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Do you have a question you would like to ask Professor Puzzler? Click here to ask your question! Or click the following link if you are looking for our Vortex Based Math expose.Veronica from Georgia asks, "A guy wants to buy a car and borrow $50,000. He gets $25,000 from his mom and $25,000 from his sister. He buys the car for $45,000. Driving home he runs into a friend that ask to borrow $3,000. You are now left with $2,000. You pay your mom and sister back $1,000 each and now you owe them only $24,000 each. After you pay them the $48,000 total to both, your friend pays you the $3000 he owes you. Do you know why the total comes out to $51,000 instead of $50,000 that you originally started out with?"
Hi Vernoica, This problem is very similar to the bellboy "missing dollar" problem, which you can read about by clicking the image below:
I'm going to use a slightly different approach in explaining this problem, by focusing on the meaning of positives and negatives in the context of transactions. Every financial transaction has a "direction." It is from one person (or organization/business, etc) and to another. We use positives and negatives to indicate the direction of a transaction. We should choose a single individual as the focus for our problem, and once we've done that, every transaction will be signed based on whether the transaction is TO or FROM that person. In this case, the choice is obvious: the guy buying the car is the focus.
So let's agree that if a monetary transaction has "the guy" as the recipient, we'll call that transaction positive. If "the guy" is the giver, we'll call it a negative transaction. So we can list every transaction as a signed number:
Recieves $25,000 from his mother: +$25,000
Receives $25,000 from his sister: +$25,000
Buys car for $45,000: $45,000
Lends $3000: $3,000
Pay mother $1,000: $1,000
Pay sister: $1,000: $1,000
Pay mother $24,000: $24,000
Pay sister: $24,000: $24,000
Friend pays back loan of $3,000: +$3,000
So now that we have all of these transactions signed, we can add them up properly:
$25,000 + $25,000  $45,000  $3,000  $1,000  $1,000  $24,000  $24,000 + $3,000 = $45,000
What does that $45,000 represent? It represents the amount by which "the guy" has decreased his bank balance (the price of the car).
What does the number $51,000 represent? Absolutely nothing. Whoever gave you the problem picked three transactions out of the list of nine transactions, gave one of them the incorrect sign, and then combined them in a way that makes absolutely no sense in the context of the problem. Why would that add up to anything interesting? You got fooled into thinking it should mean something because it was so close to $50,000, which was a number that was mentioned in the problem. But adding those three quantities means absolutely nothing. When you give the numbers the correct signs, all the transactions cancel out to a single transaction  the cost of the car.
Question: In your Gravity Train Simulator, you made the following statement: "If the train is inside the earth, the force of gravity acting on it is related not just to its distance from the center of the earth, but also on the amount of earth mass which is closer to the center of the earth than the train is." This seems to imply that all the mass outside the train's radius somehow cancels out. So if someone was in outerspace, inside an enormous steel shell  does that mean they would be weightless wherever they were in the shell? ~H. Borash
Answer: Yes, that's correct, H. To rigorously prove this, we would need to use a bit of Calculus, but I can show you a rough geometry approximation that'll help you see why it's true. First, a diagram. Here's a spherical shell. We'll assume it's absolutely enormous  big enough to enclose the earth and the sun. This hypothetical structure is called a Dyson sphere. I've marked the sun and the earth inside the sphere.
The yellow circle at the center of the sphere represents the sun, and the smaller blue circle to the left represents the earth. The brown dotted line represents the earth's orbit around the sun.
Obviously, this is not drawn to scale; it's difficult to make both the sun and earth to scale in a single diagram without the earth becoming nearly invisible. So I didn't even try.
But the question you asked is essentially equivalent to this: wouldn't the Dyson sphere throw the earth out of its orbit? That sphere has got to be incredibly massive, and planet earth is much closer to one side than the other. So wouldn't earth get dragged out of its orbit?
Or, to put it another way, if the sun wasn't there at all, and earth was motionless, would it stay motionless? Or would it get dragged toward the closer side of the sphere?
The answer to both questions is: No. No, the sphere would not drag the earth out of its orbit. No, earth would not go crashing into the sphere. All of the gravitational forces applied by the sphere on planet earth cancel each other out. Even though earth is not at the center of the sphere.
The reason is that because the earth is closer to one side, there is more mass on the other side, and the closeness of the mass on one side exactly cancels against the extra mass on the other side!
