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[scott_pilgrim]

Why would you read 3x as a single unit but not (18/3)x as a single unit?

Actually, one argument the one guy who agreed with me was making was that if you put 18/3x into a calculator, it will do (18/3)x.

So what happens with 18÷3(x)?  That would still be 3, right, since the parentheses get resolved first (doing nothing)?  And 18÷3(2) is also 3, because it becomes 18÷32 (where 32 is a "smash up" 3*2, not the number thirty-two)?

If you ever see written 18÷3(x) or, may the Lord have mercy on our souls, 18÷3(2), burn whatever book you are reading to ashes and never look back.

Well, I would think 3(x) is exactly the same as 3x, since you just do everything in the parentheses first, which outputs x, and then you're left with just 3x.  But then, I also would have thought 3x is exactly the same as 3*x, so I don't know what to think anymore.

What about 18÷xy, where x=3 and y=2?

[pacovf]

It's not about being the same or not, it's about being purposefully confusing. The parentheses do nothing, so they shouldn't be there, and if you really need the parentheses, then the expression is sufficiently complex that it deserves to be written properly (ie., unambiguously). The implicit product is a shorthand, you have to use it responsibly.

[ConMan]

I'm not trying to start an argument, I legitimately want to make sure that there's not some rule I'm somehow unaware of that says you do "smash up multiplication" before everything else.

Fair enough, but this kind of question was showing up on Facebook a lot last year (and I think it must have been in the Random Thoughts thread or something here because it's not in this thread) and it seems like the entire point of it was to get people to argue about what was "right" or "intuitive".

It is definitely true that students are taught some version of PEMDAS (or BOMDAS, or other variations) that either says that multiplication comes before division or that if you have both then you resolve them left-to-right. And that's mostly ok for arithmetic, and it seems to work for addition and subtraction, but it kind of conflicts with the kind of shorthand that we tend to use when writing more complicated algebra, especially when it's compressed onto a single line - normally we'd write something like 14/2(3+4) as a two-line fraction so you can clearly see where everything is. But, and here's the annoying bit, the accepted shorthand of omitting the multiplication sign where you have either parentheses or pronumerals is usually understood to take precedence *over* just about everything else. Like I said earlier, smooshing things together like 2(3+4) or 3x tends to mean that you have a single "unit" that should be calculated on its own before you look at other arithmetic operations happening around them.

Alternatively, you could take the approach used in some forms of arithmetic and logic, which sets out a bunch of legal constructions of these sorts of expressions, usually requiring parentheses around just about everything, and which allows you to remove the parentheses only when the resulting expression remains unambiguous. So the "formalest" way of writing the expression might be (14/(2(3+4))), or (18÷(3x)), and the rules of the system would tell you that you can remove the outermost parentheses, but no more because the expression becomes ambiguous at that point. Or you can learn Swedish notation, which removes the ambiguity in another way.

[scott_pilgrim]

It's not about being the same or not, it's about being purposefully confusing. The parentheses do nothing, so they shouldn't be there, and if you really need the parentheses, then the expression is sufficiently complex that it deserves to be written properly (ie., unambiguously). The implicit product is a shorthand, you have to use it responsibly.

Again, I don't disagree with you, and I'm very frustrated that I'm having to deal with this type of problem in the first place.  But I wanted to know what to do when I actually do run across this kind of thing again in the future.

I'm not trying to start an argument, I legitimately want to make sure that there's not some rule I'm somehow unaware of that says you do "smash up multiplication" before everything else.

Fair enough, but this kind of question was showing up on Facebook a lot last year (and I think it must have been in the Random Thoughts thread or something here because it's not in this thread) and it seems like the entire point of it was to get people to argue about what was "right" or "intuitive".

