Maths Problem

[Mr. Harley]

I am going to play this with my little girl tomorrow and get giggles and test results at the same time, so .

[Urban Ranger]

Suppose the cups are marked A, B,C and the prize is always under cup A. There are 3 posibilities:

1. You pick cup A (1/3). One of the other two cups is revealed to be empty.
2. You pick cup B (1/3). Cup C is reveealed to be empty.
3. You pick cup C (1/3). Cup B is revealed to be empty.

Only in case 1 a switch will cause you to lose. Thus, switching gives you 2/3 chance of winning. IOW, double your odds.

[General Ludd]

Originally posted by Urban Ranger
Suppose the cups are marked A, B,C and the prize is always under cup A. There are 3 posibilities:

1. You pick cup A (1/3). One of the other two cups is revealed to be empty.
2. You pick cup B (1/3). Cup C is reveealed to be empty.
3. You pick cup C (1/3). Cup B is revealed to be empty.

Only in case 1 a switch will cause you to lose. Thus, switching gives you 2/3 chance of winning. IOW, double your odds.

But that is only the chances (and the results of) the first choice. You aren't even adressing the second choice, and the final result. Which are made after a cup is revealed, which in your demonstration is shown as a result - as something to yet happen.

So that is round one. Let's say we chose 2. The possibilities are now:

1. You stay with cup A (1/2)
2. You switch to cup B (1/2)

There isn't a third choice in the second, defining, round. Only if you where required to decide to switch or stay in the first round could it be an X/3 chance, but that is not how the question was presented .

[Urban Ranger]

Originally posted by General Ludd
So that is round one. Let's say we chose 2. The possibilities are now:

1. You stay with cup A (1/2)
2. You switch to cup B (1/2)

These probabilities are only right if you don't take the intial round into account. IOW, these are right if there's only one round.

[General Ludd]

Originally posted by Urban Ranger


These probabilities are only right if you don't take the intial round into account. IOW, these are right if there's only one round.

How does the first round impact the second? You can say that there was a 1/3 chance of choosing correctly in the first round, but switching or staying is 50/50 regardless.

The first round seems completely irrelevant, because the result of it is always the same - a choice between two remaining cups, one right, one wrong. It's hard to even call it a choice.

[Petek]

Originally posted by General Ludd

How does the first round impact the second? You can say that there was a 1/3 chance of choosing correctly in the first round, but switching or staying is 50/50 regardless.

The first round seems completely irrelevant, because the result of it is always the same - a choice between two remaining cups, one right, one wrong. It's hard to even call it a choice.

If, in the first round, you chose the cup that has the prize, then the two remaining cups both contain nothing of value. In this case, switching to another cup loses.

On the other hand, suppose you chose an empty cup in the first round. In this case, if you switch, you will certainly win.

In light of this, the choice in the first round makes a difference.

[Urban Ranger]

Originally posted by General Ludd
The first round seems completely irrelevant, because the result of it is always the same - a choice between two remaining cups, one right, one wrong. It's hard to even call it a choice.

That's right, one is empty and one has the prize - but you don't know which. You might have the prize in your cup, or you might not. That's how the first round influences the second.

[Snowflake]

Originally posted by General Ludd


How does the first round impact the second? You can say that there was a 1/3 chance of choosing correctly in the first round, but switching or staying is 50/50 regardless.

The first round seems completely irrelevant, because the result of it is always the same - a choice between two remaining cups, one right, one wrong. It's hard to even call it a choice.

In the first round, after you pick a cup, the odd that it contains the ball is 1/3, and the odds that the other two cups contains the ball is 2/3. The trick is that in the second step a cup is removed from group two, knowingly, instead of randomly, in other words, information is added into the result. Essentially switching means you are choosing group two, with the 2/3 odds.

[Snowflake]

Originally posted by shawnmmcc

The case of the woman with the girl is the frequent counter-intuitive misunderstanding of causality. The next choice has no linkage to the previous one. All you have done is simplified the system when you eliminate one of the three cups. That's why I have a co-worker with four girls and no boys. Only one is sixteen families with four children end up with that (for approximation, it actually isn't quite 50-50), each time he only had a 50-50 chance of not having a girl. So with his fourth try, he still only had a 50-50 chance of having a boy. God plays mean.

