Open-Ended Math Questions?

[Francesca R]

What are the chances that the path pointed to goes to the beach?

1 in 3.

If the second native gives the same direction as the first, you know that they are either both lying [L.L] or both telling the truth [T.T]. Given [L.?] the probability of [L.L] is 2 in 3. Given [T.?] the probability of [T.T] is 1 in 3. There are no other possibilities. Therefore your probability is 1 in 3

 

[DeviousB]

Ah yes, everyone confidently giving different answers. That's why it reminds me of the MH problem.

 

[Francesca R]

So spill the beans . . .?

(The first two answers were "confidence-qualified" with their question marks. I'm risking going out on a limb)

 

[DeviousB]

"You are visiting an island far away. On this island the natives are twice as likely to tell you lie as speak the truth (they don't like foreigners much)."

Okay, so they either tell the truth or a lie, with a lie being twice as likely. I.e. P(T) = 1/3, P(L) = 2/3.

So there are 4 possible ways natives A and B could answer.

1) P(A:T,B:T) = 1/3 * 1/3 = 1/9;
A will say a direction, B will agree.

2) P(A:T,B:L) = 1/3 * 2/3 = 2/9;
A will say a direction, B will disagree.

3) P(A:L,B:T) = 2/3 * 2/3 = 2/9;
A will say a direction, B will disagree.

4) P(A:L,B:L) = 2/3 * 2/3 = 4/9;
A will say a direction, B will agree.

But we also know that B did agree with A, so we can only be dealing with scenarios (1) and (4).

So the probability they are telling the truth is

P(A:T,B:T) / (P(A:T,B:T) + P(A:L,B:L))

= (1/9) / (1/9 + 4/9) = (1/9) / (5/9) = 1/5.

--

Let the arguments comence...

 

[donq]

When presented with a problem you have to be able to make basic assumptions or else it has to be defined to the point of absurdity. The question, for me, is straight forward, but to younger minds multistep problems can be a challenge. It is easy for a 10 year old to forget about the first $10 gained when he or she computes the second transaction. Talking about inflation or other such subtleties would not only confuse a child but be a ridiculous expectation.

Strictly speaking, an open ended problem isn't defined as having more than one answer (although it is certainly possible). It is defined as having multiple solutions. How you derive at your answer can vary. Open ended problems must be defended in essay form - that is their function. Adding variables or conditions not originally addressed might change the “correct answer,” but you would have to defend your arguments with logic and solid math.

Any other answer other than $20 to the above questions is probably just a semantics game. However, there are several ways you could explain how you figured that answer out. Especially if you are young and just grappling with this new language you are learning called math.

 

[Francesca R]

Let the arguments comence...

No argument—seems right to me!

 

[drkitten]

"You are visiting an island far away. On this island the natives are twice as likely to tell you lie as speak the truth (they don't like foreigners much)."

Okay, so they either tell the truth or a lie, with a lie being twice as likely. I.e. P(T) = 1/3, P(L) = 2/3.

So there are 4 possible ways natives A and B could answer.

1) P(A:T,B:T) = 1/3 * 1/3 = 1/9;
A will say a direction, B will agree.

2) P(A:T,B:L) = 1/3 * 2/3 = 2/9;
A will say a direction, B will disagree.

3) P(A:L,B:T) = 2/3 * 2/3 = 2/9;
A will say a direction, B will disagree.

Minor typographic error:

3) P(A:L,B:T) = 2/3 * 1/3 = 2/9;

4) P(A:L,B:L) = 2/3 * 2/3 = 4/9;
A will say a direction, B will agree.

But we also know that B did agree with A, so we can only be dealing with scenarios (1) and (4).

So the probability they are telling the truth is

P(A:T,B:T) / (P(A:T,B:T) + P(A:L,B:L))

= (1/9) / (1/9 + 4/9) = (1/9) / (5/9) = 1/5

Is the (implicit) assumption of independence justified? Could B have heard A's answer and thought to himself "Yeah, that's a good idea, let's lie!"

 

[DeviousB]

Minor typographic error:

3) P(A:L,B:T) = 2/3 * 1/3 = 2/9;

Thanks, didn't notice that.

Is the (implicit) assumption of independence justified? Could B have heard A's answer and thought to himself "Yeah, that's a good idea, let's lie!"

