For argument purposes, I have an archive of selections and race results amounting to thousands of trials, and I have pretty good estimates of the Probability of certain selections winning.
Door One: the Favorite (usually wins roughly 1/3 of the time)
Door Two: My Top Non-favorite Computerized Selection (wins roughly 1/3 time)
Door Three: The rest of the field (wins roughly 1/3 time)
Ten minutes before the Race I play a Pick 6, and I decide to "single" my Computerized selection in the first Leg, I do not think there is much of a chance of the longer shots winning this race. So, I select only that "single" horse on my ticket, rather than "spreading" for the rest of the field.
Five minutes before the Race the owner of the Favorite tells me they are prepping for a stakes race and only out for the exercise today.
He never lies to me, so I can pretty much count his horse out.
If I apply Monte Hall Game reasoning, I should cancel my single in the Pick 6 because the probability of one of the horses which I left off the ticket winning is now higher than I thought it was originally: if I switch my ticket I should be correct about 50% of the time.
Will switching help me win more pick 6 wagers like this?
On the one hand, eliminating the favorite ought to mean that my top selection is even a Higher Probability to win.
On the other hand, eliminating the favorite ought to mean that everybody's chance of winning goes higher.
These important assumptions doesn't apply to your horse race, do they ?
I'm not an expert in horse racing, but if a horse does not run, bets on him are usually void and are simply returned (like a push). If a favourite does not run the odds of all other horses are too high (making the book deficit). In this scenario there is usually a rule of "non-runners", where the odds of the other horses are reduced afterwards.
Quote: montehall
Five minutes before the Race the owner of the Favorite tells me they are prepping for a stakes race and only out for the exercise today.
He never lies to me, so I can pretty much count his horse out.
This is the real value here. Being able to eliminate the favorite from contention gives you a big advantage over the other bettors in the parimutual pool
The rest of your post with the monte hall nonsense has nothing to do with it and it basically gibberish.
If you have a pick six ticket with a horse you later eliminate from contention, cancel the first ticket and bet the same picks without the eliminated horse. You've thus reduced your outlay and increased your payout odds if you win.
Quote: montehallCan the Monte Hall Game be applied to Horse Race Selections?
For argument purposes, I have an archive of selections and race results amounting to thousands of trials, and I have pretty good estimates of the Probability of certain selections winning.
Door One: the Favorite (usually wins roughly 1/3 of the time)
Door Two: My Top Non-favorite Computerized Selection (wins roughly 1/3 time)
Door Three: The rest of the field (wins roughly 1/3 time)
Ten minutes before the Race I play a Pick 6, and I decide to "single" my Computerized selection in the first Leg, I do not think there is much of a chance of the longer shots winning this race. So, I select only that "single" horse on my ticket, rather than "spreading" for the rest of the field.
Five minutes before the Race the owner of the Favorite tells me they are prepping for a stakes race and only out for the exercise today.
He never lies to me, so I can pretty much count his horse out.
If I apply Monte Hall Game reasoning, I should cancel my single in the Pick 6 because the probability of one of the horses which I left off the ticket winning is now higher than I thought it was originally: if I switch my ticket I should be correct about 50% of the time.
Will switching help me win more pick 6 wagers like this?
On the one hand, eliminating the favorite ought to mean that my top selection is even a Higher Probability to win.
On the other hand, eliminating the favorite ought to mean that everybody's chance of winning goes higher.
Wait, what?
Quote: MangoJIn Monty hall the host shows one of the remaining doors *based on your selection* (and with his knowledge of the door) . Thus, the host gives you additional information about your bet (which you can use to improve your strategy).
These important assumptions doesn't apply to your horse race, do they ?
I'm not an expert in horse racing, but if a horse does not run, bets on him are usually void and are simply returned (like a push). If a favourite does not run the odds of all other horses are too high (making the book deficit). In this scenario there is usually a rule of "non-runners", where the odds of the other horses are reduced afterwards.
Nobody seems willing to explain *why* the Monte Hall problem could not be applied.
Initially, the Doors all appear to have equal probability of hiding the Prize or the Winner - at the time that Choice of Doors (horses) is made.
I did not say that the favorite did not run in the race. The connections were entering him for prep work - conditioning - and aiming to win the *next* race he would be entered in. He was thus a non-contender, but this was not known at the time the Selections were made.
The point is that knowing that one of the horse selections (the Favorite) is not going to win is analogous to Monte Hall showing the contestant that the Prize is *not* behind Door One.
Thus, the Prize (the winner of the race) must be behind either Door Two or Door Three.
Following the analogy, anybody who picked the selection behind Door Two is better off switching to the selections behind Door Three.
Historically, the selection behind Door Two wins at about one third of the time.
So, over the long run, it should be better (+ EV) to switch one's selection from Door Two to Door Three...which ought to be the correct selection about 2/3 of the time...if the analogy can be sustained.
Quote: montehallNobody seems willing to explain *why* the Monte Hall problem could not be applied.
Simply put:
Monty *knows* where the car is hidden. When he reveals information to you, he *never* will reveal hte car. So you take opportunity of this bias.
Viewed differently: his decision depends on yours. In the horse race, the Favorite not running s an event independent from your bettings. You can use it as "inside info", but not as in the Monty Hall situation.
33% favorite
33% computer
33% longshot
If favorite is eliminated, wouldn't the 33% be split evenly among computer / longshot?
(This is assuming your percentages are correct, which is highly unlikely)
In both games, there are 3 doors named A, B, C. One door contains the prize, and the other two contains rubish.
Game 1:
Monty askes you which door you choose. You tell Monty your choice. Monty as usual reveals a door which contains rubish. You can make your choice again.
Game 2:
Python askes you which door you choose. Before you can tell Python your choice, a technical failure occures and one door opens - it contains rubish. In your head you make your choice again, and tell Python your new choice.
Game 1 is Monty's paradox, Game 2 isn't. Why ? In Game 1, Monty never show you the prize, and will never show you your original selection. Monty's new information is dependent from your first selection. In Game 2, the technical failure doesn't care if the door contains rubish or the prize. It just happened to be rubish. The technical failure is *not* dependent on your first selection.
Now to your little horse race.
There are three possible bet selections A, B, C. You bet your selection. Now the horse owner will tell you that selection A will not win. Which analog game is it, is it game 1 (Monty) or game 2 (Python) ? Should you cancel and change your bet ?
The thing is, your horse race is Python's game, not Monty's. The horse owner would have told you regardless if you had bet on horse A or not.
BTW, the correct strategy should be: Cancel all bets on A. Bet equal amounts on B *and* C with as much as you carry with you.