CATEGORY / Interesting Mathematics

Zeno’s paradox begins like this:
In order for a man to reach another, he needs to cover half the distance first. After that, he needs to cover the next quarter of the distance. Thereafter, he needs to cover the next eighth of the distance, and so on. With these small increments that he has to cover, as it is seemingly infinite, will the man never get to anywhere?

Here is the fascinating answer.

First, we realise that 1 can be split up into infinite parts – $1=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+…$

Hence an infinite parts can also be added up to 1 – $\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+…=1$

Therefore, even though the man needs to walk an infinite number of fractions to get to his destination, he will still get there.

And here is the mind-blowing nature of $\infty$.

As an auctioneer in a silent auction, to maximise profits, you should use the Vickrey auction.

A Vickrey auction allows the highest bidder to win the auction, paying only the price that the second-highest bidder chose.

Advantages:
1. Allows for the highest bidder to not feel slighted that he could have paid a much higher price than the second-highest bidder.
2. Encourages the bidder to place a bid at maximum because he does not want to lose the bid
3. Allows the bidder to place a maximum bid that he is willing to pay, knowing that he will never need to pay that maximum

Mathematically, should you believe in God?

Food for thought.

You are given 2 envelopes which are exactly the same and are told that one envelope has twice as much money as the first. You open the first envelope and find that there is \$10 inside. Should you switch or should you take the \$10, if you want to maximise your earnings?

Solution:
In this case, we think that it does not make a difference if we switch or not, but actually, it does.

Case 1:
You took the envelope with more money.
The envelope with less money has $5.

Case 2:
You took the envelope with less money.
The envelope with more money has $20.

Expected value = (Case 1 + Case 2)/2
= $12.50

Hence you should switch!

*Update: a reader spotted a fallacy in this, can you decipher what it is?

A gameshow host tests a contestant to see if he will win a car behind one of the 3 doors. Importantly, the host knows where the car is.

Assuming the contestant picks the first door, the host then opens the another door, which reveals that there is nothing behind it. The host then asks the contestant if he would like to switch to the final door or stick with the first door. What would you do?

Solution: Intuitively, we will think that it does not matter. However, Mathematics triumphs intuition here, as it is twice as likely for the car to be at the final door than the first door picked. Let us see why.

1. Assumption that the car is at the first door.
The host obviously will not open the first door and hence opens the second or third door.
If the contestant switches, he will not get the car.

2. Assumption that the car is at the second door.
The host obviously will not open the second door and hence opens the third door.
If contestant switches, he will get the car.

3. Assumption that the car is at the third door.
The host obviously will not open the third door and hence opens the second door.
If contestant switches, he will get the car.

Funny how the brain works, eh?

You showed your friend your handphone and your friend shows you his.
Both of you start to compare and you claim that your handphone is cheaper than his and he claims that his handphone is cheaper than yours.
Hence you start a bet – after you check the actual prices, the one whose handphone is more expensive will give his handphone to the one whose handphone is cheaper.
You reckon since you will either lose your cheaper handphone or gain a more expensive handphone, it must make sense to bet.
Does this make sense?

Solution:
No! Your friend has the same mentality too – that he will either lose his cheaper handphone or gain a more expensive handphone, and thus enters the bet.
But both of you cannot win, so let us examine the probabilities.

Assuming that your handphones cost either \$10 or \$20 for easy calculations.

Case 1:
Your handphone – \$10
Friend’s handphone – \$20

Your friend gives to you his \$20 handphone.

Case 2:
Your handphone – \$20
Friend’s handphone – \$10
You give your friend your \$20 handphone.

Expected value = \$20 – \$20 = \$0.

No one will stand to gain from this bet!