HSH Insider: Last Show in the Bayesian PickMaster Series

[Dave Schwartz]

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https://www2.gotomeeting.com/register/938679530

[TrifectaMike]

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https://www2.gotomeeting.com/register/938679530

Dave,

Since you mentioned Bayesian, it reminds me of your Monty Hall Method.

Have you found a model (engine) for the Monty Hall Method that your are satisfied with? If you have, I believe I have a suggestion, which can drastically improve it. If you're interested, let me know.

Mike (Dr Beav)

[Dave Schwartz]

I am always interested in learning!

PLEASE!

[TrifectaMike]

The Monty Hall problem as discussed by many (if not most has a built in assumption). The assumption is that no biases exist. That is to say that the prize is placed behind one of three doors with equal probability. And Monty, himself, is not biased in his selection of a door without a prize.

Let's look at each bias independently.

Bias #1 There is a bias on where the prize is placed (This means that each door no longer has a 1/3 chance of containing the prize). As a result of this bias we end up with each door having unequal probabilities of containing the prize. ( I believe I made mention of this in one your Monty threads)

Bias #2 Monty is biased in his selection. This bias I consider to be more important as the imapct can be great. This bias directly affects the denominator in Bayes Theorem.

Mike (Dr Beav)

[Dave Schwartz]

In my approach to Monty Hall, I explained this by saying that one could imagine a probability posted above each of the three doors. In the perfect example, each door would be labeled "33%," but in our world it could be more like "10%, 40% and 50%."

Since the process is to pick and door and then switch away from THAT door to one of the others, one should always pick the lowest probability first.


Dave

PS: Please continue.

[TrifectaMike]

The Monty Hall problem as discussed by many (if not most has a built in assumption). The assumption is that no biases exist. That is to say that the prize is placed behind one of three doors with equal probability. And Monty, himself, is not biased in his selection of a door without a prize.

Let's look at each bias independently.

Bias #1 There is a bias on where the prize is placed (This means that each door no longer has a 1/3 chance of containing the prize). As a result of this bias we end up with each door having unequal probabilities of containing the prize. ( I believe I made mention of this in one your Monty threads)

Bias #2 Monty is biased in his selection. This bias I consider to be more important as the imapct can be great. This bias directly affects the denominator in Bayes Theorem.

In the Monty Hall game Monty operates under the contraint that he can not select the door which contains the prize. In his selection of which door doesn't contain the prize, he is expected to be agnostic toward each non-prize containing door. However, what if Monty is biased?

Monty doesn't like the number 1 and only selects door 1 rarely (5, 10 or 20% of the time), unless he has to.

For example he selects door #1 only 10% of the time.

The win probability of switching in the unbiased game is 2/3
The new win probability of switching in the biased game is 10/11.

Mike (Dr Beav)

[Dave Schwartz]

Absolutely. I am following you perfectly.

[TrifectaMike]

Absolutely. I am following you perfectly.

I can show you the math contrasting the biased and unbiased case. I can do it in a post or privately in a PM. It's your product, your choice.

Mike (Dr Beav)

[Handiman]

You two do know that Monty is Dead.... Right? :lol: So he is not picking any doors.


Handi:)

[Dave Schwartz]

Actually, he is alive and well.

[sjk]

I am pleased to hear Monty is well.

When is it that you have partial knowledge of the race outcome and still are able to bet on the race?

[Dave Schwartz]

When is it that you have partial knowledge of the race outcome and still are able to bet on the race?

Perhaps I do not understand your question. My simple answer would be "always."

If you are comparing to Monty Hall's game show, consider that the information Monty gives you is perfect. But what if it wasn't?

Those who understand the concept of the Monty Hall problem know that switching doors will result in a 67% hit rate. But what if Monty was sometimes wrong?

Specifically, what if Monty did not actually open the door but just told you that the door contained a goat? Further suppose that 10% of the time he was right.

In that case, instead of winning 67% of the time you would "only" win 57% of the time. Still a big gain, right?

In fact, even if Monty is wrong 32% of the time there is still a gain by switching, isn't there? The two doors combine for 67%. If Monty steers you wrong 32% of the time you still wind up with 35% wins by switching - a gain over not switching.

So, the information that Monty gives you does not have to be perfect. It just has to be "good."

[Dave Schwartz]

Ah... now I understand your question...

Nobody ever said we had "knowledge of the race outcome." We have perceived "knowledge of the race outcome."

That is what all handicapping is, isn't it?

[sjk]

Perhaps it is hubris but I don't think Monty knows any more about it than I do

[Dave Schwartz]

I can show you the math contrasting the biased and unbiased case. I can do it in a post or privately in a PM. It's your product, your choice.


