Riddles

[Firemeboy]

Here is another way to look at it.  If you picked a card, then I took away 50 cards, we are left with two cards, one of them is the Jack of Diamonds.  If I took back both cards, mixed them up, they you are correct, there is a 50/50 chance that one of them is the Jack of Diamonds.  

But that is only if I mixed them up.  When you picked your card you had a 1 in 52 chance, or a slightly less that 2 percent.  By taking away all of the other cards except for the Jack of Diamonds, this remaining card actually has a  .98 percent chance of being the Jack.  

In other words, if we played this 'game' 100 times.  On average you would pick the Jack of diamonds twice.  The other 98 times it would serve you right to switch to the card that I picked, assuming that I always eliminate the cards that are not the Jack of diamonds.

[Archon]

No, that is not the way probability works. You don't seem to be following what I am saying. If you have a deck of cards with any number of the cards in the deck (as long as the jack of diamonds is still in the deck)  the probability that each card is the jack of diamonds is ALWAYS the same. If you have 12 cards, the chance for each is 1/12. If you have two cards, the chance for each is 1/2. Mixing the cards up does nothing but changes their geographical position, which does not tie into probability in any way.

[Firemeboy]

Yes, that is the way probability works.  As long as the person knows which card is the Jack, only takes out cards that are not the jack, then it is just as I stated.  

Are you telling me that if you pick a card at random, and then I take out the other 50 cards, then 50 percent of the time the original card you pulled out will be the Jack of diamonds?

Only 2 percent of the time will your card be the Jack of Diamonds, and 98 perecent of the time the card left over will be.  It is not 50/50, as long as the 'host' knows where the card is, and removes the cards that are not the right one.

Check out the article, it's all explained in there.

[Archon]

Ok, you are explaining this wrong. What the article says is that by not picking a different door, it is the same as not having the choice to pick another door. Therefore, between the two other doors, they have a 66% chance of containing the door, whereas the door you picked has a 33% chance. If the door is in the other group, which it will be 66% of the time, then the host can't eliminate it, which means that the other door represents the chances of both the other doors, which is still 66%. Therefore, staying with the door you have would have a 33% chance, whereas changing would have a 66% chance. So, in conclusion, you are right with the answer to the original question, but you were wrong as far as the math went.

[Firemeboy]

I knew I was me who was wrong all along.  Although I thought that is what I was waying with the cards...

Anyway, I don't know who got that one.  Who wants to throw another riddle out?

[The Holy Saint, Grand High Poobah, Master of Monkeys, Ehlers]

look, here's where the problems in this analysis are:

You have three doors (let's keep the numbers really simple). You choose one, which has a 1/3 chance of being right. The host opens one of the others, knowing WHICH door holds the right one, he opens a wrong one. Then he offers you another chance. This becomes an ENTIRELY new probability. You are effectively choosing all over, do you switch or keep. it's the SAME, in terms of meaning, as saying "Do you want door #1 or door #2. The only confusion that comes from that is when you change it semantically. you are treating it like STAYING is NOT a choice. But it is.

Plus, let's look at numbers. You say switching has a 1/2 chance of being right. that's only HALF of the probability, and it's HALF the choices. where's the other half? Staying. If you stay, you pick the OTHER 1/2 chance.

The same goes if you increase the original number to 10 or 52. Yes, when you made the FIRST choice, you only had a 10% chance, or a (slightly less than) 2% chance. But once the others are removed, the person knowing the right answer has evened the odds to 50/50. ONE of those two is the right answer. If SWITCHING makes it 50%, then STAYING is also 50%. Otherwise, in the case of the cards, you're saying that there's a 50% chance the other card is right, slightly less than 2% chance the only you picked is right, and a 48.whatever% chance that one of the cards ALREADY REVEALED TO BE WRONG is the correct answer. An incorrect answer CANNOT be correct, by definition. and we know those answers are wrong. Thus there being  48% chance they are right is patently impossible. Thus the entire other 50% chance must lie in the card you chose already.

Switch or stay, it doesn't matter.

[Archon]

SE, look at the problem from the beginning. There are two doors that you didn't pick, and one door that you did pick. Since there is only the prize in one of the doors, that means that there is a 33% chance that the door you picked has the prize, and a 66% chance that one of the other two doors contains the prize. Remember that the door with the prize in it can't have been eliminated. So, since the prize was most likely not in the door that you picked, and it wasn't in the door that was eliminated, it is most likely in the door that you didn't pick and wasn't eliminated. There is a 66% chance that it is in the door that you didn't pick, and a 33% chance that it was in the door you picked.

[The Holy Saint, Grand High Poobah, Master of Monkeys, Ehlers]

What you say is ONLY true of the initial choice. Each choice is NOT dependent on the information already gained.

Look at this example. You are given a challenge to throw heads at least once during two flips of a coin. What are the possible outcomes of two flips?
Heads, heads
tails, heads
heads, tails
tails, heads.
Thus there are four possible outcomes, and three of them have at least one heads in it. Thus you have 3/4=75% chance of succeeding
You throw once, and get tails. You still have another throw. At this point, you will never get heads/heads, so there are only three results left. So 2/3 = 66%. Right?
Wrong. The chances of getting heads ON THIS THROW are still 50%. The fact that it came up heads on the last throw has no effect whatsoever on this throw. THis throw has the same chance of coming up heads or tails, regardless of the outcome of the entire scenario. To suggest otherwise is to propose some supernatural force that tallies such contests and ensure probability over the entire series of events. Perhaps Hermes, God of Gamblers watches over you?

