Re: Monty Hall problem

José Carlos Santos a écrit :
On 13-04-2008 11:34, Denis Feldmann wrote:

It is a nasty riddle because you are messing things up. Consider the
possibility of you being killed by a tiger within the next hour. You
don't whether that will happen or not right?! Do you deduce from that
that the odds are 1/2 that you will be eaten and 1/2 that you won't? I
guess not. But you made a similar mistake, when you assumed that for the
second contestant the odds are 1/2 for each of the remaining doors. (Of
course, he thinks that, but that is another matter.)
I don't think you understand, the second contestant is shown 2 doors and
there is a car behind one of them. I don't assume the odds are 1/2 for
him, I calculate it.
Would you please display your calculations here?
How does this relate to the tiger scenario, who is not constrained to
be behind any doors?
I used that example to make cleat the logical fallacy in your argument.
I thought that you were assuming that the chances were 1/2 for each
outcome. But it seems that I was wrong. It seems that you computed those
values. Well, like I said, please display your calculations here.
Sigh...it's the probability of a coin toss...
Reread the post...remember the second contestant knows nothing of what
has gone before, he is simply presented with 2 doors, one with a car
behind it. What is the probability he will select the door with the
car? What do you think it is?
I see no calculations in your reply.
The second contestant will choose the door with the car 1/2 of the times
because he assumes _wrongly_ that the odds are 1/2 for each door. But
they aren't.
Wow...how do I begin to describe to you how messed up you have things?
At least you now agree the probability for the second contestant is
50%
Not at all. Read again all my previous posts. I wrote *twice* that the
second contestant _believes_ that the odds are 1/2 for each door... but
he is wrong about it.

Michael has just answered the same for me, odds are not something that
exist independently, they only exist for people for their situation.
The 2nd contestant believes the odds are 1/2 _for him_ and he is not
wrong about it.

Fascinating. First you wrote that you _calculated_ the odds for the
second contestant. When I asked you to display your calculations, you
just decreed that "it's the probability of a coin toss" without any
justification. Now you were able to write two contradicting consecutive
sentences. In the first one, you write that "odds are not something that
exist independently" and in the second one you write that the second
contestant "is not wrong about" the odds being 1/2 for him. How can you
hold simultaneously the beliefs that something does not exist
independently at *at the same time* that someone can be wrong (or right)
about it?

Exactly like he wrote it. Look : the odds are 1/3 for the first player (before Monty makes his proposal), before *and* after he makes his choice.

That is obvious for me.

What do you suppose they are for Monty, before and after the choice?

I do not understand the meaning of this question.

What I mean is that information changes probabilities

Now, if you read what he wrote, instead of trying to show off,

I was not trying to show off, but you are being so assertive that I
suppose that you will not believe me.

Sorry, I had underestimated the misunderstanding, see below


you would realize he speaks of an independent player, looking at the show after Monty has opened the goat door, * but not kknowing* what was the initial choice of the player.

You are right: I had not realized that that was what the other poster
was saying from the start. But then I see no puzzle here. The two
contestants have different amount of information. Therefore, it is not
surprising that they have different strategies.

ok, the communication problem is here : you use a notion of objectve probabilities, and strategies in situations where they are not (or incompletely) known; we are speaking of subjective (or Bayesian) ones


For him, indeed, the odds are 1/2, even if he knows that the right choice is to switch, and that the prob of success is 2/3

I don't see what you mean by "if he knows that the right choice is to
switch". Since he knows nothing about what happened before, what does
this possibility mean?
No, he knows the rules of the show, and can easily calculate that the good strategy is to shift, but as he doesn't know which door was initially chosen, this knowledge is useless to him : if you ask him which door hides the car , he has a prob of 1/2 to guess right.



I am always surprised that, in those discussions, never is mentioned the 1000 doors variant : you chose randomly door 137, Monty opens 998 other doors, showing you 998 goats, and you are left with your initial door and door 754. Do you switch ?

Speaking for myself, I don't need this approach to see why the right
strategy is to switch.

But the problem is with the many people wgo have trouble seeing that :-)



Best regards,

Jose Carlos Santos
.

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