Three Interesting Problems on Statistics
- Oct 05, 2016
- Sudipto Das
In this article, we have made a selection of 3 classic statistical problems that are pretty interesting from the point of view of mathematics. Let’s begin without further ado.
Monty Hall Problem
We have discussed this one in greater detail on our website. The Monty Hall problem is one that’s based on probability. Imagine yourself being on a gameshow. In front of you there are 3 doors. You are aware of the fact that behind one of them is a bucket full of cash and behind the other two there’s nothing. You have to make your choice among those 3. If you are right, you go home with a pocket full of cash. If you aren’t, you go home with nothing. So you make your choice (say the 1st door). The game host then asks you to reconsider the choice but you stay adamant. The host then opens the 3rd door and shows you that it’s empty. Now he asks you whether you would change your mind and switch your choice to the 2nd door or not. Here comes the Monty Hall problem. According to this problem, you should always shift your choice. If you want to know about the reasons for which you should shift your choice, you can go through the hyperlinked article where the entire process is described in greater detail.
Imagine the fact that you are running an office that employs twenty-three people. Now the question is- “what’s the probability that 2 employees have the same birthday?”
For the sake of the problem, you can ignore the fact that February consists of 29 days in a leap year.
Say for example, your office has a total of 366 people. In this case, it’s an absolute certainty that a minimum of 2 employees share the same birthday.
Choose any 2 people in your office.
Probability that the 2nd person’s not sharing a birthday with the 1st= 364/365.
Probability that the 3rd person’s not sharing a birthday with the 1st or 2nd= 363/365.
So, now you can easily notice a pattern here. Do remember that your office is supposed to have 23 employees according to the question. Probability that 23 people do not share the same birthday=
365/365 x 364/365 x 363/365 x 362/365 x … x 343/365 = 0.4927.
Therefore, probability that at least two people share the same birthday= 1-0.4927= 0.5073= 50.7%.
Say, for example, a gambler has a specific amount of money (B) and is involved in a game of chance where the probability of winning is below 1. The gambler is playing with a specific tactic. When he wins, he raises the stake by a certain fraction. The gambler will not reduce the stakes when he loses but will not raise it either at the time of the loss; he will keep it as it is.
Every time he wins it, he’s raised the stake by B/N.
For the sake of the problem, we are assuming,
B= $1000, N=4.
(He will gamble with $250 each time and if he wins, he’ll increase the amount. If he loses, he will keep it as it is.)
So the question is- what will be his winnings?
If the gambler bets 1/N of his bank-balance each time and then maintains the amount if he loses, the gambler’s N losing bets away from bankruptcy. Well, of course, the gambler will have to keep on betting for this problem to come true. So basically the gambler will lose everything. For more details on the problem, you can refer to this article.
So you see that the 3 problems highlighted above are quite interesting indeed. Hope you had a good read. Ciao!
Sudipto writes technical and educational content periodically for wizert.com and backs it up with extensive research and relevant examples. He's an avid reader and a tech enthusiast at the same time with a little bit of “Arsenal Football Club” thrown in as well. He's got a B.Tech in Electronics and Instrumentation.
Follow him on twitter @SudiptoDas1993
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