So let's go back to our original scenario. You have three doors. What are the possible outcomes?
door 1 is correct, door 2 is wrong, door 3 is wrong
Door 1 is wrong, door 2 is correct, door 3 is wrong
door 1 is wrong, door 2 is wrong, door 3 is right
You pick door number 2. The host reveals door3, which turns out to be wrong.
So, you have two possible remaining scenarios
Door 1 is correct, door 2 is wrong
OR
door 1 is wrong and door 2 is correct.
You are asked to make a SECOND CHOICE. This choice is between the previous two possible scenarios. We already know the third scenario is not possible. Thus by choosing between changing to the first door or staying with the second door, you have a fifty fifty chance. The original percentage has no bearing on this choice, only on the probable outcome of the entire series of events.  To suggest otherwise says that door 3 having nothing behind it indicates something about door 1 or door 2. This is not the case.

Observation may affect the observed in quantum physics, but the unobserved in a game show, unless specific provisions are made for it in the rules and the people make changes after the first door is revealed (which would be difficult to adjucate as fair, incidentally) has no affect on the location of the prize.

[Peter Ahlstrom]

http://www.grand-illusions.com/simulator/montysim.htm (only works on internet explorer)

It still doesn't make sense to my brain, no matter what the results say!!!!!!!!!

[The Jade Knight]

Okay, I think we're dealing with a different problem, here.

Here's an interesting way of looking at it:

You make a choice (boxes represent chance of winning):

Your choice = [33%]

What's left = [66%]


Now, they're going to reveal one WRONG box, and it wont be yours.  66% of the time, the right answer is NOT your box.

For that 66% of the time

They reveal one box [wrong], the other box left is automatically right.  [100%]

So, you'll win 100% of 66% of the time.

The other 33%:

They reveal one of the other boxes [both of which are wrong], and the other is automatically wrong.  If you switch, you have a 0% of winning, that 33% of the time.


The whole snag to the equation, my friends, is the fact that they NEVER reveal your INITIAL choice.  So if we are to assume truly random distribution, switching will cause you to win 2/3 of the time.  It is important to note that the door being opened is not random; it will always be a) WRONG, and b) NOT YOURS.


The interesting thing is that it's not actually a game of three choices of 33% each, or two choices of 50%.  It's a game (by its "rules") where you have two choices, one is [33%], the other is [66%]

The reason for this is that revelation is NOT random in this game - whatever you pick, they wont reveal it.  If they randomly revealed any wrong box (including yours), your odds would be 50/50, thus:

Your choice [33%]

What's left [66%]

If the correct answer is in "What's Left" [66%]:
50% of the time they reveal your box [wrong], and so you have a [50%] chance of getting the right other box.

The other 50% of the time they reveal one of the other boxes [wrong], which leaves you with a [100%] chance that it's the one you pick.

If the correct answer is in "Your Box" [33%]:
They reveal one of the other boxes [wrong].  Switching would leave you a [0%] chance of winning.

So, if you always keep your box, you're going to win 33% [your box] + 16% [50% of 50% of 66%] of the time. = 50%

If you always switch, you're going to win 33% [50% of 66%] + 16% [50% of 50% of 66%]

If you always want to switch, you're automatically wrong a third of the time, and it's completely random another third (whenever they reveal your box).

If you never want to switch, you're automatically right a third of the time, and it's completely random another third (whenever they reveal your box).

What the rules of this game do, by never revealing your box, is that they take that last random third out, so it's now you have a 33% chance of winning if you keep your box, and a 33% chance of losing if you don't.


Phew.  I hope that clears things up.  Remember, they NEVER reveal the

answer.  If they did, you'd win 66% of the time.

[Firemeboy]

What you say is ONLY true of the initial choice.

 

It's all one choice.  You either stick with what you have, or go with the other 2 (9, 51, etc).  You are right, it's either one or the other, but by revealing what is behind the other door (cards), you haven't eliminated them from the equation.  They are still a part of it.

[The Jade Knight]

Bah, read my explanation.


You didn't give us enough information.  You didn't mention that a) they will NEVER reveal the door you chose, and b) they will NEVER reveal the right answer.

It's 50/50 if there's a chance they'll reveal your door, it's 66/33 (automatically) if there's a chance they'll reveal the correct door, and it's 33/66 if they'll never do either (depending on whether you switch or not).

[Legion]

Yes, but think about it.  I know where the Jack of diamonds is.  You don't.  You pick 1 in 52 cards.  That card has a 1 in 52 chance.  I, knowing where the card is, remove the other 50 cards and leave one card left.  You can't say that the original card has the same chance of being the Jack of Diamonds as the 1 card that is left over...  

Here is an article describing how it works.  

http://www.straightdope.com/classics/a3_189.html

But you might know where it is, and by the time you ask us to switch we have a 50% chance of being right....So it may be the card you have not flipped over, but it could (equally) be the card that we have picked.  Each has as likely chance of being true.  So if we switch or stay with our original card we still have the same chance of being right.  Hence it is not nessasary "in your best intrest" to switch

[The Jade Knight]

Legion, go read my explanation, also.


You people need to stop ignoring me.  I've already explained why and how these work the way they do.

[Legion]

WOW I read your reply Jade and it is filled with misleading facts....

OK you have 3 boxes
[1] [2] [3]

You pick box 2.....of course the host is going to eliminate a wrong answer first..... So box 3 is eliminated

[1] {2}

Now you only have box 1 and 2 left.....You had a 33% chance of winning in the beginning....now since one of them is eliminated that was wrong, your chances increase to 50% no matter if you switch or not.  So is it in your best interest to switch...no because regardless of which box is removed you still have the same chances of winning at this point,  because the host is NEVER going to look at yours or (if its different) the correct box first, they will always eliminate one of the wrong ones.  You have at this point that same amount of likelihood of being right on your first guess then if you change.  Remember you are, after eliminating 1 box still just picking from 2 boxes that both have an equal chance of having the prize in it.