The Birthday Paradox Problem of Probability🍰🎈🎉

The concept of probability may seem like a very straightforward and obvious if it is seen from the surface level. However, if you go to the depth of the concept, you can find a little level of difficulty due to several mathematical operations. Nevertheless, if you succeed in grasping the basic concept of probability, it is no longer a tedious task. Once you understand it, it may seem tricky and challenging but quite interesting to you. You might have solved problems generally relating to rolling a die, deck of cards, colored balls, tossing a coin etc. Here, we are going to explain something different yet interesting.

            

In this blog, we will discuss a “paradox” of birthday problem probability which will precisely make you familiar about the compounding power of paradox which is strange, counter-intuitive but factually true! Generally, this problem is viewed as counter-intuitive because it goes against our intuition as our brains are bad at figuring out the power of chance. Our brains have trouble estimating the things that grow exponentially because we are not accustomed to such calculations in our daily life. We generally expect probabilities to be linear as we consider the conditions only through which we are familiar. We don’t think of other possibilities. This limits our thinking and we find exponential probabilities to be counter-intuitive.

            

Many of you wait the whole year for your birth date, right? But if you come to know that your colleagues share the same birthday as you, you become even more excited to bloom your birthday candles and celebrate the day. So, let’s begin!

The birthday problem (also called the birthday paradox) deals with probability that in a set of $$n randomly selected people, at least two people share the same birthday. It states 3 points-

·       a)The birthday problem states that the probability of at least 2 people sharing the same birthday is 50% when total number of people are 23.

·        b)Probability of at least 2 people sharing the same birthday is 99% if total number of people are 70.

·        c)Probability of at least 3 people sharing the same birthday is 100% when n >365.

 We will elaborate on first point by which you can easily by yourself understand the rest two points. The following explanation will explain you about correct way of thinking about the problem and the reasoning behind it.

 Standard assumptions

To make things simpler we make some standard assumptions –

There is no leap year.

There is no twin born.

Every event is equally likely .i.e.  Seasonal, or weekday variations are disregarded.

At first glance, many people guess 183 for not so obvious reason that it is half of the number of days in the year, which seems quite intuitive. Unfortunately, such intuition doesn’t take mathematical tool in solving this problem and rather you will in the birthday paradox TRAP.

 

Let's find why?

With 23 people in a room, the number of pairs in the room can be taken out through combination,

Where n = 23 and r = 2

We know that, the chance or probability of 2 people having different birthdays in a year of 365 days is

This seems easy, right? This is because when we consider 2 people, we have only 1 pair.

But things get complicated with 253 pairs because of extraneous information. So we make use of exponents to find the probability:

P (at least two people share same birthday) = 1 — 0.4995= 0.50005

                                                         =  50.005% APPROX

The easy intuitive trick that solves the birthday problem!

Instead of using the not “so easy” exponents rule, we can correct but the intuitive way to easily solve the birthday paradox.

P (At least one pair shares birthday) = 1 – P (All birthday are on different days)

So, now we focus our energy to find the answer to this: “What’s the probability of people NOT any birthday in the group?”

Unique birthdays DENOMINATOR

Intuitively we know, the denominator is 365ⁿ (where n = 23). The denominator is unscathed.

Unique birthdays NUMERATOR

This is where the MAGIC BOOM happens!

The first folk in the group has 365 options for a birthday (GREEDY BOAR). This leaves the second folk with only (365-1) option, as we force them to select a different birthday.

In this correct but initiative manner, we can say that the N person has (366— N).

If N is 23, we put in the multiplication signs to get the numerator,   365 x 364 x 363 x 362 x 361 x 360 x 359 x 358 x 357 x 356 x 355 x 354 x 353 x 352 x 351 x 350 x 349 x 348 x 347 x 346 x 345 x 344 x 343 = …

P(All 23 birthdays unique) = (365 x 364 x 363 x 362 x 361 x 360 x 359 x 358 x 357 x 356 x 355 x 354 x 353 x 352 x 351 x 350 x 349 x 348 x 347 x 346 x 345 x 344 x 343) / (365 ^ 23)

= 0.4995 (rounded off to the capacity of my attention span)

Finally, All’s well that ends well

P (at least two people share same birthday) = 1 — 0.4995= 0.50005

                                                                      = 50.005% approx.

The unusual conclusion thus derived seems far from what our initial instinct would have us believe….

There can be yet another dimension to analyze and hence conclude the above paradox. This time we use conditional probability to aid thus through the problem!

We have taken a set of 23 people and the aim is to find the probability that at least 2 people have the same birthday

v  P (Someone shares the birthday with at least someone else)?

Let P(S) be the required probability and P(D) be the probability that no one shared their birthday with anyone. The two probabilities combined together constitute the SAMPLE SPACE.

This can therefore be expressed as:

P(S) + P(D) = 1 = 100% and P(S) = 1- P(D)

In order to avoid complications we will instead focus our attention on calculating the probability that everyone has distinct birthdays. This can be another way of looking at the paradox statement. Our initial reaction to this possibility will be a guess that the probability of everyone having different birthday will be very high but…..the result is COUNTER-INTUITIVE!!

For simplicity and to derive the pattern we take 2 people

The probability that the two people will have different birthday or the conditional probability that the person B has distinct birthday given the birthday of person A,

                                                           Person A       Person B

                                        P(D)      =   365/365    x    364/365   =   365* 364 / (365)²

                                                                                 OR

365! / 363! * (365)²

Similarly for three people the probability that person B has a distinct birthday then person A and person C has a distinct birthday given the birthdays of A and B. No one shares birthday.

Person A  * Person B *  Person C

                                        P(D)   =      365/365  *  364/365  *  363/365 = 365* 364 *363 / (365)³

For the above results we can conclude that for n = 23

P(that no one shares birthday with anyone) or P(D) =

365! / (365-n)! * (365)ⁿ : where n is the number of random people

Since n= 23 therefore

P(D) = 365*364*363*…….343 / (365)²³

 OR

365!/ {(365-23)! * (365)²³} = 365! / {342! * (365)²³} = 0.4927 = 49.27%

Hence,

P(S) = 1- P(D) = 1 – 0.4927= 0.5073

P(S) = 50.73%

Therefore, the probability in the room of 23 people, of at least two having the same birthday is 50.73% that is more than half of the total chances. This conclusion is clearly an unexpected one since one could never through usual logic imagine that if he or she were to enter any room with other 22 random people, there will be more than fifty percentage chance of that person having the same birthday with at least one among the slot maybe more.

HOW COMMONLY SHOULD WE HAVE COMMON SENSE!

After an empirical analysis, we are now in a position to draw the relevant information that sometimes we may be presented with situations or cases where logic borne out for our immediate consciousness may be less accurate or even simply incorrect as we saw in the undertaken paradox.These situations can be deconstructed through various algebraic manipulations that enable us to understand the paradox in more detail. Chance problems sometimes requires us to be more technical and statistical in approach rather than being usual about it.