The Solution to The Missing Dollar Paradox, The Monty Hall Paradox, and Other Famous Paradoxes

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What is a paradox? Simply put, a paradox is an argument whose conclusion cannot exist because of the prevailing conditions set forth in the argument but does exist because of those same conditions. While not all paradoxes have Earth-shattering implications, many can have interesting implications for our lives. If anything else, they provide a new viewpoint and perhaps spark the inquisitive nature of the mind.

The Missing Dollar Paradox

I remember being told this one several times in my youth and it got me every time. Let's see if it gets you too. Three people go to a hotel and check out a room for $30. As they go up the elevator, the desk clerk realizes he made an error and should have charged them only $25. He sends a bellhop up to give the people back the $5 they were overcharged. But as the elevator goes up, the bellhop realizes that they don't know they are getting a refund and pockets $2 of the dollars for himself. Upon arriving at the hotel door, he explains the supposed $3 overcharge and hands the money over. That means that each person paid just $9, for a total of $27. But the bellhop has $2, giving us a total of $29. Where is that missing dollar?

The answer lies in number sense: we shouldn't do 27+2 because we actually did 27-2. That is because we had 30 and took away a total of 5 but the people are only away of the 3 being removed from the toal, yet the wording of the problem has us thinking that we had 27 and put 2 back into this. We really have a new total in the 27 and we got to that by adding two to the 25 that was supposed to be charged to them. I know this explanation seems fishy, but look over it again. It is all a matter of properly translating the situation (Al 3).

The Monty Hall Paradox

This paradox was known before Monty Hall became associated with it on his game show Let’s Make a Deal, but it has such a counterintuitive result that even the Mythbusters tackled it! Originally called The Three Prisoners Problem by Martin Gardner, it was first published in the “Mathematical Games” Column of Scientific American in 1959. The version we will discuss here is the one many were exposed to: the basic setup from the game show (Al 9)

Imagine that you are a contestant on a game show and the host presents to you three doors. Behind one of those doors ia a fabulous prize but behind two other ones is nothing. The host asks you to pick one of the doors, which you do, then reveals an empty door to you. The host will now offer you the ability to stay with your pick or switch to the other remaining door. Most people think that it shouldn’t matter or that it is just a 50/50 shot and therefore stick with their original choice. But as it turns out, by switching doors you have a 2/3 chance of winning! How is that possible?

The key is in the breakdown of the choices and the outcomes you have. If you picked door 1 and did not switch, then one of three things could happen. If the prize is behind 1, then you win. If the prize is behind 2, then the host revealed 3 and you lost with 1. If the prize is behind 3, then the host revealed 2 and you lost with 1. Therefore, you have a 1/3 shot at winning. But what if you pick door 1 and then switched? If it was indeed behind 1, then shucks, you lost. If it was behind 2, you switched from 1 and the host revealed 3, so you won. If it was behind 3, you switched from 1 and the host revealed 2, so you won. It is indeed not a ½ chance but a 2/3 chance. Keep that in mind next time you play such a game! (Al 16-7)

Russell's Barber Paradox

This one was developed by the famous mathematician Bertrand Russel, whose work in logic and set theory is far-wide and still in use to this day. He came up with this paradox to describe an interesting facet of set theory.

Imagine a town full of people who need a haircut. No one is allowed to cut their own hair but fortunately a barber is in the town. He does in fact cut everyone's hair as promised, but now he needs a haircut, for is a member of the town and they all need a haircut. But he cannot cut his own hair, for no one is allowed to cut their own hair. What does he do?

In this example, something is a member of a set but by being a member of that set cannot exist in it. If a special set for barbers was made then he would be fine but since he is concluded in the town then he cannot violate that rule.

Bertrand's Box Paradox

This a paradox that I frequently see given to children, so one may think it’s a breeze. But pay close attention. Imagine you have three boxes and each has 2 objects in it. All the boxes are sealed, and one has two object As in it, another has two object Bs in it and the final one has an object A and an object B in it. The probability of picking the AB box is simply 1/3, as expected. But what if I pick a box at random and only look at one of the things inside? What is the chance that I have indeed picked the AB box? Well, if it is A then you don’t have the BB box, so a ½ chance. Similarly, if you picked a B then its not the AA box and so also a ½ chance. How can knowing one fact change the chance that it was 1/3 and now is ½? After all, I still have three options left, right? Well, think about it this way: there are 3 A and 3 B choices out there. If you pick one of those three, there is still a 2/3 chance that the remaining one isn’t inside the box, but only half of the remaining options can have a dual object status. Half of 2/3 gives me…1/3, all is well! (Al 4-5).

Birthday Paradox

While this is pushing the bounds of what makes a paradox, it is certainly a counter-intuitive result that is worth taking a hard look at. Imagine we have a room full of people and I want to know how many need in order to have a 50/50 shot of 2 people having the same birthday. Intuition makes us feel we need more than 366, for all the days of the year. But, as it turns out, we only need 23 people! But it gets weirder! If I wanted a 99% chance, almost a sure thing, I only need 57 people. What is going on here? (Al 6-7)

Think about the probability of bringing in a person to the room and them not having the same birthday. The chance that two people do not have the same birthday would be 364/365, for only 1 day out of the year would be the day they have in common. Now, bring in a third person. Now the odds of none of them having a common birthday would be (363/365)*(364/365), for now we have to eliminate the 2 days that could be in common. Taking that number away from 1 (100%) gives us the probability of them having the same day. If we extend this technique, we find that around the 23rd person, we finally get to a better than 50/50 chance. Simply amazing (7-8).

Works Cited

Al-Khalili, Jim. Paradox: The Nine Greatest Enigmas in Physics. Broadway Paperbacks, New York, 2012: 3-9, 16-7. Print.

© 2014 Leonard Kelley

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