It is definitely true that students are taught some version of PEMDAS (or BOMDAS, or other variations) that either says that multiplication comes before division or that if you have both then you resolve them left-to-right. And that's mostly ok for arithmetic, and it seems to work for addition and subtraction, but it kind of conflicts with the kind of shorthand that we tend to use when writing more complicated algebra, especially when it's compressed onto a single line - normally we'd write something like 14/2(3+4) as a two-line fraction so you can clearly see where everything is. But, and here's the annoying bit, the accepted shorthand of omitting the multiplication sign where you have either parentheses or pronumerals is usually understood to take precedence *over* just about everything else. Like I said earlier, smooshing things together like 2(3+4) or 3x tends to mean that you have a single "unit" that should be calculated on its own before you look at other arithmetic operations happening around them.

Alternatively, you could take the approach used in some forms of arithmetic and logic, which sets out a bunch of legal constructions of these sorts of expressions, usually requiring parentheses around just about everything, and which allows you to remove the parentheses only when the resulting expression remains unambiguous. So the "formalest" way of writing the expression might be (14/(2(3+4))), or (18÷(3x)), and the rules of the system would tell you that you can remove the outermost parentheses, but no more because the expression becomes ambiguous at that point. Or you can learn Swedish notation, which removes the ambiguity in another way.

Thanks, this is basically the kind of answer I wanted.  Probably it can go either way, and it doesn't come up very often because people know better than to write things like that anyway.

I do agree that there's something that feels intuitively right about putting 3x together as one unit, but it really goes against my intuition of 3x being identical to 3*x.

---

Anyway, I feel bad about starting another one of these discussions, so here's a fun little problem to (hopefully) make up for it.

You have an hourglass that measures 4 minutes and an hourglass that measures 7 minutes.  How do you measure exactly 9 minutes (using only these hourglasses)?

Clarification 1: The nine minutes must be all consecutive (for example, you can't measure 3 minutes, take a break, then measure out 6 more minutes).
Clarification 2: The nine minutes starts the instant you have the hourglasses (which both start out with all the sand on one side), so you can't get things set up beforehand.

[sudgy]

You can just do the 4 minute one until there's an equal amount on both sides, then do the seven minute one, right?

[scott_pilgrim]

You can just do the 4 minute one until there's an equal amount on both sides, then do the seven minute one, right?

You can't tell when they're equal.  Maybe the radius of the top half is bigger than the radius of the bottom half.

[skip wooznum]

start with both hourglasses flowing, or whatever the word is. After four minutes, turn the 4 minute hourglass over. After seven minutes, turn the 7 minute hourglass over. After eight minutes turn the 7 minute hourglass over.

[skip wooznum]

I've been wondering something myself, maybe the guys around here can help me out with it.

Is everyone familiar with the Monty Hall problem? To review: You're on a game show. Let's make a deal. You must choose one of three curtains and you win what is behind that curtain. One curtain is hiding a brand new converible, the other two are hiding toasters. You pick curtain number one. Monty, the host, says "wow, great choice! But I'll tell you what, I'll show you what's behind curtain number two." He opens up curtain number two and it's a toaster. He then asks you if you want to switch to curtain number three or stay at curtain number one. What do you do? (Monty always opens up a curtain regardless of whether or not you picked the correct one originally. He never opens the one you picked, and he never opens the one with the car.) The answer is you switch of course, because it's twice as likely that it's in curtain number three.

My question: There are 100 curtains, one with a car, 99 with toasters. You choose curtain number 1, he shows you that curtains 4-100 all are hiding toasters. So you now obviously should switch to curtain two or three. Let's say you switch to curtain two. Now he shows you that curtain three is also hiding a toaster and gives you the option to switch again. Do you go back to curtain one or stay at curtain two?

My understanding, btw, is you should switch back to curtain one for 50.5% chance of winning

[qmech]

Not switching is correct the second time.  Curtain 1 still only has a 1/100 chance of being the car, so curtain 2 has the car almost every time.

[terminalCopper]

I think qmech is right.

