odds line

[imofe]

Couldn't you set the non contenders percentage at what they win at as a group? So if your contenders win 80%, the non contender group would be assigned 20%. In the two contender race, the odds would be 40-40-20.

[Overlay]

Originally Posted by thaskalos

To assign odds of even money on them would mean that you expect this duo to prevail in this scenario 100% of the time.

That's what I thought (i.e., that a "fudge factor" was being employed), but I wanted to be sure.

Originally Posted by imofe

Couldn't you set the non contenders percentage at what they win at as a group? So if your contenders win 80%, the non contender group would be assigned 20%. In the two contender race, the odds would be 40-40-20.

That's what Dick Mitchell used to advocate -- allocating 80% of the total winning probability in any given race to contenders, and assigning the remaining 20% to the non-contenders as a group (which he called his "stuff happens" pile), without doing further individual analysis on them.

[Tom]

Check out Steve Fierro's book.

http://www.paceadvantage.com/forum/a...hp/t-1840.html

[lamboguy]

Originally Posted by Tom

Check out Steve Fierro's book.

http://www.paceadvantage.com/forum/a...hp/t-1840.html

is that right, the book is $173?

i think i am going to order it, i have been trying new things lately.

[Capper Al]

A lot of good replies. Let's do the math first. Going with the 5 horse field and narrowing it down to only 2 contenders.

The new random isn't 1/5 or 4/1 odds. It will 1/3 or 2/1 odds. The reasoning is that we have 2 contenders and we add one for that always possible unknown contender. Easy enough for the paper and pencil capper to use.

For those using software then it shouldn't matter. Run your odds line like normal and ignore the non- contenders. What typically happens is that, let's say, out of a some of 1200 the two contenders will have 800 points between them. The other three non-contenders would have 400 between them to split. The odds on the non-contenders will be much higher in proportion to the contenders. Just ignore the non-contenders as having a chance.

Now what might happen is that a 4/5 favorite might get eliminated. Here the two contenders might only have 300 points between them. The 4/5 favorite might have 300 points on it's own, and the remaing field might have 600 points. The odds for your two contenders will be higher in this case of betting against a favorite which they should be. But just ignore the non-contenders at any price which they might be receiving while going against what you think is a false favorite.

[cj]

Originally Posted by Capper Al

A lot of good replies. Let's do the math first. Going with the 5 horse field and narrowing it down to only 2 contenders.

The new random isn't 1/5 or 4/1 odds. It will 1/3 or 2/1 odds. The reasoning is that we have 2 contenders and we add one for that always possible unknown contender. Easy enough for the paper and pencil capper to use.

For those using software then it shouldn't matter. Run your odds line like normal and ignore the non- contenders. What typically happens is that, let's say, out of a some of 1200 the two contenders will have 800 points between them. The other three non-contenders would have 400 between them to split. The odds on the non-contenders will be much higher in proportion to the contenders. Just ignore the non-contenders as having a chance.

Now what might happen is that a 4/5 favorite might get eliminated. Here the two contenders might only have 300 points between them. The 4/5 favorite might have 300 points on it's own, and the remaing field might have 600 points. The odds for your two contenders will be higher in this case of betting against a favorite which they should be. But just ignore the non-contenders at any price which they might be receiving while going against what you think is a false favorite.

Here is a really simple method I've posted before that helps me a lot with regard to contenders and field size. Take the number of contenders you select, divide by field size, then average the result with the value 1. That should be the percentage of time you identify the winner in your contending group. A few examples:

3 contenders, 6 horse field. 3 / 6 = .5 + 1 = 1.5 / 2 = .75.

So, assign 75% to the contenders, 25% to the rest.

6 contenders, 10 house field. 6 / 10 = .6 + 1 = 1.6 / 2 = .8

4 contenders, 12 house field. 4 / 12 = .33 + 1 = 1.33 / 2 = .665

My experience and what I've seen from others is that we underestimate the non-contenders... this helps with that.

[davew]

Originally Posted by TJDave

Why?

What if you could eliminate all but two contenders?

so what would be your total probability of one of those 2 winning? 100%, 90%, 80%?


I can't remember who wrote it and in which book (Mitchell?), but they figured they could pick the contenders and set them at 80% combined chance of winning and the other 20% for others on most races (field sizes averaged larger when written)

[cj]

Originally Posted by davew

so what would be your total probability of one of those 2 winning? 100%, 90%, 80%?


I can't remember who wrote it and in which book (Mitchell?), but they figured they could pick the contenders and set them at 80% combined chance of winning and the other 20% for others on most races (field sizes averaged larger when written)

That was a big part of my post. That 20% won't hold up, even with the smaller field sizes of today.

[Capper Al]

Originally Posted by cj

Here is a really simple method I've posted before that helps me a lot with regard to contenders and field size. Take the number of contenders you select, divide by field size, then average the result with the value 1. That should be the percentage of time you identify the winner in your contending group. A few examples:

3 contenders, 6 horse field. 3 / 6 = .5 + 1 = 1.5 / 2 = .75.

So, assign 75% to the contenders, 25% to the rest.

6 contenders, 10 house field. 6 / 10 = .6 + 1 = 1.6 / 2 = .8

4 contenders, 12 house field. 4 / 12 = .33 + 1 = 1.33 / 2 = .665

My experience and what I've seen from others is that we underestimate the non-contenders... this helps with that.

You have formulized the paper and pencil method. Good job.

I agree with the unknown horse surprising about 20% of the time.

Thanks

[TrifectaMike]

Two contenders by field size...

5........ .72
6........ .65
7........ .60
8........ .55
9........ .51
10........ .48
11........ .45
12........ .44
13........ .41
14........ .39
15........ .37
16........ .36

Mike

[Capper Al]

Originally Posted by TrifectaMike

Two contenders by field size...

5........ .72
6........ .65
7........ .60
8........ .55
9........ .51
10........ .48
11........ .45
12........ .44
13........ .41
14........ .39
15........ .37
16........ .36

Mike

Mike, I thought I was following you until the percentage increased and then decreased. Please explain. Thanks.

[Capper Al]

Originally Posted by cj

That was a big part of my post. That 20% won't hold up, even with the smaller field sizes of today.

Does a 70 to 80% range for a hit rate with your top contenders?

[TrifectaMike]

Originally Posted by Capper Al

Mike, I thought I was following you until the percentage increased and then decreased. Please explain. Thanks.

If you have a reasonable, sensible ranking algorithm (best to worst), then

5........ .72
6........ .65
7........ .60

in a list of 5 items, you would assign a .72 percentage of whatever... money, winning probability, girlfriends, kisses, etc ( a list of 6 items - .65)

Mike

[cashmachine]

Originally Posted by TrifectaMike

If you have a reasonable, sensible ranking algorithm (best to worst), then

5........ .72
6........ .65
7........ .60

in a list of 5 items, you would assign a .72 percentage of whatever... money, winning probability, girlfriends, kisses, etc ( a list of 6 items - .65)

Mike

I got lost. It seems that you assigning 0.72 of winning probability to two top horses in 5 horses field. How did you come up with this number "0.72"? What it means? I don't understand how can you assign probability? I understand when you used your quantitative algorithm and estimated probability to be 0.72, but just assign???? How about just assigning 100%, wouldn't you sleep better?

[imofe]

Maybe it's me and I am reading this wrong. But if long term I considered 2 contenders in 5 horse fields and my 3 throwouts won 28% of the time I would not be happy.