MORE Riddles a la Batman

[TonyC]

--- CONTINUED ---

NOW, what you might notice is a seeming contradiction to what I've stated all along. Apparently the odds went from 1 in 3 with the initial choice to 1 in 2 with the second choice. An important distinction between what I've maintained and what I've just demonstrated is RANDOMNESS. The rules of your game remove randomness by eliminating all of the sequences above that resulted in AUTOMATIC win or loss. In other words, NEVER, according to your rule, will the host reveal the car, or my first choice. NEVER. Consequently, the only possible sequences of choices are below:

CAR 1, Pick 1, Reveal 2, NO SWITCH = WIN
CAR 1, Pick 1, Reveal 3, NO SWITCH = WIN
CAR 1, Pick 2, Reveal 3, NO SWITCH = LOOSE
CAR 1, Pick 3, Reveal 2, NO SWITCH = LOOSE
CAR 2, Pick 1, Reveal 3, NO SWITCH = LOOSE
CAR 2, Pick 2, Reveal 1, NO SWITCH = WIN
CAR 2, Pick 2, Reveal 3, NO SWITCH = WIN
CAR 2, Pick 3, Reveal 1, NO SWITCH = LOOSE
CAR 3, Pick 1, Reveal 2, NO SWITCH = LOOSE
CAR 3, Pick 2, Reveal 1, NO SWITCH = LOOSE
CAR 3, Pick 3, Reveal 1, NO SWITCH = WIN
CAR 3, Pick 3, Reveal 2, NO SWITCH = WIN
CAR 1, Pick 1, Reveal 2, SWITCH = LOOSE
CAR 1, Pick 1, Reveal 3, SWITCH = LOOSE
CAR 1, Pick 2, Reveal 3, SWITCH = WIN
CAR 1, Pick 3, Reveal 2, SWITCH = WIN
CAR 2, Pick 1, Reveal 3, SWITCH = WIN
CAR 2, Pick 2, Reveal 1, SWITCH = LOOSE
CAR 2, Pick 2, Reveal 3, SWITCH = LOOSE
CAR 2, Pick 3, Reveal 1, SWITCH = WIN
CAR 3, Pick 1, Reveal 2, SWITCH = WIN
CAR 3, Pick 2, Reveal 1, SWITCH = WIN
CAR 3, Pick 3, Reveal 1, SWITCH = LOOSE
CAR 3, Pick 3, Reveal 2, SWITCH = LOOSE

This looks a lot like your first list, but it also includes the sequences where the contestant SWITCHES his initial choice. 24 sequences, 12 result in a win, 12 result in a loss. Furthermore, the win/loss ratio is equally distributed among the SWITCH and NO SWITCH sequences. Odds of winning with a SWITCH are 6 in 12, or 1 in 2. Odds of winning with NO SWITCH are 6 in 12, or 1 in 2.

AND, because we knew ahead of time what the possible sequences are, and what the odds of winning with each sequence is, we knew from the very beginning that the odds of winning with any choice is 1 in 2, and we know that switching doesn't affect those odds.

Tony, the host is absolutely giving us new information. He is revealing (i.e. eliminating) a losing door!

We always knew he would reveal a losing door. We always knew there could only be 24 possible sequences, and we knew all along that the odds of winning with a SWITCH sequence equal the odds of winning with a NO SWITCH sequence. We knew that the host's choice of doors to reveal would not be random.

Knowing that "some stocks will go up on Monday" is far, far less useful than knowing WHICH stocks will go up!

Knowing that 12 of these 24 stocks will go up on Monday is just as useful as knowing that 1 of these 2 stocks will go up on Monday.

Consider that 100 mailboxes game -- ... The host isn't opening mailboxes randomly --

My point exactly.

"You have but two choices" is correct. That is not at all the same as saying that your two choices are equally likely to produce a winner!

But it IS... it is...

 

[Snoopy58]

Tony,

We're getting close here. The crux of the matter is disagreeing about how many "scenarios" there are:

Car behind door 3; contestant picks 3; switch = LOSE (door 1 or 2 revealed, machts nichts)
[once again, this is 2 situations and should be counted as such]

If you are correct that this is 2 situations & should be counted as such, the number of scenarios where you win by switching is equal to the number of scenarios where you lose by switching, hence your 50/50 point, and no value in switching. If I am correct that the two scenarios (car behind 1, contestant picks 1, host reveals 2 and ... host reveals 3) [edit] really ARE functionally the same, [end edit]then the same math supports my conclusion.

That question is the crux of the matter. If the host CAN reveal the car, or the goat behind the door you picked, then it's a different game (much like the test question -- where there is NO value in switching... the proctor's statement was in no way based upon the answer just marked). But I think we agree that the host is constrained in what he reveals.

So, to explore the question:

A) Car is behind door #1, I pick #1, host does his thing, to switch is to lose.
B) Car is behind door #1, I pick #2, host does his thing, to switch is to win.
C) Car is behind door #1, I pick #3, host does his thing, to switch is to win.

That is 2 out of 3 "times" that switching produces a win. Should we give double weight to A), turning the problem into

A1) Car is behind door #1, I pick #1, host reveals #2, switch=lose
A2) Car is behind door #1, I pick #1, host reveals #3, switch=lose
B) Car is behind door #1, I pick #2, host reveals #3, to switch is to win.
C) Car is behind door #1, I pick #3, host reveals #2, to switch is to win.

Which gives even odds?

I will contend that the proper consideration is A, B, and C as the "universe" of probabilities, rather than A1, A2, B, and C, for the following reason:

The location of the car is one random event, uninfluenced by anything else. There is a 33.333% chance that the car is behind each door. The door picked is a random event, uninfluenced by anything else. There is a 33.333% chance that I pick each door. You now have a 9 possibilities:

Car 1, Pick 1: 11.111%
Car 1, Pick 2: 11.111%
Car 1, Pick 3: 11.111%
etc.

Now, of the times that I pick the door with the car behind it, half of those times, the host will reveal door 2, and half of those times he will reveal 3. Of the times that I pick 2 with the car behind 1, ALL of those times he will reveal 3, and if I pick 3 with the car behind 1, ALL of those times he will reveal 2. So:


Car 1, Pick 1, reveal 2: 11.111% * 50% = 5.555% (switch = lose)
Car 1, Pick 1, reveal 3: 11.111% * 50% = 5.555% (switch = lose)
Car 1, Pick 2, reveal 3: 11.111% * 100% = 11.111% (switch = win)
Car 1, Pick 3, reveal 2: 11.111% * 100% = 11.111% (switch = win)

I'm running short of time, but you can see that the pattern would repeat for Car behind 2 and car behind 3. The final result, though, would be that the sum percentages of switch = lose would be 5.555% x 6 = 33.333%, and the sum of (switch = win) would be 11.111% x 6 = 66.666%.

I stand by my initial conclusion. Make sense?

If I have 2 choices, and 2 outcomes, my odds aren't necessarily 50/50. If one choice is very strong and the other is a long shot, the odds aren't equal!

a slightly rushed Snoopy

 

[enigma]

Ya'll must have relished statistics. You are sick!



cheers,
enigma

 

[crjdxr]

Ya'll must have relished statistics. You are sick!



cheers,
enigma

Amen!