I missed your question, Mike.

Please, by all means, go ahead.

[sjk]

Say you have a fortiori odds and then get a piece of information that you know to be true. Unless you have a fine understanding of how well your odds already incorporate that information you can't really act on the news.

[GameTheory]

Perhaps I do not understand your question. My simple answer would be "always."

If you are comparing to Monty Hall's game show, consider that the information Monty gives you is perfect. But what if it wasn't?

Those who understand the concept of the Monty Hall problem know that switching doors will result in a 67% hit rate. But what if Monty was sometimes wrong?

Specifically, what if Monty did not actually open the door but just told you that the door contained a goat? Further suppose that 10% of the time he was right.

In that case, instead of winning 67% of the time you would "only" win 57% of the time. Still a big gain, right?

In fact, even if Monty is wrong 32% of the time there is still a gain by switching, isn't there? The two doors combine for 67%. If Monty steers you wrong 32% of the time you still wind up with 35% wins by switching - a gain over not switching.

So, the information that Monty gives you does not have to be perfect. It just has to be "good."
My question with this stuff is why wasn't the information incorporated in the first place? (i.e. there is no "new" information before the race is run other than the tote board and I don't think your method is about that, or is it?) So what is this new information and why is it so special so as not to include it in the first place along with the dozens or hundreds of factors that give you the probabilities you have before you start with the doors and goats business?

[Dave Schwartz]

Perhaps it is hubris but I don't think Monty knows any more about it than I do

I could not agree more.

Not sure what your point is, though.

[Dave Schwartz]

So what is this new information and why is it so special so as not to include it in the first place along with the dozens or hundreds of factors that give you the probabilities you have before you start with the doors and goats business?

Do you use ALL information in your handicapping?

If not, then there is information that is new to the handicapping of this race.

[sjk]

I do use all the information in my handicapping. That is my point and sorry if I was not explaining that.

[Dave Schwartz]

I do use all the information in my handicapping. That is my point and sorry if I was not explaining that.

Well, anyone who uses everything on the face of the earth in their handicapping has nothing to add. This approach obviously does not work for you.

[GameTheory]

Do you use ALL information in your handicapping?

If not, then there is information that is new to the handicapping of this race.I use all the information about this race that is useable (i.e. that I know how to use). At least in theory.

It sounds like you are saying you:

-- Handicap the race fully, but save one bit of information. So maybe you have 12 factors, but you only use 11 of them.

-- Use that one bit of information you saved for the Monty Hall procedure to give you your final selection or adjusted probabilities.

If you are able to blend 11 factors to come up with probabilities, then you can do 12. Why the extra procedure at the end? (I could see an iterative process whereby we process each of the 12 factors one-at-a-time in a Bayesian updating fashion, but again I don't get the impression that your method does that.) So what bit of info do you save until the end, and why?

[Dave Schwartz]

GT,

Sorry, I cannot help you with this discussion. You are too advanced for me. I am trying to take lessons from Mike.






For those who might need a dictionary to follow along:

Definition of A FORTIORI : with greater reason or more convincing force —used in drawing a conclusion that is inferred to be even more certain than another <the man of prejudice is, a fortiori, a man of limited mental vision>

[TrifectaMike]

My question with this stuff is why wasn't the information incorporated in the first place? (i.e. there is no "new" information before the race is run other than the tote board and I don't think your method is about that, or is it?) So what is this new information and why is it so special so as not to include it in the first place along with the dozens or hundreds of factors that give you the probabilities you have before you start with the doors and goats business?

Isn't "new" information a relative term?

Mike (Dr Beav)

[GameTheory]

GT,

Sorry, I cannot help you with this discussion. You are too advanced for me.Ok, although I don't think so. I've just never been able to figure out what this Monty Hall stuff of yours is about. Just trying to determine where we are in the handicapping process when we apply it and what the inputs are. It has always sounded like you apply it after you have your contenders. But in my process (and most, I would think), if you've got your contenders, then except for any adjustments that are going to be influenced by the toteboard, then your handicapping is done. You've exhausted all the information you are capable of using -- or else you'd still be handicapping and your contenders not set. Therefore, you've got no information to work with for the Monty Hall procedure because again, you've used it all up for this race. It doesn't make any sense to me that you've finished handicapping and still have "new" information to use. So, I remain confused.

[TrifectaMike]

I use all the information about this race that is useable (i.e. that I know how to use). At least in theory.

It sounds like you are saying you:

-- Handicap the race fully, but save one bit of information. So maybe you have 12 factors, but you only use 11 of them.

