Consider the Monte Hall Game Discussed in Class Again Suppose You Choose Door 1 and the Host

Overview

• Go to know what the Monty Hall Problem is.
• Sympathize provisional probability with the apply of Monty Hall Problem.

Introduction

I was indulged in a project where we aim to predict the IPL auction prices for cricket players in such a manner that every franchise gets maximum of their choices in their team and every role player gets an optimized toll according to his caliber. Here we used dominion-based classification which was based on "conditional probability".

Since then I started exploring conditional probability and a few days back I confronted a very interesting puzzle called "Monty Hall Trouble".The Monty Hall Problem became world-famous in 1990 when "Marilyn Vos Savant" gave a simple yet logical solution in the popular weekly column "Ask Marilyn" in Parade magazine equally a respond to the following question-

"Suppose you're on a game testify, and you're given the choice of three doors: behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's backside the doors, opens some other door, say No. three, which has a goat. He then says to yous, "Exercise y'all want to pick door No. ii?" Is information technology to your reward to switch your choice?"

And as usual, many statisticians taunted her saying that women don't understand statistics but later all those critics ate apprehensive pie and appreciated "Marilyn Vos Savant". They all forgot the fact that at that time she was the person with the highest IQ every bit per the Guinness Book of Records

So in this article, I volition explicate to you the concept of conditional probability in particular using the Monty Hall Problem.

In case you want to learn these concepts and start your journeying in data science, check out the following grade- Introduction to Data Science

Table of Contents

1. The Monty Hall Problem
2. Intuitive Explanation
3. Situational Explanation
4. Detailed Caption
5. Explanation with Provisional Probability

The Monty Hall Problem

Consider this scenario – Suppose y'all are in a game evidence and they requite you three doors.

They accept been caged.

Behind 1 door is a car and backside the other two doors are goats. (But you lot don't know which door has what).

At present the host of the game show asks you to selection i door. Note that he knows what is behind which door.

Suppose you choose one of the doors among these three. Now, out of the remaining two doors,  the host reveals one door which has a goat behind it.

And makes a very interesting proposal for yous.

He asks you if you want to stick to the choice of your door or want to cull the door that has not been revealed however?

What do you recollect? Do you want to switch the door as your choice or want to stick with your earlier selection? Y'all must be thinking that it doesn't matter. Because there are only 2 doors left and each of them has a 0.five probability of having a car? Right?

Just that's not right, allow's figure out how.

Intuitive Explanation

Let's recall about it Cognitively first.

Let'due south say that instead of three doors, you now accept g doors, of which merely one has a car behind information technology while 999 will have a goat. You are asked to choose again, and you are select a door. The host then reveals 998 doors (of the remaining 999 doors) which have goats behind them. Two doors still remain airtight – one of them is the one you lot picked previously. What would you do?

Exercise you desire to switch now? Probably the answer is aye, merely what is the reason for switching now?
Is information technology self realization or some cognitive thought occurring in the statistical mind.

The reason you lot want to switch at present is that you have realized that probability of choosing the door which has the auto at a offset guess is one/grand, So if you don't switch, yous only have 1/1000 gamble of winning

Situational Explanation

Have a expect at the image beneath, which shows iii possible scenarios. Suppose door one has a car backside it and the other two doors take goats.

• In the first instance, you selected the outset door
• While in the second instance you selected the second door
• And in the third case, you lot selected the 3rd door.

State of affairs one: Y'all pick door one at your outset estimate. The host can reveal door 2 or three because both of them accept goats. Suppose the host reveals door 3 which has a goat behind it, Now if yous switch to door ii you lot will lose.

State of affairs two: What if yous picked door 2 initially? Door one has a machine then the host can non reveal it, he has to reveal door iii which has a goat. At present if you switch to door 1 from door 2 y'all volition win.

State of affairs three: Similarly what if you picked door three initially? Again in this case, since door 1 has a machine so the host tin not reveal it, he has to reveal door two which has a goat. Now if you switch to door 1 from door iii you volition win once again.

So in case of switching you are winning 2 out of 3 cases. So Probability of winning afterwards the switch is ⅔ while the Probability of winning sticking with the initial choice is ⅓.

To summarise nosotros can say that If you don't switch, yous volition win the car just if you were correct in guessing right door initially Prob = ⅓ = 0.33
If you switch, you lot will win the car only if yous were not right in guessing the right door initially = ⅔ = 0.66.

Detailed Explanation

Permit's understand this problem now with the concepts of Probability. A typical random experiment involves several randomly-adamant quantities which are-

1. The door concealing the auto.
2. Door that is initially chosen by the player.
3. The door that the host opens to reveal a goat.

As you lot tin can see above, this Tree Diagram represents all possibilities of this Puzzle. Car location, Player's guess, and doors revealed making the outcome. All possible values of the outcome make the sample space. Let's understand what each of these outcomes represents. For example, BAC hither represents that car is located behind B, your initial guess is A and Host revealed the door C.

Out of this sample space, how volition you define the events? For case, the event that the prize is behind door C is the ready of outcomes:

{(C,A,B),(C,B,A),(C,C,A),(C,C,B)}

Because the offset letter of the alphabet represents the Car Location in our Tree Diagram.

The event that yous initially picked the door concealing the prize is the set of outcomes:

{(A,A,B),(A,A,C),(B,B,A),(B,B,C),(C,C,A),(C,C,B)

Because the first Letter represents the Automobile Location and the second Letter represents your pick which needs to be same for this issue (initially picked the door concealing the prize), what we're actually after is the is the set up of outcomes of event that the actor wins by switching:

{(A,B,C),(A,C,B),(B,A,C),(B,C,A),(C,A,B),(C,B,A)}

Why? Allow's look at whatever of these outcomes,

CBA represents that C has the Car, B was picked past you, and A is revealed. So switching volition lead you to C door from B door which actually has the Automobile. Hence, you lot win. Similarly, all these outcomes make you win in example of switching.

