Philosophy Now Forum

dionisos wrote:

The possibilities are 12 in the first case because the dice rolls for 6 options of which the case in the even add 3 (one for each even number.)

6+3=9 no ?

dionisos wrote:

I understand your reference that some things are not equiprobable

Show me a simple example where the possibilities are not equiprobable, for me to confirm that you understand what it mean.

Scott Mayers wrote:

dionisos wrote:

The possibilities are 12 in the first case because the dice rolls for 6 options of which the case in the even add 3 (one for each even number.)

6+3=9 no ?

The 12 comes from respecting the in-between act of picking an EVEN number, and then counting the two events after that. In the program experiment it is only intended to actually count the results without concerning ourselves with those 12 odds as you intended it to be constructed simply to see if a computer should actually guess close to your odds in multiple events regardless.

The possibilities are 12 in the first case because the dice rolls for 6 options of which the case in the even add 3 (one for each even number.)

dionisos wrote:

Scott Mayers wrote:

dionisos wrote:
6+3=9 no ?

The 12 comes from respecting the in-between act of picking an EVEN number, and then counting the two events after that. In the program experiment it is only intended to actually count the results without concerning ourselves with those 12 odds as you intended it to be constructed simply to see if a computer should actually guess close to your odds in multiple events regardless.

I am sorry but this is nonsense.
"12 odds" mean nothing, "in-between act of picking an EVEN number" mean nothing.
If there was 12 possibilities, you would be able to enumerate them, like i did with the 9 possibilities. (d1, d2-head, d2-tail, …)
You can’t enumerate the 12 possibilities, because there are only 9, this is why, you totally confuse yourself with english sentences.

The possibilities are 12 in the first case because the dice rolls for 6 options of which the case in the even add 3 (one for each even number.)

If i add 3 possibilities, for each even number as you propose, to the 6 options of the dice rolls, i get 6+3+3+3=15
Outside the fact this calculus is totally meaningless, you don’t even get 12.

I am sorry but i stop this discussion, because either you don’t try to understand me, or you are just trolling me, or you are way too much delusional.

dionisos wrote: I will try to explain what i mean with a more simple problem, tell me if you understand, and if you see the link with our actual problem:

Imagine i roll a dice, i will have 6 possible cases: d1(i get 1 on the dice), d2, d3, d4, d5, d6.
A important property here, is that these cases are equiprobable, that mean, they have equal probability to happen: P(d1)=P(d2)=P(d3)=…
Now imagine i win the game if i got a 2 or a 3.
To calculate my probability to win, i will (please keep it mind here the goal is to show the general method):
1) See that P(d1)=P(d2)=P(d3)=… and that P(d1)+P(d2)+P(d3)+… = 1
2) Deduce that P(d1) = 1/6
3) Sum the probabilities of each cases where i win: P(d2)+P(d3) = 2/6

Note that what most people will do, is to count the number of cases where they win, and divide it by the total number of cases, and get 2/6, the good thing is that it is more simple, it include the 3 steps i just give, the bad thing, is that it hide the real reasoning.

Now let me show you a problem where cases aren’t equiprobable:

I roll a dice, if i got a even number i toss a coin, i will have 9 possible cases: d1, d2-head, d2-tail, d3, d4-head, d4-tail, d5, d6-head, d6-tail
Now imagine i win the game if i got 2 on the dice and a head on the coin, or if i got a 3 on the dice.
The naive way to calculate my probability to win would be to count the cases where i win, and divide it by the total number of cases: P(win) = 2/9
But this is false, because here, the cases are not equiprobable.
The real way to get the good result is to do:
1) P(d1)=P(d2)=P(d3)…=P(d6) and P(d1)+P(d2)+…=1
2) P(d1)=1/6
3) P(d2-head) = P(d2-tail) and P(d2-head) + P(d2-tail) = P(d2) = 1/6
4) P(d2-head)=(1/6)/2 = 1/12
5) Sum of the probabilities of each cases where i win: P(win) = P(d2-head)+P(3) = 1/12+1/6 = 3/12 = 1/4

You never said that P(option1 ligne 1) = P(option1 line 3), but the calculation you did, implied it, because without it, it was wrong.