In RL possibly not, but I can't see how to reconcile an increased P(L) for B if A lies with an overall probability of B lying of 1/3, which is a condition of the problem.

 

[69dodge]

[...]

= (1/9) / (1/9 + 4/9) = (1/9) / (5/9) = 1/5.

--

Let the arguments comence...

No argument from me either, assuming independence of A's and B's answers.

In RL possibly not, but I can't see how to reconcile an increased P(L) for B if A lies with an overall probability of B lying of 1/3, which is a condition of the problem.

B might sometimes decide to agree with A when he tells the truth too, which could even things out. To take an extreme example, B might always agree with A. Then, obviously, if A tells the truth 1/3 of the time, so will B. Yet in that case, the answer to the original question is 1/3, not 1/5.

 

[69dodge]

1 in 3.

If the second native gives the same direction as the first, you know that they are either both lying [L.L] or both telling the truth [T.T]. Given [L.?] the probability of [L.L] is 2 in 3. Given [T.?] the probability of [T.T] is 1 in 3. There are no other possibilities. Therefore your probability is 1 in 3

I know you've already agreed with the right answer, but maybe it would still help to point out what was wrong with your original reasoning. Namely, you didn't take into account the fact that your two possible "givens" don't have the same prior probability as each other. Before we know what the question mark is, [L.?] already has a probability of 2/3 while [T.?] has a probability of only 1/3. This reduces the probability of [T.T] vs. [L.L], compared to the answer you got.

 

[Francesca R]

Holy Jee3us you sure know about kicking a girl when she's down!

[only kidding]

 

[DeviousB]

B might sometimes decide to agree with A when he tells the truth too, which could even things out. To take an extreme example, B might always agree with A. Then, obviously, if A tells the truth 1/3 of the time, so will B. Yet in that case, the answer to the original question is 1/3, not 1/5.

In fact, if B always agreed when A told the truth, he would tell the truth 1/3 of the time, hence in order to lie twice as often he would necessarily always have to lie when A lied.

Which raises another interesting problem. Everything is as before, except B and A are old friends, so when A lies B doesn't like to contradict him, and only does so half the time. What now is the probability that A pointed out the right path?

 

[Beleth]

Which raises another interesting problem. Everything is as before, except B and A are old friends, so when A lies B doesn't like to contradict him, and only does so half the time. What now is the probability that A pointed out the right path?

I am assuming that when A tells the truth, B doesn't mind contradicting him, since you didn't say otherwise.

P(A:T, B:T) = 1/3 * 1/3 = 1/9

P(A:T, B:L) = 1/3 * 2/3 = 2/9

P(A:L, B:T) = 2/3 * 1/2 = 2/6 = 1/3

P(A:L, B:L) = 2/3 * 1/2 = 1/3

(Sanity check: 1/9 + 2/9 + 1/3 + 1/3 = 1. Good.)

P(A:T, B:T) / (P(A:T, B:T) + P(A:L, B:L)) = (1/9) / ((1/9 + 1/3) = (1/9) / (4/9) = 1/4.

 

[69dodge]

Holy Jee3us you sure know about kicking a girl when she's down!

[only kidding]

My apologies. No offense was intended.

Of course, had I known you were a girl, I'd have been more careful . . .

 

[69dodge]

Which raises another interesting problem. Everything is as before, except B and A are old friends, so when A lies B doesn't like to contradict him, and only does so half the time. What now is the probability that A pointed out the right path?

Interesting.

I interpreted "everything is as before, except..." differently from how Beleth did, so I got a different answer from his.

I figured that in the new problem, B still tells the truth 1/3 of the time, and lies 2/3 of the time, overall; the only difference from the original problem being that his choice of when to do which is no longer independent of A's.

So if A lies 2/3 of the time, and B lies half of that 2/3, then B tells the truth the other half of the 2/3, which is 1/3 of the total time. But this uses up all his truth-telling. So, whenever A tells the truth, B has to lie, thereby disagreeing with A. So, if the two agree, A can't be telling the truth.

So, the answer is, if they agree, the probability is 1 that they're both lying.

