Probability Paradoxes

A paradox is essentially a statement that leads to a circular and self-contradictory argument, or describes a situation which is logically impossible. This is what we call a true paradox, for the word ‘paradox’ is thrown around a lot. A ‘perceived paradox’ on the other hand is a puzzle that seems like a true paradox at first but has hidden tricks or perspectives which allow us to solve the ‘paradox’.

Many ‘perceived paradoxes’ require knowledge about physics principles in order to solve them, but there are a few which can be solve with the clever and appropriate use of logic.

The case of the missing dollar

Where did the dollar go? The mystery continues...

Where did the dollar go? The mystery continues...

3 travelers check into a hotel and pay $10 each for a $30 room with 3 beds. After a few minutes of settling in, the receptionist realizes she made a mistake and should have charged them $25 because of a special offer. So she takes 5 $1 bills and goes to correct her mistake. She realizes that she can’t split $5 evenly between 3 people, so decided to keep $2 and give them $1 each. Therefore each of the travelers has contributed $27, the receptionist kept $2, which makes $29. Where is the last $1 of the $30?

This puzzle only sounds paradoxical because of the way it’s phrased. The answer indeed is quite simple; the error comes from adding the $2 to the $27 when instead you should be taking it away from $27, because $25 needs accounting for not $30.

Bertrand’s box paradox

This perceived paradox comes from the mind of French mathematician Joseph Bertrand.

You have 3 boxes, each divided into 2 halves by a partition, and each side can be opened separately meaning you can’t see what’s in the other half. One box contains 2 gold coins (GG), one contains 2 silver coins (SS), and the last contains 1 silver and 1 gold coin (GS). If you pick a box at random and open one of the lids and find a gold coin; what are the chances of this being the GS box? Because it can’t possibly be the SS box, surely the probability is ½? Had you opened the lid to find a silver coin, you could rule out GG, and so the probability of the box being the GS is still ½. There are 3 coins of each kind, and since you can only find G or S when you open a lid, there is therefore a ½ chance you have found the GS box, whatever coin you find. After taking a peek inside ½ of a box, the overall probability that you have picked the GS box goes from 1/3, which it was at the beginning, to ½. How can seeing one coin, and gaining no information because you’re certain to find gold or silver, does the probability go from 1/3 to ½?

The answer is that the probability of the box being GS is always 1/3 and never ½. Consider the possibility that you find a gold coin, one of three in total, G1 G2 and G3. Let’s say that the box GG has G1 and G2, and G3 is in box GS. If you then find a gold coin, the probability that you’ve picked the GG box is 2/3, since the coin you find could be G1 or G2. There is only 1/3 chance that it’s the G3 coin, therefore 1/3 chance you picked box GS.

 

The birthday paradox

This perceived paradox is one that has no tricks or hidden clues. The solution makes complete mathematical and logical sense but seems completely wrong and unconvincing.

How many people would there have to be for the chances of any two of them sharing a birthday, to be larger than 50-50; that is to be more likely to not have any two sharing a birthday?

Let’s start of by applying some common sense (which of course won’t work). Imagine a hall with 365 seats and 100 students walk in. Some people of course may prefer to sit next to friends, at the front, at the back etc. No matter how they distribute themselves, there will always be more than 2/3 of the seats empty. The chances of two students also wanting to sit in the same place is quite thin because they can spread out anywhere. Using this idea and applying it to the birthday problem, we can assume that the any of the students sharing a birthday is quite slim. But things start to get interesting when we reduce the numbers.

As incredible as it may seem, you only need 57 people in the room for the probability of any 2 sharing a birthday to be 99%. But the puzzle asked for the probability to be over 50%, and for that you only need 23 PEOPLE! Many people are uncomfortable with this answer because it’s difficult to understand.

The first step to getting to the figure of 23 is to look at two people. For people to share a birthday we need pairs of people, not individuals. If you start with 3 people you can make 3 pairs; AB, AC and CB. With 4 people you can make 6 pairs; AB, AC, AD, CB and CD. With 23 people you can make 253 different pairs. The way to work out probability is to start with one pair and keep adding people to see how probability of birthday sharing changes. This is done by working out the probability of each new person avoiding all birthdays so far; thus the probability of the 2nd person avoiding the birthday of the 1st is 364/365. The probability of the 3rd person avoiding the first two birthdays is 363/365. But the probability of the 3rd person avoiding the two birthdays and the 2nd person avoiding the first birthday is (364/365) x (363/365) = 0.9918. Therefore the probability of any 2 of the 3 sharing a birthday is 1-0.9918= 0.0082. This is quite small, exactly what we would expect with 3 people.

We now have to carry on this process until we get an answer which drops below 0.5; meaning that the probability of any pair sharing a birthday is above 50%. We find we need 23 fractions, hence 23 people.

(364/365) x (363/365) x (362/365) …. {23 fractions multiplied together} = 0.4927

1-0.4927= 0.5073= 50.73% that any two people of 23 in a room will share a birthday.

 

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