It is you who is concerned about the 12, not me.

Scott Mayers wrote:Then prove that the case in the Monty Hall Problem to which you agree or disagree rather than your own.


OR attempt this:
In the simple two-box case, if I am the host of as simple game to which I said,

"There are two boxes in front of you. I have one ball that fits into it any one of them. I MAY place a ball in one of them."

In this simple game, there are three 'types' of possibilities (arrangements).

Box A....Box B
...Ball.......0
....0........Ball
....0.........0

The question here is without knowing anything, when you pick any box without looking into it, can you determine if it has a ball or not?
Select one: Determinate or Indeterminate

Is the probability 1/2 for each box or 1/3, or 1/4, considering a fair game?

If the host restarts the game and decides to initiate the it by showing you an empty box leaving you the other to have to pick, what is the probability that it will have a ball? 1/2 or 1/3 or 1/4? OR...indeterminately both?

The question here is without knowing anything, when you pick any box without looking into it, can you determine if it has a ball or not?
Select one: Determinate or Indeterminate

Scott Mayers wrote:Is this not similar in part to the second half of the Monty Hall game when trying to determine the case or cases where the remaining options are empty?

When the host reveals the empty box in the first two arrangements above, he is 'forced' to choose in a determinate way; in the last arrangement, since both are empty, he has real options to select either Box A OR Box B to reveal.

Since both have no ball in them, whether you think of the the probability as a Result (what is not in the boxes in both collectively) OR if you consider them independent separate cases as the Act of him choosing Box A or B, exclusively, are they not equiprobable as '0' as one possibility OR two?

So is it:

Box A....Box B
...Ball.......0....
....0........Ball..
....0.........0.... <--- Probability = 0 ? [Respecting the Result value of '0' only]

or

Box A....Box B
...Ball.......0....
....0........Ball..
....0.........0.... <--- Probability = 0? [Respecting the Host Actions to reveal one of these two options independently?]
....0.........0.... <--- Probability = 0?

dionisos wrote:

Scott Mayers wrote:Is this not similar in part to the second half of the Monty Hall game when trying to determine the case or cases where the remaining options are empty?

No it is not.

Scott Mayers wrote: Do you not wonder how or why other mathematicians even with full degrees question the original presumptions of the puzzle? Or are they just imagining things?

?? Here is the diagram I want you to scrutinize from above again:

Fixed Solution.png

I already understand the traditional position and how it is derived (as per the way they treat rows one and two in that diagram as merely one where I see it as distinctly separate events. )

dionisos wrote:

Scott Mayers wrote: Do you not wonder how or why other mathematicians even with full degrees question the original presumptions of the puzzle? Or are they just imagining things?

I don’t wonder it, because it is false, about no mathematicians question the puzzle.
Some mathematicians do a mistake when first faced with the puzzle, because there is a trap, and anybody could fall in a trap, even very simple trap.
But i never see a mathematicians question the answer to the problem, after we show them the trap.

?? Here is the diagram I want you to scrutinize from above again:

Fixed Solution.png

Here in you diagram, in 2 cases you win by switching, and in 2 cases you lose by switching (win by staying).
Then, what ?
You can’t do anything with that, if you don’t know the probabilities of each 4 cases.
You can’t conclude that P(win when switching) = P(win when staying).

dionisos wrote:

I already understand the traditional position and how it is derived (as per the way they treat rows one and two in that diagram as merely one where I see it as distinctly separate events. )

It is two distinctly separate events, i said it from the start. (you could also consider it to be one event, it change nothing to the final result)
What you don’t seem to understand, is that probability is not just about counting the number of events.