However, under certain assumptions the first switch  to curtain two was already horrible.
if we assume that Monty has to leave us a final choice after opening all but two curtains, we should simply stick to Curtain 1 until then, and we are guaranteed a 99% chance by switching at the last occasion. If we switch immediatly, Monty might open Curtain 1, leaving us with a 50-50 chance between Curtain 2 and Curtain 3. That would have been horrible.

[skip wooznum]

Not switching is correct the second time.  Curtain 1 still only has a 1/100 chance of being the car, so curtain 2 has the car almost every time.

why is this true?

[skip wooznum]

I think qmech is right.

However, under certain assumptions the first switch  to curtain two was already horrible.
if we assume that Monty has to leave us a final choice after opening all but two curtains, we should simply stick to Curtain 1 until then, and we are guaranteed a 99% chance by switching at the last occasion. If we switch immediatly, Monty might open Curtain 1, leaving us with a 50-50 chance between Curtain 2 and Curtain 3. That would have been horrible.

you dont know he's going to give you another chance to switch

[Watno]

After you choose door 1, there is a 99/100 chance you're wrong
Revealing that 4-100 are wrong makes 2 and 3 have 99/200 to be right. So in 200 cases, 2 will have 1 as the right door, and 99 each have 2 and 3.
Revealing that we can't be in one of the cases where 3 is right means that the probability 2 is right is 99/(99+2)=99/101

[skip wooznum]

After you choose door 1, there is a 99/100 chance you're wrong
Revealing that 4-100 are wrong makes 2 and 3 have 99/200 to be right. So in 200 cases, 2 will have 1 as the right door, and 99 each have 2 and 3.
Revealing that we can't be in one of the cases where 3 is right means that the probability 2 is right is 99/(99+2)=99/101

but out of the 99 cases that have 2 as the right door, most of them would have the host eliminating door 1. In fact, if we apply your reasoning to the original monty hall problem, we'd get an answer of 50/50.

[Watno]

Why do most of the cases where 2 is right have the moderator eliminate 1?

[skip wooznum]

Why do most of the cases where 2 is right have the moderator eliminate 1?

do you mean as opposed to eliminating 1 half the time and 3 half the time? Well because if he only eliminates 1 half of the cases that it's in 2, then when he eliminates 1, you can switch to 3 for 2/3 probability of being correct. And when he eliminates 3, you can stay with 2 for 49.5/51.5, or around 96% probability of being correct. So eliminating door 1 and door 3 equally is a terrible strategy for Monty. His best strategy is one that gives you the same chance of winning no matter what he does. (This was true in the original problem as well. It's just that there the best thing he could do in cases where you happened to guess right would be to open each other door an equal number of times.) That optimal strategy is:
In 20,200 cases, there are 202 cases having door 1 as correct, 9,999 cases having door 2, and 9,999 cases having door 3.  In the 9,999 cases having door 2 (which is the one you chose) he should eliminate door 3 only 198 times (or around 2%) and eliminate door 1 the other 9801 times (or around 98% of the time). If he sticks to this strategy, then whichever door he eliminates, you should switch to the other for a 50.5% probability of being correct. if I'm wrong can someone show me where I went off?

[Watno]

I assumed that the moderator had no motivatio in this whoel thing and just opens a random door out of the possible ones. If he wants you to not win,  I think this gets very complicated, because the optimal strategies you and the moderator then choose depend on each other unless I'm mistaken.

[skip wooznum]

I assumed that the moderator had no motivatio in this whoel thing and just opens a random door out of the possible ones. If he wants you to not win,  I think this gets very complicated, because the optimal strategies you and the moderator then choose depend on each other unless I'm mistaken.

fine so assuming that in situations where he can open either door he will open one at random, id say there's 99/103 chance it's in door number 2 because exactly half of the cases having it door 2 are ruled out.
And I don't see why the optimal strategies would depend on each other.

