Ramjas College Case Study Round 2

 Case:

Khullja Sim Sim is an Indian television game show which originally aired on STAR Plus for the first two seasons, premiered on July 27, 2001.The first season was hosted by popular film actor and presenter Aman Verma and the second season by Hussain Kuwajerwala. The third season of the series premiered on BIG Magic on September 17, 2012, also hosted by Aman Verma, produced by BIG Productions of Reliance Entertainment. Khullja Sim Sim is a unique game show concept that tugs at your inherent desire to trade and win. It is an Indian adaptation of the hit U.S. game show Let's Make a Deal, which aired from 1963–77, 1980–81, 1984–86, 1990–91, 2003, and currently since 2009. It was one of the largest ever shows produced in the country.
The show Khullja Sim Sim involved the strategy that contestants should take to maximize their chances of winning prizes. On the program, a contestant would be presented with three doors, behind one of which was the prize. After the contestant had selected a door, the host, Hussain Kuwajerwala (say you are watching the second season), would open one of the other two doors, showing that the prize was not there. Then he would give the contestant a choice—either stay with the door initially selected or switch to the “third” door, which had not been opened.
For many viewers, common sense seemed to suggest that switching doors would make no difference. By assumption, the prize had a one-third chance of being behind each of the doors when the game began. Once a door was opened, it was argued that each of the remaining doors now had a one-half probability of hiding the prize, so contestants gained nothing by switching their bets.
Dig into your statistics textbooks and use one of the fundamental theories of probabilities to show that common sense would leave you astray here. Tell us how you can increase your chances of winning in this game. Use a typical case to prove your argument.
Note: The analysis of this game would require you to insert formulae in the text. This option is available in the ‘insert’ tab. The solutions however must be submitted in the pdf format.
Twist 1
In this round participants are allowed to ask for a hint. In order to know the ‘one-word’ hint you can call [information redacted] and ask for the hint by quoting the unique team ID that has been provided to you in the email. Only, when the unique ID quoted by you matches, will you be provided with a hint. But, this clause has the following sub-clauses:
a. If all the other four final teams ask for a hint, you will have an equal chance of winning the competition.
b. If anyone of the other four final teams does not ask for a hint and is able to submit the correct answer to this case, that team will be declared the winner.
c. If anyone of the other four final teams does not ask for a hint, but is unable to submit the correct answer, the teams who request for a hint would have an equal chance to win this event.
d. If none of the teams ask for a hint, then all the teams would have an equal chance of winning the event. 

 Solution:

(By Sumit and Arjun)

This is actually a very interesting case and a rather counter intuitive one! To analyse this we must think strictly in terms of probability & make some important assumptions.

An important feature of the game will be to not consider the processes of choosing a door and then the choice of switching the door as independent events. The game is essentially sequential.

Assumptions:

1.      Actually as the host has opened the empty door only (given), to take this fact into account we have to assume that the host knows which door has the prize and he will only open the door which he knows does not contain the prize.

2.      We have to consider that the prize is randomly distributed among the doors

 

Using conditional probability and Bayes’ theorem we can attempt to clarify the situation.
Events:

A1: Prize is behind door 1

A2: Prize is behind door 2

A3: Prize is behind door 3

X1: player picks door 1

B3: Host opening door 3 (empty)

We know that P (A1) = P (A2) = P (A3) =1/3

Let us assume that the host opens door 3 (B3 event), then we have:

P (B3|A1, X1) =1/2

P (B3|A2, X1) =1

P (B3|A3, X1) =0

The above statements imply that the probability that the host opens door 3 if the prize is in door 1 given the player has selected door 1 initially is 0.5 because the host can then open door 2 or 3 (at random). And if the prize is behind door 2 then the host will have to open door 3 only because door 2 has the prize and door 1 is chosen by the player. If the prize is behind door 3, then he surely would not open it, so zero probability.

So if the player has selected door. 1 and the host opens door 3, the probability of winning by switching the doors is given by

P(A2|B3,X1)= [ P(B3|A2,X1) P(A2|X1)]/P(B3|X1)

  = [ P(B3|A2,X1) P(A2|X1)] / [ P(B3|A1,X1)P(A1|X1)+ P(B3|A2,X1) P(A2|X1) +       P(B3|A3,X1)P(A3|X1) ]

  = [1*(1/3)] / [(1/2)*(1/3) + 1*(1/3) + 0*(1/3)]

   =2/3

Whereas if he does not switch, his probability remains 1/3 (calculated using 1 – 2/3, as there are only two possible choices)

Recommendation:

It can be concluded that switching the door after being given the choice is the better option as the next door will have double the probability of winning. So this is when we would have to ignore the call from our common sense and rather rely on impartial analysis and switch our choice after being given the offer.