Analyzing our Tree Diagram, we observe that exactly one-half of the outcomes are marked, meaning that the histrion wins by switching in one-half of all outcomes. You might be tempted to conclude that a player who switches wins with probability 1/2.

This is wrong. The reason is that these outcomes are non equally probable. As you can see, outcome ABC and result BBA accept different probabilities why?

Case 1

Permit's analyze, ABC first which ways :

1. Car is backside Door A
2. You Picked Door B
3. And Host revealed Door C.

1. The car is behind door A,  has probability ⅓ because Car could be behind any of those 3 doors with equal probability.
2. Y'all picking door B also has probability ⅓ because you could have picked any of the doors.
   Thus this outcome has probability ⅓ * ⅓ = one/9.
3. At present Host has door A and C, Host tin can not reveal door A because it has a machine behind information technology and then he must cull door C to reveal. Thus, the probability is 1 here.

Thus this result has probability ⅓ * ⅓ = i/ix.

Case two

In Case of BBA, BBA  means :

1. Car is behind Door B
2. You lot Picked Door B
3. And Host revealed Door A

1. The car is behind door B,  has probability ⅓ because Automobile could be behind any of those iii doors with equal probability.
2. Y'all picking door B besides has probability ⅓ considering you could have picked any of the doors.
3. At present Host has doors A and C, Host can reveal any of the doors A and C considering they both take a goat behind it so choosing door A by the host is Probability 1/ii.

So this outcome has the Probability ⅓ * ⅓ * ½ = 1/eighteen every bit mentioned in the Tree Diagram. Thus Now it is articulate that non all these outcomes are equally likely.

Permit'due south summate the probability of winning in instance of switching. Thus using the concepts of Sample Space, Issue and Tree diagram

Nosotros conclude that:-

Probability (Switching wins)  = (P{A,B,C} + P{A,C,B} + P{B,A,C} + P{B,C,A} + P{C,A,B} + P{C,B,A})
= ane/9 + i/9 + one/ix + 1/9 +1/9 + 1/nine = ⅔

The reason why it is non ½ is considering of the condition that every fourth dimension afterwards your option host is revealing a door that has a goat behind it and this particular condition impacts the likelihood of final outcomes in our sample space.

Explanation with Conditional Probability

The provisional probability is the probability of any event A given that some other event B has already occurred. The thought here is that the probabilities of an event "peradventure" afflicted past whether or non other events have occurred. The term "provisional" refers to the fact that nosotros will have additional conditions, restrictions, or other information when we are asked to summate this type of probability.

Information technology is denoted in the following way – "representing the Probability of A given B has occurred"

Conditional Probability can be calculated as Probability of A intersection B, divided by the probability of result B

P(A | B)  = P(A ∩ B) / P(B)

Permit u.s.a. start to analyze this trouble when the contestant has chosen door one. We assume that P(prize door i) = ⅓, for i = 1, 2, 3

If the prize is behind door 1 then the host show volition open door 2 or door iii each with probability 1/two.

Then we have P(prize door ane and host door 2) = ane/3 × 1/2 = 1/half-dozen

P(prize door ane and host door 3) = 1/3 × one/ii = 1/6

On the other mitt, if the prize is behind door 2 or door 3, then the host has just one door that he tin open, namely door three or door 2.

P(prize door 2 and host door 3) = one/iii × 1 = 1/3

P(prize door 3 and host door 2) = 1/three × ane = 1/3

We have described all possibilities starting from the fact that the contestant has already chosen one door. Since the prize can be behind any door with the same probability, it does not matter which door is called. Given that the host opens door 3 the probability to win the prize by keeping the door is the conditional probability

P(Keep and win) = P(prize door 1 | host door iii)

= P(prize door i and host door 3)/ P( host door 3)

= (1/vi) /P( host door three)

while

P(Proceed and loose) = P(prize door two | host door three)

= P(prize door 2 and host door 3)/ P( host door3)

= i/3 P/( host door 3)

It is therefore twice as likely to win past switching and so we accept:-

P(Proceed and win) = ane/3

P(Go along and loose) = ⅔

If ane wishes to compute the probability that the host opens door 3 then one tin detect it by conditioning on the location of the prize:

P( host door 3) = P(host door three| prize door 1)P(prize door 1)
+P(host door 3|prize door two)P(prize door ii)
+P(host door three|prize door 3)P(prize door 3)

= i/two × one/3 + 1 × 1/3 + 0 × 1/3 = 1/2

With this, nosotros conclude the Monty Hall Problem Explanation using Conditional Probability.

End Notes

To summarize, in this article we explained the concept of provisional probability using the Monty Hall Trouble. Information technology is an imperative concept that all aspiring data scientists need to understand. Not only this, but there are likewise several other concepts that you should exist well versed with. The post-obit are some of the articles on statistics and probability that yous should sympathise-

• Statistics for Information Science: What is Normal Distribution?
• Statistics for Data Science: Introduction to the Key Limit Theorem (with implementation in R)
• vi Common Probability Distributions every data science professional should know

I hope this article was fruitful to you. Allow us know in the comments in instance you have any queries.

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Source: https://www.analyticsvidhya.com/blog/2020/08/probability-conditional-probability-monty-hall-problem/

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