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View Full Version : Breakage - what am I missing?


2low
12-07-2008, 11:56 PM
I had the winner in the final race at GG today at 3/1.

The win pool total was $93,493. The winner had 18,278 wagered on him, leaving 75,215 wagered on the other horses.

I think the odds are figured: 18,278/(18,278+75,215*.846) as there is a 15.4% takeout. This is 22.3148% or 3.48/1, shown as 3/1 on the tote board. The win paid $8.60. Why doesn't it pay 3.48 rounded down to 3.40 * 2 + 2 = $8.80?

What did I do wrong?:bang:

Imriledup
12-08-2008, 12:15 AM
I had the winner in the final race at GG today at 3/1.

The win pool total was $93,493. The winner had 18,278 wagered on him, leaving 75,215 wagered on the other horses.

I think the odds are figured: 18,278/(18,278+75,215*.846) as there is a 15.4% takeout. This is 22.3148% or 3.48/1, shown as 3/1 on the tote board. The win paid $8.60. Why doesn't it pay 3.48 rounded down to 3.40 * 2 + 2 = $8.80?

What did I do wrong?:bang:

93, 493 X 0.846 = 79,095.078 which is giveback

79, 095.078 divided by 18,278 = 4.3273377

4.3273377 times 2 = 8.6546754 (rounds down to 8.60)

You lose 5 cents breakage, or 2.5 cents per dollar.

2low
12-08-2008, 12:57 AM
93, 493 X 0.846 = 79,095.078 which is giveback

79, 095.078 divided by 18,278 = 4.3273377

4.3273377 times 2 = 8.6546754 (rounds down to 8.60)

You lose 5 cents breakage, or 2.5 cents per dollar.

Thanks much:ThmbUp: I knew I had to be doing something wrong.

rrbauer
12-08-2008, 08:37 AM
Consider breakage a form of reverse-rebate because when you win you lose.