Oldie but Goodie

[Daniel]

"On the "Let's Make a Deal" game show, Monty Hall gives a contestant his choice of three closed doors. One of the doors has a fabulous prize behind it; the other two have worthless prizes. Only Monty knows which door has the good prize.
The contestant chooses a door. Before the chosen door is opened, Monty opens one of the other doors, showing that it has a booby prize behind it. At this point he tells the contestant that he may, if he wishes, change his choice to the remaining closed door.

Is it better for the contestant to switch his choice to the remaining door or to stick with his original choice?"

Answer.

Any lessons for traders here?

 

[SOCRATES]

No, Daniel, sadly, none.
The reason for this is that this problem involves tactical decisions dependent upon conditions of uncertainty in which the outcome cannot be estimated. The choice is pure guesswork and hinges on luck.
In proper trading this kind of attitude has to be avoided because it tinges the activity with an element of gambling instead of being prepared for a forseeable outcome and being prepared to deal with that outcome. In trading there is the use of the stop loss placed in advance in the event that the outcome is not the one expected, and the opportunity to run the position if the outcome develops as expected and then the additional choice of trailing the stop to lock in profits, which is a prudent strategy to adopt. It is prudent because the trader is offered choice. In the game the contestant is not offered choice in the same way, because the outcome is already fixed and there is no redress.

 

[TheBramble]

No, Daniel, sadly, none.

So the ability to respond within a dynamic environment (the market/an instrument being actively traded) to new information and to tailor one's position is not of use to traders?

In trading there is the use of the stop loss placed in advance in the event that the outcome is not the one expected

I'm stretching, but the contestant's stoploss is set to ZERO (he/she loses nothing) at the start of the contest. They have a 1:3 chance of winning. A rather favourable R:R.

They choose.

A different (non-winning) door is excluded from the equation.

They now have a 1:2 chance of winning.

The dynamics of the situation have changed.

A successful trader will recognise that a change has taken place and will either utilise the information to amend or confirm his/her position or, if unsure, get out.

So, yes, I think there are lessons to be learnt from this example.

 

[DaveJB]

Did this at a brainstorming type session - you should always swap from your original choice... 'logic' suggests there'd be no advantage to it - however, stick with original choice means no difference to having one chance in three of being right, so you'd expect to get 1 in 3 good calls. If you 'rechoose' on the second round you have 1 in 2 chances and odds improve.
Tried it in practise and won the car 7 times out of 10 on the day <g>

Lesson for traders? Commonsense is a bad guide to anything that involves money, better to do the research/work and check what 'is obviously right'. No lesson on stock picking, however, I'd suggest.
Dave

 

[frugi]

We've kinda done Monty to death on another thread but it's always a pleasure to discuss

Socrates,

I'm afraid to say the eventual outcome does not depend on pure luck; perhaps one might call it conditional probability.

Tony,

So the ability to respond within a dynamic environment (the market/an instrument being actively traded) to new information and to tailor one's position is not of use to traders?

It surely is and it applies here too.

They choose.
A different (non-winning) door is excluded from the equation.
The dynamics of the situation have changed.

Yes.

They now have a 1:2 chance of winning.

No.

Once the "clue" door has been opened the probability of success changes. Originally 1:3 it becomes 2:3 as long as the correct choice (i.e to change original choice of door) is made by the contestant, follwing a logical deduction. If their logic is faulty and they stick with their original choice of door, the probability of success remains at 1:3. Either way, the chance is never 1:2! [Although if a contestant arrived late at the game and had not made an original choice there would of course be a simple 1:2 chance of guessing right. But this is not the case here as the contestant has previously chosen a door and then been given information by Monty.]

Random quotation from t'internet, best explanation I can find :~

"When you chose Door A, the probability that you chose the Grand Prize was 1/3 and the probability that it was behind one of the other doors was 2/3.
By showing you which of Doors B and C does not hide the Grand Prize (Door B, say), the host is giving you quite a bit of information about those two doors.
The probability is still 2/3 that one of them hides the Grand Prize, but now you know which of the two it would be: Door C. So, the probability is still only 1/3 that the Grand Prize is behind Door A, but 2/3 that it is behind Door C. "

 

[DaveJB]

Right,
got the 1:2 wrong, but 2/3 worked out in practise provided you swapped - 1:2 would work out if, having had a door open, you then flipped a coin whether to change or not.... I remembered that the 'obvious' answer (odds don't change) was incorrect, but not the mechanics of it.
Dave

 

[frugi]

Hi Dave,

If you flipped a coin it would make you change door half the time and not change the other half.
The flip would decide whether you had a 2/3 or 1/3 chance of winning, as before.
So with a coin you would have a 1/2 chance of changing, but not a 1/2 chance of winning. I think.