-- Use that one bit of information you saved for the Monty Hall procedure to give you your final selection or adjusted probabilities.

If you are able to blend 11 factors to come up with probabilities, then you can do 12. Why the extra procedure at the end? (I could see an iterative process whereby we process each of the 12 factors one-at-a-time in a Bayesian updating fashion, but again I don't get the impression that your method does that.) So what bit of info do you save until the end, and why?

I don't think that is what Dave is doing. He isn't adding a new factor. He is handicapping the race again with a model different than the first.

You do agree that two models can differ and not necessarily be incorporated to work as one?

Mike (Dr Beav)

[GameTheory]

Isn't "new" information a relative term?Meaning?

I'm saying "new" in the Bayesian sense, i.e. "not yet incorporated into our probabilities". As soon as we have accounted for it, it is "old". Therefore "new" simply means we haven't accounted for it yet.

[TrifectaMike]

Meaning?

I'm saying "new" in the Bayesian sense, i.e. "not yet incorporated into our probabilities". As soon as we have accounted for it, it is "old". Therefore "new" simply means we haven't accounted for it yet.

Let me ask you? Are tote odds "new" information or prior probabilities?

Mike (Dr Beav)

[GameTheory]

Let me ask you? Are tote odds "new" information or prior probabilities?

Well, if you did all your handicapping and then added them last, they'd be "new". If you started with them and added your handicapping, they'd be priors.

[TrifectaMike]

Well, if you did all your handicapping and then added them last, they'd be "new". If you started with them and added your handicapping, they'd be priors.

"New" is relative, isn't it?

Mike (Dr Beav)

[GameTheory]

"New" is relative, isn't it?Ok. So what? (Please no more quizzes -- if you have something to say, just say it.)

[Handiman]

Here's a possibility. The horses are near the gate and the announcer comes across the speaker at Golden Gate saying that Jockey Billy Bob will be replaced by Jockey Russel Baze and it's a 5 horse race. I suspect that NEW info would change things a bit and become relevant. I don't know where that fits into the 'Bay...' thing but it sure would be NEW info.

Just a thought,

Handi:)

[TrifectaMike]

So, if Dave has two disjointed, possibly two contradictory methods given rise tto two sets of subjective probabilities, can not one set or the other be considered "new" information relative to the subjective probabilies he decides on using as his prior?

Mike (Dr Beav)

[GameTheory]

So, if Dave has two disjointed, possibly two contradictory methods given rise tto two sets of subjective probabilities, can not one set or the other be considered "new" information relative to the subjective probabilies he decides on using as his prior?
I don't know if he does though. The sales pitch goes something like "take your contenders and be able to pick the one that actually wins more often than you usually do". The implication is that *whatever* the method you use already (and whatever information went into it), this Monty Hall procedure is going to improve it.

[Handiman]

Ok I will concede Monty is still alive at 90. But I bet just barely. Do hope he is doing well as I always liked him. :)

[TrifectaMike]

I missed your question, Mike.

Please, by all means, go ahead.

Three door Monty Hall problem... Door A, Door B, and Door C.

There is a prize behind a door. You select a door. The host, Monty, opens a door that has no prize and gives you the opportunity to exchange your door for the remaining door.

Here an intuitive way of understanding the problem:

1. Assume you select Door A. If the prize is behind Door C, the Host opens Door B, so if you switch you win.....1 way to win

2. If the the prize is behind Door B, the host opens Door C, so if you switch you win....1 way to win

3. If the prize is behind Door A, the Host opens Door B or Door C, so if you switch you lose....1 way to lose

We win twice as often if we switch than not.

The above is the unbiased version. When Monty opens a door not containing a prize he does so with equal probability (when he has a choice between opening a door from two doors).

However, let's assume or you have inferred that Monty is biased. He doesn't like to select Door C, unless it is absolutely necessary (it is only absolutely necessary when his other choice contains the prize).

He selects Door C only 20% of the time (when he has a choice).

The question is what effect does this bias have on the odds?

Let's describe the probability of Monty selecting Door C,

P(Door C) = 1/3*1 + 1/3*0 + 1/3*1/5 = 6/15

The first term (1/3*1) is if Monty has no choice, but to select Door C
The second term (1/3*0) is if Monty can't (prize is behind Door C)
The third term is if Monty has a choice, but will only select Door C one in five times.

So, Monty selects Door C

The probability that the prize is behind Door A = (1/5*1/3)/(6/15) = .1667
The probability that the prize is behind Door C = (0*1/3)/(6/15) = 0
The probability that the prize is behind Door B = (1*1/3)/(6/15) = .8333


Mike (Dr Beav)