Bridge jumper bets on the Academy Awards

Quote: Wizard

Sandra Bullock is about a 2 to 1 favorite to win best actress. I didn't bet it because she is up against Meryl Streep, who I just learned was nominated more times for Oscars than any other actress. That is another one besides best picture that could go either way.

Quote: Nareed

I think [Sandra Bullock] was as near a shoo-in as you can get at the Oscars. She'd won the Golden Globe, the Critics Association and the SGA awards this year. That's a sweep.

Quote: wildqat

Don't forget, she won a Razzie this year, too (Yeah, it was for All About Steve, but still :^P ).

Quote: Wizard

This is getting off topic, but in the event a huge favorite does show, can the track take from the show pools on the other two horses to cover the 10 cents on the big favorite? For example, let's say the total show pool is $300,000, the take-out is 17%, and show bets on the three winning horses are divided as follows:

A 97%
B 2%
C 1%

How much would a $2 show ticket on each horse pay?

Quote: wizard

That is an interesting story. The terminology is a bit different in Australia. As I understand it, in Australia there are not separate bets for place and show, but just a place bet. The place bet will pay bettors on the first two dogs in races with seven or less dogs, and the first three dogs with eight or more total in the race. In the race in question there were eight dogs, two of whom where strong favorites. Following is the general way the winning odds are calculated on a three-dog place pool in Australia.

1. Take out the track cut from the total pool of place bets. For the sake of argument, let's use the usual American take-out of 17%.
2. Divide the rest into three pools.
3. Pay the winners on each dog a pro-rata basis according to the size of the pool and the amount bet on the dog. If the amount bet on the dog exceeds its share of the pool, then bettors will get a refund.

Let's look at an example. Suppose $100,000 is bet on the a pool with three dogs. Assume bets on the winning dogs totaling $5,000 on dog A, $10,000 on dog B, and $15,000 on dog C. First the 17% would be deducted, leaving $83,000. That would be divided by 3, leaving $27,667 to pay the winners of each dog. Winning bets on dog A would be paid $27,667/$5,000 = 5.53 for 1, before any rounding (I'm not sure how they round down under). Likewise, winning bets on dog B would be paid 27667/10000=2.77 for 1, and on dog C 27667/15000 = 1.84 for 1.

The bettor in this case exploited the rules by betting such huge amounts that he pretty much controlled the odds. For the sake of simplicity, let's assume he was the only bettor. The article said he bet $350,000 on the two favorites, and $5,000 on each additional dog. With six underdogs (pun intended) that resulted in a total pool of $730,000. After the takeout and split, there was 201,997 to the winners of each dog. The rule about getting at least a push resulted in bets on the two favorites being refunded. However, the share of the pool on the third dog was huge compared to wagers on it. The winning odds would have been 201,997/5000 = 40.4 to 1. So the profit on the third dog was $5,000 * 39.4 = $197,000. He actually only won $170,000, probably because of other bets on the third dog.

This technique would not work in the U.S., by the way, because here we deduct the original wagers made on each winning dog from the total show pool, and then add them back in after dividing by 3. This deduction would have caused the pools to be negative, resulting in just small winnings of the minimum $0.10 per $2 bet.