Rags To Riches - What will her odds be?

[SentToStud]

Originally Posted by Dunbar

The answer obviously depends on how you see Curlin and Hard Spun's chances relative to the rest of the field. I think even money for your bet is about right, so you got the worst of it IMO.

To get a feel for the answer, I made some simplifying assumptions.

1. Curlin and Hard Spun together have a 62% chance to win the race. (that's what I come up with in my line)
2. They are interchangeable (I know that sounds like a horrible assumption, but as long as you don't think Hard Spun is WAY out of it, I don't think it prevents us from getting a ballpark-type answer.)
3. If Curlin doesn't win, then the chance that he comes in 2nd is the chance that he would have won a race without the horse that came in first. (Harville formula). Same holds for Hard Spun.

If Curlin or Hard Spun finishing in the top 3 is "Y", and finishing off the board is "N", then your friend can win in these 3 ways.

YYN
YNY
NYY

Chance of (YYN) = 62% * 43% = 27% (where "43%" is the chance that Curlin would win a race without HardSpun, and vice versa)

Chance of (YNY) = 62% * (1-43%) * 46% = 16% (where "46%" is the chance that Curlin would win a race without HardSpun and without another horse--I chose a 20-1 longshot for the "other horse", because that gives a more conservative fig.

Chance of (NYY) = (1-62%) * 66% * 46% = 11% (where "66%" is the chance that either Curlin or HardSpun would win a race without the 20-1 shot in there, and the "46%" is same as for YNY.)

Adding those up, I get 54% as a rough estimate of the chance of both Hard Spun and Curlin being on the board. This is very dependent on my first assumption, that the 2 horses have a combined 62% chance to win the race.

If you think there's a huge diff between Curlin's and Hard Spun's chances to win, AND you don't think one or the other even deserve to be favored over the rest of the field, then my 2nd assumption above is a poor one.

My 3rd assumption would be a poor assumption if you think one or both of these horses will either win the race or fall apart completely. I happen to think these 2 will stay on even if they don't win.

--Dunbar

I follow and thanks. My methodology was far less academic. I figured Curlin 80% likely to hit the board and Hard Spun 50%. .8 x .5 = .4 = 3/2.

We started bidding down from 3/1. My last bid was to take it at 3/2. He went 7/5.

[Dunbar]

Originally Posted by SentToStud

I follow and thanks. My methodology was far less academic. I figured Curlin 80% likely to hit the board and Hard Spun 50%. .8 x .5 = .4 = 3/2.

We started bidding down from 3/1. My last bid was to take it at 3/2. He went 7/5.

I like the "auction" approach, STS!

Here's an example that shows you can't do the calc like you did it above. Say we have 3 horses, except I'm going to identify them by the cards J, Q, and K. After shuffling, you will pick them 1 at a time for win and place.

What is the chance a J will “finish” either 1st or 2nd? Common sense says that the J has 2/3 chance of finishing 1st or 2nd in a 3-card "race". Same for Q or K.

What’s the chance that J and Q BOTH finish in the top 2? By your methodology, it would be 2/3 * 2/3 = 44.4%. But it’s easy to show that 44.4% is wrong. There are only 3 possible pairs that can hold the top 2 places: JQ, JK, and QK. Each of those 3 pairs must have the same chance, namely 1/3 chance. So the chance of J AND Q finishing in the top 2 spots is 33.3%, not 44.4%. Similarly, your ".8 x .5 = .4" is an over-estimate of the combined chance of both horses finishing in the top 3. So, given your 80%/50% assumptions for Curlin and Hard Spun, the bet is even better for you than you thought.

--Dunbar

[SentToStud]

Originally Posted by Dunbar

I like the "auction" approach, STS!

Here's an example that shows you can't do the calc like you did it above. Say we have 3 horses, except I'm going to identify them by the cards J, Q, and K. After shuffling, you will pick them 1 at a time for win and place.

What is the chance a J will “finish” either 1st or 2nd? Common sense says that the J has 2/3 chance of finishing 1st or 2nd in a 3-card "race". Same for Q or K.

What’s the chance that J and Q BOTH finish in the top 2? By your methodology, it would be 2/3 * 2/3 = 44.4%. But it’s easy to show that 44.4% is wrong. There are only 3 possible pairs that can hold the top 2 places: JQ, JK, and QK. Each of those 3 pairs must have the same chance, namely 1/3 chance. So the chance of J AND Q finishing in the top 2 spots is 33.3%, not 44.4%. Similarly, your ".8 x .5 = .4" is an over-estimate of the combined chance of both horses finishing in the top 3. So, given your 80%/50% assumptions for Curlin and Hard Spun, the bet is even better for you than you thought.

--Dunbar

I follow. What if I said I thought it was 80% and 50% after accounting for the fact that they might both run in the $? I think that's the difference. In any event, I enjoyed and appreciate your responses.

There's a few of us from high school who have been doing this reverse auction stuff for years, mostly conducted during Thanksgiving/Christmas and usually related to people surviving through the following year. I have a pretty big exposure on Ted Kennedy this year. One of the other guys has him checking out at 14-1.

[alysheba4]

pletcher gets his first win..........i hope