[TrojH]

I assumed that the moderator had no motivatio in this whoel thing and just opens a random door out of the possible ones. If he wants you to not win,  I think this gets very complicated, because the optimal strategies you and the moderator then choose depend on each other unless I'm mistaken.

fine so assuming that in situations where he can open either door he will open one at random, id say there's 99/103 chance it's in door number 2 because exactly half of the cases having it door 2 are ruled out.
And I don't see why the optimal strategies would depend on each other.

skip, under the usual assumptions that the host never opens your current door or the door with the car, and otherwise has an equal probability of choosing each door... I would say that this final answer of yours is correct.

(Take what I say with a grain of salt, though. I'm just a math prof.  )

[skip wooznum]

I assumed that the moderator had no motivatio in this whoel thing and just opens a random door out of the possible ones. If he wants you to not win,  I think this gets very complicated, because the optimal strategies you and the moderator then choose depend on each other unless I'm mistaken.

fine so assuming that in situations where he can open either door he will open one at random, id say there's 99/103 chance it's in door number 2 because exactly half of the cases having it door 2 are ruled out.
And I don't see why the optimal strategies would depend on each other.

skip, under the usual assumptions that the host never opens your current door or the door with the car, and otherwise has an equal probability of choosing each door... I would say that this final answer of yours is correct.

(Take what I say with a grain of salt, though. I'm just a math prof.  )

thanks. And what about if the host doesn't have equal probability of choosing each door, what if the host is trying to win?

[TrojH]

If the host can play strategically, then you'll have to be more specific about what the host is allowed to do. If you initially pick the wrong door, can the host just open that door and show that you've lost? Then you only have a 1/100 chance to win, and you might as well stick with your original door no matter what the host does.

[skip wooznum]

If the host can play strategically, then you'll have to be more specific about what the host is allowed to do. If you initially pick the wrong door, can the host just open that door and show that you've lost? Then you only have a 1/100 chance to win, and you might as well stick with your original door no matter what the host does.

the host must open doors and allow you to switch. But he does not have to  choose which door to open in a completely random fashion. So in the situation at hand, he can for example, open door one 2% of the times it's in 2, and open door three 98% of the time it's in 2, which is roughly what I'd imagine he'd do.

[singletee]

I've been wondering something myself, maybe the guys around here can help me out with it.

Is everyone familiar with the Monty Hall problem? To review: You're on a game show. Let's make a deal. You must choose one of three curtains and you win what is behind that curtain. One curtain is hiding a brand new converible, the other two are hiding toasters. You pick curtain number one. Monty, the host, says "wow, great choice! But I'll tell you what, I'll show you what's behind curtain number two." He opens up curtain number two and it's a toaster. He then asks you if you want to switch to curtain number three or stay at curtain number one. What do you do? (Monty always opens up a curtain regardless of whether or not you picked the correct one originally. He never opens the one you picked, and he never opens the one with the car.) The answer is you switch of course, because it's twice as likely that it's in curtain number three.

My question: There are 100 curtains, one with a car, 99 with toasters. You choose curtain number 1, he shows you that curtains 4-100 all are hiding toasters. So you now obviously should switch to curtain two or three. Let's say you switch to curtain two. Now he shows you that curtain three is also hiding a toaster and gives you the option to switch again. Do you go back to curtain one or stay at curtain two?

My understanding, btw, is you should switch back to curtain one for 50.5% chance of winning

The problem with your formulation is that it does not unambiguously determine the algorithm the host is using. Neither does the formulation of the original problem, but it is implicitly understood to be the following by virtue of the fact that it is how the real-life game show worked:

1. Contestant chooses a door.
2. Host chooses a door at random that the contestant has not chosen and that contains a goat. Host reveals this door.
3. Contestant is given a choice to switch or stay.

Now, for your new question, the algorithm could be either of the following:
A.
1. Contestant chooses a door.
2. Host chooses 97 random doors that the contestant has not chosen and that contains toasters. Host reveals them.
3. Contestant chooses to switch.
4. Host chooses a random door that contestant has not currently chosen and that contains a toaster. Host reveals it.
5. Contestant chooses to switch or stay once more.
In this case, you should switch back to door number 1, with a 50.5% chance of winning.

