given in the table, compute the probability implied by a fair bet for each team. Do these numbers you have computed satisfy the rules for probability?

An arbitrage opportunity exists if there is a sequence of gambles at the posted odds that never loses money and wins a positive amount with positive probability. Betting houses try to arrange the bets they accept in order to create and exploit an arbitrage opportunity.

2. Consider several schemes the betting house could use:
(a) accept exactly 32 bets, each of size £1 on each of the thirty two listed teams;
(b) accept exactly 32 bets -- £32 on New England, £31 on Green Bay, ... £1 on New York Jets.

Does either of these schemes represent an arbitrage opportunity for the betting house? If not, does an arbitrage opportunity exist, and what might one look like?

(Hint: Consider a scheme that accepts exactly 32 bets, with the amount accepted on each team proportional to the implied fair-bet probability you computed above.)

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Reference no: EM131595030

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inf1595030

9/9/2017 4:45:58 AM

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inf1595030

9/9/2017 4:35:27 AM

When you provide the solution, please include the calculation steps to show how you have reached to the final answer. I have a few below concerns: For Q1 - The table over the word document should be managed. Please provide excel with formula used to get the value. Also, question has one more part: "Do these numbers you have computed satisfy the rules for probability? " - Please include solution for this one as well. For Q2 - One part of the question ask - Is there any more arbitrage opportunity exist? Can you please put some light on this as well.

len1595030

8/9/2017 3:29:02 AM

Does either of these schemes represent an arbitrage opportunity for the betting house? If not, does an arbitrage opportunity exist, and what might one look like? (Hint: Consider a scheme that accepts exactly 32 bets, with the amount accepted on each team proportional to the implied fair-bet probability you computed above.)

len1595030

8/9/2017 3:28:33 AM

If a bet on the Los Angeles Chargers at these odds were a fair bet, then on average you would neither win nor lose by making this bet. Suppose that you bet £1 on the Chargers. Let p denote the probability of winning implied by a fair bet; hence, (1-p) is the probability of losing. Then, £40p is the anticipated amount to be won, £1(1-p) is the anticipated amount to be lost, and, if the bet is fair, 40p = 1(1-p). Solving for p yields p = 1/41. More generally, a fair bet with odds of 1 to b implies that the fair bet probability p of winning is p=1/(b+1).

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