Mathemagics 2: Solving the Birthday Paradox in two ways

Only 23 people are needed to have 50% probability of any two people sharing the same birthday. It may seem to be weird, but it is in fact true. And that’s because a human mind is habituated to think linear. We expect probabilities to be linear. Now what do you mean by that? Well, take this example. What is the probability of getting all heads in 6 coin flips?

Mathemagics 1: Challenging Ramanujan's Summation

One may think like getting one head in 1 flip is ½. So probability of getting 6 heads is (1/2)/6 or 1/12. Well, probability goes on exponents. Hence it will be (1/2)⁶ which is something around 0.0156. So now let’s come at the birthday problem. So we will solve the birthday problem in two ways.

Here is the graph showing the chance of the Birthday paradox:

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First method

Now if I ask you what’s the chance of getting one or more tails in 6 coin flips. It could be anywhere – in the 1st flip, 2^nd 10^th or 23^rd. So we will approach it from the other side. What is the probability of getting all heads? We know this is 0.0156. So the answer to the above question is 1-0.0156 = 0.9844. Now the formula that we applied – (1/2)⁶ has three parts 1,2 and 6.

· 1 is the number of desired outcomes – we want 1 head in all the flips

· 2 is the number of possible outcomes – they are head and tail

· 6 is the number of events

Similarly we want the probability that two people share the same birthday – P(A). So we will first calculate the probability that two people do not share the same birthday – P’(A). For that take any one person with some particular birthday. Now we want the other person not to have the same birthday – so he can have birthday on any of the 364 days. Here 365 is the no. of total outcomes and 364 is the no. of desired outcome. So the probability that two people will not share the same birthday in one occasion = 364/365.

Now between 23 people there can be ²³C_n matches which is (23 x 22)/2 = 253. Putting n = 253 in the formula, we have P’(A) = (364/365)²⁵³ = 0.4995

Therefore, P(A) = 1- 0.4995 = 0.5105 = 51.05% which is higher than 50%.

So this is the first method of solving the problem.

Second method

In this method also, we will find the probability that no two people will share the same birthday.

In this case we will take 23 people – numbered from 1 to 23. Now consider any one particular person with any particular birthday. Now what is the probability that the person will have some birthday? Obviously 1. Now the probability that the 2^nd person have a different birthday is 364/365 – because he can have birthday in any one of the other 364 days. Similarly the probability of the 3^rd person is 363/365 and so on. The 23^rd person will have the probability of the same as 343/365.