horse odds analogy to roulette
August 14th, 2010 at 5:17:30 PM
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HELP I SAW THIE IN RACING ARTICLE TALKING ABOUT MARKET PERCENTAGE
can anyone work out the maths how did he work the roulette odds given the horse odds IN the second part??
Imagine a four horse race where each horse has the same chances of winning. That means each horse has exactly a 25% chance of winning. You simply divide 100% by four.
To obtain the true dollar odds for each horse you divide 100 by each horses percentage chance of winning. So 100 divided by 25 equals $4.00. This market percentage is also exactly 100% as 4 x 25 = 100.
The higher the total market percentage is above 100%, the worse it is for the punter.
For example, if the odds of each of the horses was $3.00, instead of $4.00, then by dividing 100 by three, each horse is given a 33.33% chance. As there are four horses each given a 33.33% chance you get a total market percentage of 4 x 33.33% = 133.32%. Now that is a bad market for punters.
To obtain the starting price market percentage, as in the example above, you simply divide 100 by each horse's starting price dollar odds and add up the totals.
Now how often will you find a big race with a field of say 17 runners where the starting price market percentage is a very tight 137% or even higher?
Quite often.
What is even more amazing is in such races you will most likely be told by some in the racing media that you can get plenty of value on anything you like as it is $4.80 the field!
Here's the reality:
That starting price market of 137% is equivalent to a game of roulette, but with an important change:
Instead of the 37 roulette numbers being 18 even, 18 odd and one zero for the house, there are now 12 even, 12 odd and 13 zeros for the house!
Alternatively, keep the 18 even numbers, 18 odd numbers and one zero for the house. But all the even money payouts for odds, evens, black and red are instead reduced from $2.00 to $1.50!
Now that certainly puts tight betting markets into a very easy to understand perspective.
can anyone work out the maths how did he work the roulette odds given the horse odds IN the second part??
Imagine a four horse race where each horse has the same chances of winning. That means each horse has exactly a 25% chance of winning. You simply divide 100% by four.
To obtain the true dollar odds for each horse you divide 100 by each horses percentage chance of winning. So 100 divided by 25 equals $4.00. This market percentage is also exactly 100% as 4 x 25 = 100.
The higher the total market percentage is above 100%, the worse it is for the punter.
For example, if the odds of each of the horses was $3.00, instead of $4.00, then by dividing 100 by three, each horse is given a 33.33% chance. As there are four horses each given a 33.33% chance you get a total market percentage of 4 x 33.33% = 133.32%. Now that is a bad market for punters.
To obtain the starting price market percentage, as in the example above, you simply divide 100 by each horse's starting price dollar odds and add up the totals.
Now how often will you find a big race with a field of say 17 runners where the starting price market percentage is a very tight 137% or even higher?
Quite often.
What is even more amazing is in such races you will most likely be told by some in the racing media that you can get plenty of value on anything you like as it is $4.80 the field!
Here's the reality:
That starting price market of 137% is equivalent to a game of roulette, but with an important change:
Instead of the 37 roulette numbers being 18 even, 18 odd and one zero for the house, there are now 12 even, 12 odd and 13 zeros for the house!
Alternatively, keep the 18 even numbers, 18 odd numbers and one zero for the house. But all the even money payouts for odds, evens, black and red are instead reduced from $2.00 to $1.50!
Now that certainly puts tight betting markets into a very easy to understand perspective.