If the question is what the odd a woman will have a boy when she already have a boy, then the odd is 50-50. However if we know she already have two kids, and one is a boy (could be the first, or the second), then the odd that another is a boy can only be 1/3.

[Catfish]

Originally posted by General Ludd
The question was "what is his chance of him winning if he sticks to his choice?", and "what is his chance of winning if he switches?"

That's right. Two separate questions, two separate methods. What you are suggesting is totally random.

Originally posted by General Ludd
And why the emphasis on always - are you going at this problem as though the choice to switch or stay is predetermined?

Of course! You predetermine the strategy. These are the 3 possibilities:

1. Stick with your original choice (1 in 3 chance of success),
2. Switch to the other cup (2 in 3 chance of success) or
3. Choose randomly from the last 2 cups (essentially what you're suggesting; 1 in 2 chance of success).

Using methods 1 and 2, you do not get the option of choosing between the last 2 cups – by definition the choice of your final cup has been made in the first instance, ie, there is only one "round". By using either of the first two selection methods you are reducing the number of possible outcomes. Maybe the attached diagram will help you.

Originally posted by General Ludd
There isn't a third choice in the second, defining, round. Only if you where required to decide to switch or stay in the first round could it be an X/3 chance, but that is not how the question was presented .

Yes it was. See previous.

Originally posted by Snowflake
If the question is what the odd a woman will have a boy when she already have a boy, then the odd is 50-50. However if we know she already have two kids, and one is a boy (could be the first, or the second), then the odd that another is a boy can only be 1/3.

I'll second that.

[LulThyme]

Originally posted by General Ludd


How does the first round impact the second? You can say that there was a 1/3 chance of choosing correctly in the first round, but switching or staying is 50/50 regardless.

The first round seems completely irrelevant, because the result of it is always the same - a choice between two remaining cups, one right, one wrong. It's hard to even call it a choice.

See the nice thing about this problem, is that its just not a thinking exercise...
You CANNOT disagree forever...
Do like Azazel did and get your cups out.
Try switching everytime, and try staying everytime, or whatever strategy you want, and come and tell us about it, and why you now understand we were right...

[Kuciwalker]

Originally posted by Azazel

Az: if you switch always, you have a 2/3 chance of winning each time. If you don't ever switch, you have a 1/3 chance.

Once again, how come?
whatever you choose at first, you're left with two cups: one of them has a prize, the other has no prize. "switching or not switching" is just like saying "choosing this cup, or choosing that cup". The fact that you chose some cup before that is irrelevant.

Two thirds of the time, you pick the wrong cup. The OTHER wrong cup is removed, and so if you switch, it must be to the right cup.

One third of the time you pick the right cup. One wrong cup is removed, and the other is still there. You switch to it.

See?

[Az]

Dude, I came to it on my own. Stop beating my poor dead horse.

[Kuciwalker]

I hadn't read the thread yet

[General Ludd]

Originally posted by LulThyme


See the nice thing about this problem, is that its just not a thinking exercise...
You CANNOT disagree forever...

Yeah, I thought about it some more last night and I can see now how the second choice is tied to the first - through the language used.

Having to "stay" or "switch" ties the second round to the first, so that you are betting on the odds of the first round and aren't actually choosing a new cup. So, there's a 1 in 3 chance that you'll need to stay, and a 2 in three chance that you'll need to switch.

In order for the second round to be totally reset and 50/50, your first choice would have to be removed, so that it's no longer 'contaminating' the second choice. Either the cup you chose would have to be taken away itself, or the two remaining ones shuffled without the choice of switching or staying with the choice you originally made.


It's an interesting problem. I don't see how this can be applied to the odds of having a boy, however. What is the connecting factor in that regard? EDIT: I see now, after having actually read the question - just more word games. It's only warped odds if you're asking about how many boys of this and that, not what are the chances of a birth specifically.