Topic: Mind & Probability Paradox bitcoin betting (Read 1431 times)

In a true 50/50 game, even if you've just seen 10 reds in a row, the next turn is still 50/50.

The Monty Hall problem, which you're referring to, isn't and never was 50/50.  You picked a door, you have 1/3 probability of having picked the car.  The other two doors have a 2/3 probability of having the car behind one of them.  Since the host knows where the car is, he can always open one of the remaining two doors and show that it doesn't hide a car.  By offering you the choice to switch, he is effectively letting you chose both the other two doors, which doubles your chance of success.  As you can see, this is in no way similar to a new independent 50/50 dice roll.

me: "I'm thinking of a number between 1 and a million.  Guess what it is"
you: "665566"
me: "I'll give you a clue.  the number is either 888888 or 665566.  it's nothing else.  want to switch?"

There's a 1 in a million chance that your first guess was right.
There's a 999999 in a million chance that your first guess was wrong, and that my number was 888888.
If you switch, you have 999999 times more chance of winning.
If you still reason "well, there are two numbers it could be, so it's 50/50" then I don't know how to help you...

Does that clear things up?  I've tried to explain the Monty Hall problem to people before.  Some of them really don't get how it isn't a 50/50 choice even given the above exaggerated situation.  I've never been able to understand why; I don't see what it is they can't understand, but would like to.

The three-door problem you're pointing to isn't the same problem you're imagining it is. It's rigged. The host isn't picking 50-50 which door to show you, he will *always* show you a door with nothing behind it, and that changes the game.

As an illustration, expand it to the 100-door problem.

You choose one door out of 100. Say, door 71. The game show host, rather than open that door right away first opens up 98 other doors, and they're all empty. Only doors 12 and 71 remain closed.

Now, do you stick with door 71, or switch to the glaringly obvious door 12?

The game show host didn't affect the probabilities... he just gave you extra info with which to make your decision.

You don't understand probability, which is common, many mathematician were/are perplexed by 3-door problem, anyway simplest explanation is that it is not about past outcomes, you are changing the game, e.g. if you load the dice will outcome change or remain 1/6 as change to dice was in past

After getting two reds, the chances of getting a third is 0.5.  But you'll only get two initial reds 0.25 of the time.  The product is that one time in eight you get three reds, exactly what you'd expect before seeing the first two colors.

If the host doesn't know where the car is, he'll pick it half of the time if you haven't already picked it, and never pick it when you have.  When the host doesn't know, you have a 1 in 3 chance of picking the car correctly the first time, and switching doesn't change your odds.

There are 12 possible outcomes, including some duplicates.  Say we rename the doors (after the game) to ABC, where A is whichever door had the car, and B and C had goats.  The outcomes are (the first letter is your choice, the second letter is the hosts choice, the third letter is which door you end up with):

 1   ABA   see goat, stay, win
 2   ABC   see goat, switch, lose
 3   ACA   see goat, stay, win
 4   ACB   see goat, switch, lose
 5   BAB   see car, lose
 6   BAC   see car, lose
 7   BCB   see goat, stay, lose
 8   BCA   see goat, switch, win
 9   CAC   see car, lose
10   CAB   see car, lose
11   CBC   see goat, stay, lose
12   CBA   see goat, switch, win

5, 6, 9, and 10 are moot because once you see the car, you've already lost.  You can still switch, but there isn't any point.  1/3 of the time (4 of 12), you'll lose when the host opens the door with the car.  When the host opens a goat (8 of 12), in 2 of those cases, you win if you switch (7/8 and 11/12), in 2 you lose if you switch (1/2 and 3/4).  Your odds are the same either way, 50% from when you get to that point, and you get there 2/3 of the time, for total odds of 1 in 3, exactly what you'd expect up front.

When the host knows where the car is, he reveals part of that information to you when he opens a door, which changes the odds.  When the host doesn't know, he doesn't have any information to reveal, so the odds can't change.

Past results don't influence a future random result, whether they are known or not.  (not getting into entangled particles and Bell's inequality here)

In the three door problem, the actions of the host are not random.  Therefore, the outcome of choices made after his interference are not random.  That's why his knowledge allows him to influence the odds.  In the scenario you suggest, where he doesn't know what is behind the door, his actions become random again.

It isn't that knowledge of past choices or lack thereof influences random events.  It's that knowledge enables the host to create a system that isn't truly random.

Mind & Probability Paradox
-------------------------

Can a game such as Satoshidice Bitcoin site, and can you influence the outcome?
The answer is obviously no. Well lets go through the states, and several
odd bending games and problems, it should apply to all equally right?

If someone were to hypothetically play a game where it is truly 50-50 Odds (red/black)
If you were to guess beforehand and say "The odds of 3 reds coming up now is 1/8?"
Is that statement true?

However lets say we proceed with the roll and two reds came up.
Would the third red be 1/8? with the series? Would as the game goes
on the odds of it keep hitting red decrease? Or would it always remain 50-50% from
that moment on?

Now lets say your faced with two options:
1) "You see 10 reds came up"
2) "Someone claims 10 red came up, but hes actually lying"
3) "You just print red 10 times down, and look at that 10 times.

Wouldn't the outcome for all these scenarios make absolutely no difference for the next roll?
Then if you guess like okay I guess three reds to come up, then the odds of it matching
your guess is 1/8 right? But after two has came up, it quickly becomes 50%?

It seems to conclude, that odds at that moment remain 50-50% and all past influences
make absolutely no difference? This should apply to all games in a similar matter right?
That works with odds?

Now since this is true and all past influences make no difference,
then how come in the three door problem, when the host presents
you with the choice of switching and/or staying with your original choice,
it influences the outcome?

When you switch you are presented with two doors left after he reveals a goat.
Now since past influences make absolutely no difference, and the above statement
should apply to all games, then it should be 50-50%?

Well no it isn't, its actually 66% chance of winning, if you switch, and 33% chance of losing
if you don't switch?

The paradoxical thing is if lets say you change the game, and make it that
the host does in fact not know where the car is, if he picks that on the
first round, sorry you lose the car. But if he picks the goat,
in this scenario the odds will be 50-50%, I tested this out myself.

It seems that knowing and not knowing can influence the game outcome,
and thus it seems that mind can influence odds,
but I thought this was fallacy and false? Isn't this a paradox?
Or just a optical illusion inside a illusion?

The host has to know where things are, if he didn't know, and lets say he revealed a car,
then it goes destroyed in this scenario it should remain 50-50

It seems mind and knowledge can influence the outcome of a game?
Or its just a illusion?

Thoughts?

http://www.sciencedaily.com/releases/1998/02/980227055013.htm
Quantum Theory Demonstrated: Observation Affects Reality

http://theobservereffect.wordpress.com/the-most-beautiful-experiment/
The double slit observer effect