B.
1. Contestant chooses a door.
2. Host chooses 97 random doors that the contestant has not chosen and that contains toasters. Host reveals them.
3. Contestant chooses to switch.
4. Host chooses a random door that contestant has never chosen this game and that contains a toaster. Host reveals it.
5. Contestant chooses to switch or stay once more.
In this case, you should stay with door number 2, with a 99% chance of winning.

[skip wooznum]

I've been wondering something myself, maybe the guys around here can help me out with it.

Is everyone familiar with the Monty Hall problem? To review: You're on a game show. Let's make a deal. You must choose one of three curtains and you win what is behind that curtain. One curtain is hiding a brand new converible, the other two are hiding toasters. You pick curtain number one. Monty, the host, says "wow, great choice! But I'll tell you what, I'll show you what's behind curtain number two." He opens up curtain number two and it's a toaster. He then asks you if you want to switch to curtain number three or stay at curtain number one. What do you do? (Monty always opens up a curtain regardless of whether or not you picked the correct one originally. He never opens the one you picked, and he never opens the one with the car.) The answer is you switch of course, because it's twice as likely that it's in curtain number three.

My question: There are 100 curtains, one with a car, 99 with toasters. You choose curtain number 1, he shows you that curtains 4-100 all are hiding toasters. So you now obviously should switch to curtain two or three. Let's say you switch to curtain two. Now he shows you that curtain three is also hiding a toaster and gives you the option to switch again. Do you go back to curtain one or stay at curtain two?

My understanding, btw, is you should switch back to curtain one for 50.5% chance of winning

The problem with your formulation is that it does not unambiguously determine the algorithm the host is using. Neither does the formulation of the original problem, but it is implicitly understood to be the following by virtue of the fact that it is how the real-life game show worked:

1. Contestant chooses a door.
2. Host chooses a door at random that the contestant has not chosen and that contains a goat. Host reveals this door.
3. Contestant is given a choice to switch or stay.

Now, for your new question, the algorithm could be either of the following:
A.
1. Contestant chooses a door.
2. Host chooses 97 random doors that the contestant has not chosen and that contains toasters. Host reveals them.
3. Contestant chooses to switch.
4. Host chooses a random door that contestant has not currently chosen and that contains a toaster. Host reveals it.
5. Contestant chooses to switch or stay once more.
In this case, you should switch back to door number 1, with a 50.5% chance of winning.

B.
1. Contestant chooses a door.
2. Host chooses 97 random doors that the contestant has not chosen and that contains toasters. Host reveals them.
3. Contestant chooses to switch.
4. Host chooses a random door that contestant has never chosen this game and that contains a toaster. Host reveals it.
5. Contestant chooses to switch or stay once more.
In this case, you should stay with door number 2, with a 99% chance of winning.

it's certainly not option B. The two options in my mind for step 4 are:
A. Host chooses a completely random door that isn't the contestant's current door or the car door.I'd say this is 99/103 and TrojH concurs.

B. Host chooses a door that isn't the contestant's current door or the car door but he doesn't choose equally between  the available doors, rather he chooses with a modified rate that will maximize his chances of winning. I'd say this is 50.5%

And in your option B, I'd say the odds are 99/101, not 99/100, but that's neither here nor there.

[qmech]

It doesn't matter what strategy the host uses, as long as he follows the rule that the prize is always behind one of the

Not switching is correct the second time.  Curtain 1 still only has a 1/100 chance of being the car, so curtain 2 has the car almost every time.

why is this true?

It's very easy to get very confused by this sort of problem.  As I write this I'm not sure whether my or Watno's number is correct.  I think terminalCopper's observation might have something to do with explaining the difference.

PPE: will read 4 new replies now.