As I said above, a full, rigorous proof of this involves some Calculus, but I'm going to try to give you a noncalculus explanation that may help you visualize what's happening. I want to focus on two disks of the sphere  one close to the earth, and the other on the opposite side, furthest away from earth. To simplify, we're going to treat these as flat disks, so they don't exactly match the contour of the Dyson sphere:
Notice how much bigger the disk on the right is, compared to the one on the left. That disk is much further away, so each cubic inch of that disk exerts less force on earth than a corresponding cubic inch of the closer disk.
On the other hand, you can easily see that the disk on the right contains many many more cubic inches than the disk on the left.
So maybe they cancel? Let's take a closer look.
If we knew the thickness (h, for height) of the disk, as well as the density (D) of the material the disk is made of, we could calculate the masses (m_{1} and m_{2}) of the two disks.
m_{1} = hπr_{1}^{2}D
m_{2} = hπr_{2}^{2}D
However, we have more variables than we need. Since we have some similar triangles, we can say that r_{2} = r_{1}(x_{2}/x_{1}). This changes our second equation to: m_{2} = hπr_{1}^{2}(x_{2}^{2}/x_{1}^{2})D.
Now this is where the magic happens: we're going to plug both of those masses into the gravitational formula:
So even though one disk is much larger than the other, they both exert the same gravitational force on earth, canceling each other out!
Again, this is not a rigorous proof. Most proofs I've seen of this involve rings instead of disks, and doing integrals (calculus) to show that all the rings cancel. But this geometric demonstration hopefully helps you to see that it's a reasonable conclusion!
Judah from Maine asks for an explanation of how to negate a compound statement.
We'll explore two possibilities. One is that the compound statement contains an "AND" and the other that it includes an "OR." Let's make up one statement for each possibility.
 It's snowing or it's raining
 Christmas is almost here, and the elves are hard at work.
To help you visualize these, I've made Venn Diagrams for each. Let's start with the weather example:
In this example, the red region represents the condition of snow, and the green region reporesents rain. Thus, the greenishyellow region represents both rand and snow, and the blue region (outside the two circles) represents the condition that it is neither raining nor snowing. This covers all the possible conditions (that is  all the conditions we're concerned about here  there are, of course, other weather conditions like sleet, freezing drizzle, etc, but these are not our concern right now). The possible conditions are:

There is neither rain nor snow.
 There is snow, but not rain.
 There is rain, but not snow.
 There is both rain and snow.
Which of those four matches our initial condition? The statement "It's snowing or it's raining" is covered by the red, green, and greenishyellow regions. That's regions 2, 3, and 4. Therefore, the the logical NOT of "It's snowing or it's raining" must contain everything except those three regions. Or, to put it more simply, it has to be only region 1  neither rain nor snow.
So the negation of "It's snowing or it's raining" is "It's not snowing and it's not raining." Interesting, isn't it? When you negated an "or" you got an "and." Let's see what happens with our other example:
In this case, the red region represents Christmas being almost here, the green region is elves hard at work, and the greenishyellow region is both Christmas is almost here and elves are hard at work. The light blue region represents neither of the statements being true.
So which region(s) represent our original statement? "Christmas is almost here, and the elves are hard at work." This is only covered by the overlapping greenishyellow region.
So if we want to negate this one, we need to figure out a way of expressing everything except the greenishyellow region. It's tempting to say that the negative of the statement is "Christmas is not almost here, and elves are not hard at work," but that is only the blue region; it doesn't include the red or the green. So instead, we need to say "Christmas is not almost here, or elves are not hard at work." Using an "or" instead of "and" allows us to include all the regions in the Venn diagram.
So our general rule of thumb is, if we're trying to negate a compound statement, we need to make sure that the original statement and it's negation cover every single region of the diagram, without any overlap.
If you've got that, you can go a little crazy with compound statements:
(It's raining and Christmas is almost here) or (it's snowing or the elves are hard at work).
Can you negate that statement?
Yesterday, I answered a question about the Monty Hall ThreeDoor Game, which you can read about in the previous blog post. After posting the article, I shared it on social media, and commented that talking about this problem feels like wading into a murky swamp, because everyone brings their own assumptions into the problem, and it's tough to guess what those assumptions are.