Now I'm confused

 

[roguetrader]

The whole logic of the equation is flawed, the probabilities existing in the first choice have no relevance now. In the first instance the contestant picks between 3 doors, no need to bore anyone going through the odds. After one door is removed the contestant is offered the ability to switch, in other words he is offered a coice of 2 doors, it is irrelevant what happened before he is now choosing between 2 doors. Plain and simple. Suppose for example he unchooses the door he first chose and then chooses it again. Were he not offered the choice his odds of having the right door would remain the same. but by offering him the choice to stick or change he is in effect entering a new game.

 

[donaldduke]

The easiest way of looking at this is to imagine a 10 doors instead of three. You chose one and
then Monty opens 8 of the 9 others, it then becomes obvious that its probably wise to switch your door for the one he didnt open.

 

[roguetrader]

The easiest way of looking at this is to imagine a 10 doors instead of three. You chose one and
then Monty opens 8 of the 9 others, it then becomes obvious that its probably wise to switch your door for the one he didnt open.

Thats a similar idea that if you toss a coin 3 times and I guess tails correctly, that for the next coin toss I should chamge my call to heads, when in fact the probability of you tossing a head or a tail is exactly the same 50/50

 

[RogerM]

Thats a similar idea that if you toss a coin 3 times and I guess tails correctly, that for the next coin toss I should chamge my call to heads, when in fact the probability of you tossing a head or a tail is exactly the same 50/50

You can play the "Monty Hall" game online here

http://math.ucsd.edu/~crypto/Monty/monty.html

In the explanation section it says that the key to the paradox is the fact that Monty knows where the prize is and so never picks that door. If he didn't know, then 1/3 of the time he'd uncover the prize and the game would be void and you'd have to play again.

Still doesn't explain why the toss of a coin example above doesn't work tho'. I think I'm off to the kitchen to play with 3 upturned cups and a sweet!

 

[TheBramble]

If you take the current situation (after MH has taken the 3rd door out of the equation) that you are faced with 2 doors - and you have a choice - the situation now is that you have a 1:2 chance.

 

[roguetrader]

If you take the current situation (after MH has taken the 3rd door out of the equation) that you are faced with 2 doors - and you have a choice - the situation now is that you have a 1:2 chance.

That's exactly my take on it Tony, just the same as if he had never done the 3 door gig. There are 2 doors and he has a choice, simple.

 

[TheBramble]

Thanks RT.

I'm only labouring this point as an aide to the originator of this thread's question 'is it useful to traders' in that YES, any change in the dynamics of any environment need to be evaluated in their own light as well as within the context of the background.

 

[frugi]

He has a 50:50 choice of doors, yes, but one of them has a greater chance of hiding the prize, regardless of whether it is chosen.

Anyway I'll leave it there - I guess we'll just have to disagree on this one

 

[roguetrader]

He has a 50:50 choice of doors, yes, but one of them has a greater chance of hiding the prize, regardless of whether it is chosen.

Anyway I'll leave it there - I guess we'll just have to disagree on this one

Back to the coin flip logic frugi, it's always 50/50 no matter what went before.

 

[jimvt]

Let's say your policy was to switch regardless.

The probability of your first choice being wrong is 2/3.
However in that instance Monty is forced to indicate the only other empty room so the probability of switching being correct in this scenario is 1.

The probability of your first choice being right is 1/3, so the probability of switching being correct is obviously 0.

So the overall probability starting from the begining is (2/3 * 1) + (1/3 * 0) = 2/3.

 

[Edster]

He has a 50:50 choice of doors, yes, but one of them has a greater chance of hiding the prize, regardless of whether it is chosen.

Anyway I'll leave it there - I guess we'll just have to disagree on this one

So expanding from your diagram, if there's 100 doors and the contestant changes his mind then he's got a 99% chance of winning.
Now where can I get on a gameshow like that?.....

 

[roguetrader]

The probability of your first choice being right is 1/3, so the probability of switching being correct is obviously 0.

So the overall probability starting from the begining is (2/3 * 1) + (1/3 * 0) = 2/3.

There in lies the problem, the first choice bears no relevance now except in the nind of the contestant.
Let me try another way, lets say everthing is as described so far, and you do all your probability calculations for contestant X.
Now let me add a twist, at the point that we get to 2 doors I bring in contestant Y. He is presented with a choice of 2 doors, pick one. Do the probability calculation for him, and then tell me that contestant Y has a worse chance of picking the right door. How do I work that out simple, most people seem to think that if contestant X changes to the other door his chance of being right is 2/3, but what if contestant Y picks that door on his arrival, they both can't have different odds in the same 2 door senario.

 

[jimvt]

RT

The key point is that your initial choice influences the behaviour of Monty.
If you're right initially, he has a choice of 2 doors to open.
If you're wrong he has only 1 door.
If you think it's 50:50 you're not considering that.
It has nothing to do wityh the mind of the contestant but is does have a lot to with the mind/choices of Monty.