However, it's odd to describe this new version of the problem by saying, "B and A are old friends, so when A lies B doesn't like to contradict him", because in this version, when A lies, B contradicts him more often than in the original version where their choices were independent---namely, half the time instead of only a third of the time.

 

[DeviousB]

Well done 69dodge.

My next question was going to be "What is wrong with B's 'friendly' strategy?" but you got that one too.

Care to answer my next three questions now? Or are you too busy filling in your application to the MDC?
;-)

 

[Francesca R]

Of course, had I known you were a girl, I'd have been more careful . . .

Cool. My post on that topic (which is kinda gender implicit) is #46

 

[DeviousB]

My next question was going to be "What is wrong with B's 'friendly' strategy?" but you got that one too.

Well, nearly.

When independent, B contradicts A 2/3rds of the time when A tells the truth, 1/3rd of the time when A lies. I.e. (2/3 * 1/3) + (1/3 * 2/3) = 4/9ths of the time overall.

By following his 'strategy' (while keeping within the rules), B now contradicts A one half the time when A lies, and all the time when A tells the truth. I.e. (1/2 * 2/3) + (1 * 1/3) = 2/3rds (6/9ths) of the time overall.

 

[69dodge]

Cool. My post on that topic (which is kinda gender implicit) is #46

Ok, now let me get this straight here, once and for all. What you're saying is, you expect me to pay attention to what I read? Is that it? Huh? Huh?

...

Oh. Wait. That's quite reasonable, actually. Hmph.

 

[Ben Tilly]

Well here are a few

First of all an amusing piece of trivia. In the 50's in a random sample of 5th grade teachers, more than half could not correctly decide whether 2/3 is larger than 3/5. (1) Based on anecdotal evidence, I don't think that teachers today would do any better than they did 50 years ago. So innumeracy is not a shock for the general public.

However here is a simple but truly ambiguous problem.

An advanced being shows you two boxes. In box A there is $1000. In box B there is $0 or $1000000. You are allowed to take either box B or both boxes, but the being will put the million in box B only if it thinks you will only take box B. Assuming that you have never seen the entity make a mistake on this problem, what should you do to make as much money as you can?

This is called Newcomb's problem. Most people find the answer obvious. They just disagree on what the answer is. In fact the answer is not well-defined. (For the record I'd only take box B.)

Here is a stupid one. Which weighs more? A pound of gold or a pound of lead? (2)

Here is a tricky one. A boat is in a lake. A man in the boat takes a big piece of lead out and drops it overboard. Does the water in the lake go up or down? Be warned that most Harvard graduates got it wrong. (3)

Not a highschool problem, but here is the strangest math result that I know. Suppose you have 2 envelopes with numbers in them that are different. We will perform the following experiment. You randomly hand me one of your envelopes, I open it, and hand it back to you. Would you believe that I now have sufficient information to, with guaranteed better than even odds, to correctly tell you that you gave me the larger number? Note that my odds of being correct depend on my technique and what both of your numbers are. I cannot, therefore, figure out my odds. All that I can do is prove that they are better than 50%. Be warned, the answer is significantly worse than the problem! (4)

That should do for a start.

Cheers,
Ben

(1) 2/3 = 0.66666... > 0.6 = 3/5

(2) Lead, because gold is always measured in Troy weight.

(3) When the lead is in the boat it displaces its weight of water. When it is in the water it displaces its volume of water. Therefore after it is thrown overboard it displaces less water and so the lake goes down.

(4) My algorithm is to randomly pick a number with a method that has some chance of landing in any interval, pretend it is the number that you didn't give me, and pick accordingly. My odds of being correct will be 50% + the odds that I picked a number between your two numbers, which is always greater than 50%.

You can demonstrate this using algebra, or else with a picture. Most people find the picture clearer. Draw the line, and mark your 2 numbers as x and y (with x < y). Draw an arrow from x to the right. If I'm given x, this is the range I have to pick my number in to be correct. Draw an arrow from y to the left. If I'm given y, this is the range that I have to pick my number in to be correct. If I pick my number in a section where there is only one arrow, I have even odds of being right. If I pick in the overlap, I'm right. Since I always have a chance of picking in the overlap, my odds are better than even.

You can find an excellent explanation of this at spookymathexplained.notlong.com (you'll have to add http, I can't post links).