But I realized, too, that it's not just this problem; it's probability in general. I love probability, and I hate probability. Whenever a district/county/state asks me to write competition math problems for their league or math meet, I know that I need to give some probability problems  everyone expects it! But there's no kind of math problem that I'm more afraid that I'll mess up. I'm grateful for a proofreader to give my problems a second pair of eyes (although my proofreader shares my feelings about probability). More often than not, I'll figure out a way to write a program to function as an electronic simulation of my problem, to verify empirically that I have arrived at the right solution.
Anyway, one of the reasons that probability feels so murky to me is that gut reactions can lead you astray. The ThreeDoor game is a prime example of how those gut reactions can mess you up. Monty asks you to pick a door, knowing that one of those doors has a prize behind it, and the other two have nothing. After you pick, Monty opens one of the other doors to show you that it's empty, and asks you if you'd like to keep your original guess, or switch to the other unopened door.
Most people's gut reaction is that it doesn't matter whether you keep your original guess or switch to a different guess. This gut reaction is wrong, as you can verify by playing the simulation I built here: The Monty Hall Simulation. For those who are still struggling with this, I'd like to offer a couple different ways of looking at the problem.
The "Not" Probability
We tend to focus on the probability of guessing correctly the first time. Instead, let's focus on the probability of NOT guessing correctly. If you picked door A, then the probability that you were correct is 1/3. This means that the probability you did not guess correctly is 2/3. But really, what is that? It's the probability that either door B or door C is correct.
So if P_{B} is the probability that door B holds the prize, and P_{C} is the probability for door C, we can write the following equation:
P_{B} + P_{C} = 2/3.
Now Monty opens one of those two doors (we'll say C), that he knows is empty. This action tells you absolutely nothing about the door you opened, but it does tell you something about door C  it tells you that P_{C} = 0.
Since P_{B} + P_{C} = 2/3, and P_{C} = 0, we can conclude that P_{B} = 2/3. This is exactly the result which the simulation linked above gives.
Two Doors vs. One Door Choice
Related to the above way of looking at it, try looking at it with a set of slightly modified rules:
 You pick a door
 Monty says to you: "I'll let you keep your guess, or I'll let you switch your guess to both of the other two doors."
Under these circumstances, of course you're going to switch. Why? Because if you keep your guess, you only have one door, but the choice Monty is offering you is to have two doors, so your probability of winning is twice as great. In a sense, that's actually what you're doing in the real game, even though it doesn't appear that way. You're choosing two doors over one, and the fact that Monty knows which one of those two doors is empty (and can even show you that one is empty) doesn't change the fact that you're better off having two doors than one.
Before getting to the question for this morning, I'd like to direct your attention to our simulation of a game show with three doors. It's called the "Monty Hall Game." Understanding this game will be important to thinking about Tracie's question below.
The short summary of the game: There are three doors, and only one of them has a prize behind it. You pick a door. Then Monty opens another door to show you that it's empty, and asks you if you want to switch your guess. What do you do?
Tracie from South Dakota asks, "So if the game show host opened 2 doors for me, would I still have 1/3 probability of winning?"
Hi Tracie, your question highlights a couple of the things about the Monty Hall game that often causes confusion.
 Monty Hall knows in advance where the pot of gold (or new car, or whatever) is. The rules of the game are that he opens an empty door. But if he always opens an empty door, that means he knows where the empty doors are. That's an important concept to understand. He's not being entirely random in his choice of doors to open. There are two empty doors, so if you pick one of them, he picks the other. The only time he's being random is when you pick correctly; in that circumstance he randomly picks one of the empty doors to open.
 This is not a static problem  the circumstances change, and because the circumstances change, the probabilities could change as well. Let me give you an example. Suppose that during Christmas vacation, we decide we want to take our kids sledding. Since I'm not teaching that week, I can pick any day of the week to go. So we would say that my probability of picking Tuesday is 1/7 (there are seven days in a week, and Tuesday is one of them). Now suppose that my wife looks at a weather forecast, and says, "Oh! Thursday, Friday and Saturday are supposed to be cold, with a wind chill of 20 F!" If I use that information in making my choice, what is my probability of picking Tuesday now? It's 1/4, because we've eliminated 3 of the 7 days. That doesn't mean the original 1/7 is wrong  it means that the conditions of the problem have been changed, and so we have to calculate a new probability. This is not, by the way, a perfect analogy to the Monty Hall problem, so please don't try to make it match. The differences are:
 We're talking about the probability that I'll pick a certain day, not the probability that the day I pick is a good one.
