Maths Problem

[Mr. Harley]

Here are your actual choices

choose - other - deleted (bold has the prize)

A-B-C no switch - win
A-B-C switch - lose
A-C-B no switch - win
A-C-B switch - lose

A-B-C no switch - lose
A-B-C switch - lose

A-C-B no switch - lose
A-C-B switch - win

I can manipulate the letters any way I want, it will always have paired dualities, since you actually really have only two conditions for two choices. The third choice is eliminated in every case.

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.

[Catfish]

@Azazel, General Ludd, shawnmmcc:

Dudes, the reason you guys are confusing yourselves is because you're adding another layer of choice. If you randomly choose between the last two cups, then the odds of success are of course 50% - but that wasn't the question. If you always switch, your odds rise to 66.7%. If you always stick, your odds remain at 33.3%.

[LulThyme]

This is called the Monty Hall problem for those who want to search for a nice explanation.
If you stick with original choice, you have 1/3 chance of winning.
If you change, you have 2/3.

Easiest explanation I can give is this :
There are two strategies for this game.
You can switch or keep.
If you keep, you win if you chose the right one at start and lose otherwise, so you have 1/3 of winning.
If you switch, you win if you chose a bad one at start and lose otherwise, so 2/3.

Simple no?

[LulThyme]

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.

Thats were you wrong.
The cup that gets removed DEPENDS on your choice.
You did the same mistake most ppl do at first.
Suppose the answer is B and you choose A, then choice C gets eliminated, but suppose you choose C, then choice A gets eliminated.

See my explanation above.

[Az]

That's a nice explanation, Lulthyme. It all made sense when I actually performed the experiment myself. As I did it, It immideately became clear.

[LulThyme]

Yeah I had to explain it to many people so over time, I got to know which explanation worked better

You can also think of the same problem with 1000 cups.
Then I promise I will turn over 998 which DO NOT have anything under them which you did not choose.
You can keep your original choice, or take the last one I didnt turn over...
Seems trivial which strategy is better in this case he?

[Az]

Actually, no, since I didn't approach the question this way. ( read beginning of thread ) According to my own very special logic, it would mean that the first stage of the choosing doesn't matter, at all. Actually, I was so convinced that I still can grasp it logically, but only accept the reality of it being true ( Like QM for some people )

[General Ludd]

Originally posted by Catfish
@Azazel, General Ludd, shawnmmcc:

Dudes, the reason you guys are confusing yourselves is because you're adding another layer of choice. If you randomly choose between the last two cups, then the odds of success are of course 50% - but that wasn't the question. If you always switch, your odds rise to 66.7%. If you always stick, your odds remain at 33.3%.

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?" It's one in two, because there are only two choices presented.

The final question asked in the problem is: Did you guess right? It could be that one, or it could be this one - this third one wasn't it, though. How can you have a 1/3 or 2/3 chance of geting a yes or no question right?

[General Ludd]

Originally posted by Catfish
If you always switch, your odds rise to 66.7%. If you always stick, your odds remain at 33.3%.

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

That would make some sense - because if you chose one out of three and stay with it regardless of the person at the table giving your a "second chance" that could be seen as 1/3 chance of winning.

But when you make a second choice after the removal of a wrong choice you are no longer choosing between three, you are only choosing between two.

The choice to switch or stay does not happen when there are three cups on the table, only when there are two.


The only other way I could think of stretching the chances like that is if you base the second choice on the likelyhood of your first choice being right. You only had a one in three chance to get it right, so if you stay with that "one in three" choice, the alternative must be two in three, right? Which almost makes sense in a round about way, but is ignoring the reality of the final choice, which is that there are only two choices - A or B, B or C, C or A - doesn't matter, it's always one possible choice of two.

[Solomwi]

Ludd, rather than thinking of it as "Which of the two is the prize under?", think of the stay/switch option as "Was my initial choice right or wrong?", since by switching, you're betting that you were wrong in the first place.

Since the odds of being wrong when you first chose a cup were 2/3, the odds of switching resulting in you getting the prize are also 2/3. The cup that gets eliminated was still a choice at the beginning, which is what keeps it from being a simple "choose A or B, and C is just a cup" problem.

[General Ludd]

Originally posted by Solomwi
Ludd, rather than thinking of it as "Which of the two is the prize under?", think of the stay/switch option as "Was my initial choice right or wrong?", since by switching, you're betting that you were wrong in the first place.

I covered that in my last post... as I said, it almost makes sense in a round about way, but it requires you to consider three options in a choice of two - ie. it's ignoring the final question. "was my initial choice right or wrong" is the same as "which of the two is the prize under?" in this case.

The cup that gets eliminated was still a choice at the beginning

But not at the end, which is when the final choice - the choice that matters - is made.

[Worthingtons]

lol, cannot believe there are still believers in the 50-50 system, then again some proffessors (sp?) of Maths did originally get this wrong too.

What about boddingtons question? anyone else want to move onto that or is it too easy?
Cheers
Matt

[General Ludd]

Originally posted by Worthingtons

What about boddingtons question? anyone else want to move onto that or is it too easy?
Cheers
Matt

Shawn already did (correctly).

[Mercator]

The final choice is indeed only between 2 cups, and indeed, in that situation there's always one cup with and one cup without a prize.

But the first choice does matter, because sometimes your first guess will be the cup with the prize and sometimes it will be an empty cup.

Since there are two "wrong" cups, you are more likely to have chosen an empty cup on your first choice.


Consider this: There's a bowl with three balls in it. Two balls are red, one ball is blue. You can take one ball out of the bowl (without looking). What's the chance that you picked the blue ball?
Now, after you took a ball I take a red ball out of the bowl (you can look again). What's the chance that you picked the blue ball now? Well, yes of course, it's still the same, namely 1/3. Me meddling with the bowl doesn't change anything about which ball you just picked.
So, if you have a blue ball in your hand, the remaining red ball is in the bowl, if you have a red ball in your hand, the blue ball is still in the bowl.
But you still have a 1/3 chance that that ball in your hand is blue. In other words, there's a 2/3 chance that the blue ball is still in the bowl.

And going back to the cups... There's a 2/3 chance that your first guess was wrong.

So the question "was my initial choice right or wrong" is NOT the same as "which of the two is the prize under?", because with that second question it's like you throw the ball in your hand back into the bowl and then try to take the blue ball out.

[General Ludd]

Originally posted by Mercator

Since there are two "wrong" cups, you are more likely to have chosen an empty cup on your first choice.

And what relevance does the chances of the first choice have on the chances of your second, different, choice?

Consider this: There's a bowl with three balls in it. Two balls are red, one ball is blue. You can take one ball out of the bowl (without looking). What's the chance that you picked the blue ball?
Now, after you took a ball I take a red ball out of the bowl (you can look again). What's the chance that you picked the blue ball now? Well, yes of course, it's still the same, namely 1/3. Me meddling with the bowl doesn't change anything about which ball you just picked.

But in that scenario you aren't given a second choice. You just choose once, and you taking a red ball out has no baring on the outcome because the choice is already made. (and if you where given a second choice in that scenario, with being able to clearly see which is which, it would be impossible to lose - because either the blue ball would be in the bowl, or you would be holding it)

So the question "was my initial choice right or wrong" is NOT the same as "which of the two is the prize under?", because with that second question it's like you throw the ball in your hand back into the bowl and then try to take the blue ball out.

How is it not the same? Either your choice was right and you picked the right one, or it was wrong and the remaining cup is the right one. That's one or the other. A choice of two.