 We don't know that there's only one good sledding day.
 My wife is not deviously hiding known information from me.
 Weather forecasts are not 100% accurate anyway.
 There is no #5, but at the end of this post, I'll  just for fun  turn this example into a better analogy for the Monty Hall Game.
Okay, so how does this relate to your question? Number two should be fairly obvious; we have a change in conditions, so there's no reason to assume that the probability will stay the same. The probability of you winning WAS 1/3, but the changing conditions mean we have to recalculate the probability.
But here's the more important issue. Monty Hall can't play the game the way you suggest. Why not? Because if the rules of the game are "Monty will open two empty doors," Monty is going to run into trouble every time you don't pick the right door. Because if you pick the wrong door, how many empty doors are there left for Monty to open? Only one! If you pick one of the empty doors, then there's only one other empty door for him to open.
If he opens two doors, that means you've picked the correct door. So even though your original choice was 1/3 probability of winning, under the new circumstances, you have a probability of 1 (100% chance) of winning.
If you're wondering how that number 1/3 fits into the solution it fits like this: The probability that Monty can open two doors is 1/3 (the same as the probability that you selected correctly). So the number 1/3 is still in there  just in a different place!
If you're wondering why the probability changes in this case, but doesn't when he opens one door, the answer is this: When he opens just one door, he has not given you any information about the door you opened. When he opens two doors, he has (indirectly) given you information about your door.
To add to this, the only way Monty could do the twodoor rule would be to have to have a pair of rules:
 If you pick the right door, he will open two doors. (this happens 1/3 of the time)
 If you pick the wrong door, he will open one door. (this happens 2/3 of the time)
The problem with this is: if you know the rules, you can do an alwayswinning strategy. If he opens two doors, you've automatically won, and if he only opens one door, that means you picked the wrong door, so you must swap. Monty would DEFINITELY not want to play by those rules!
Addendum: If you would like to look at a couple different ways of understanding the Monty Hall problem, you can find more writeup here: The Murky Swamp of Probability.
Sledding in December. Let's say my wife looked at the weather forecast and told me, "Every day but one this week is going to have a wind chill of 20 F, and there will be one day that's going to be sunny and warm. So, randomly pick a day for sledding." (My wife hates the cold, so she would definitely not do something as ridiculous as that!) We'll make the wild assumption for this example that the National Weather Service has 100% predictive accuracy.
So I pick Tuesday. I have a oneinseven chance of "winning," based on the information I have. Of course, my wife has a different perspective, because she has more information than I do. She knows with 100% certainty whether I've picked correctly or not!
Now my wife says, "Okay, I'll tell you that the following days are going to be 20 F: Sunday, Monday, Thursday, Friday, Saturday."
She's narrowed my choices down to two possibilities. But it's important to remember that this was not a completely random choice. She didn't pick 5 out of 7 days to eliminate; she picked 5 out of 6! She couldn't eliminate Tuesday, because that was the day I had picked, and if she eliminated it, I would be forced to change.
My reasoning now goes as follows: Based on my original information I was given, Tuesday had a 1/7 probability of being the best day for sledding. That means there was a 6/7 probability that one of the other days was the good sledding day. So there's a 6/7 probability that the good sledding day is: Sunday, Monday, Wednesday, Thursday, Friday or Saturday. As a probability equation, I could write:
P_{su }+ P_{m }+ P_{w }+ P_{th }+ P_{f }+ P_{sa }= 6/7
But now my wife has changed the problem. She's changed it by giving me more information: She's told me the values of all but one of those probabilities is zero!
That means: 0 + 0 + P_{w} + 0 + 0 + 0 = 6/7, which means that P_{w} = 6/7!
So do I switch my decision to Wednesday? You bet I do! The key to understanding this is that my wife's choice of information to give me is not random, and it therefore significantly alters the shape of the problem.
I hope that this is a helpful answer. The Monty Hall problem is one that repeatedly trips people up. I wonder if my blog post will increase the probability of people not getting tripped up by it!
PS  if you go to the Monty Hall game and change the game settings to 7 doors, 5 hints, that will be the same as my sledding example. Turn on the automation, and watch the percent. If you let it run long enough, you'll see that it eventually settles down to about 85.7%, which is the 6/7 probability we calculated.