{"id":1833,"date":"2014-03-16T15:19:23","date_gmt":"2014-03-16T09:49:23","guid":{"rendered":"http:\/\/ucanalytics.com\/blogs\/?p=1833"},"modified":"2016-10-13T15:12:14","modified_gmt":"2016-10-13T09:42:14","slug":"bayes-theorem-monty-hall-problem","status":"publish","type":"post","link":"https:\/\/ucanalytics.com\/blogs\/bayes-theorem-monty-hall-problem\/","title":{"rendered":"Bayes’ Theorem – Monty Hall Problem"},"content":{"rendered":"
\n

Monty Hall Problem<\/span><\/h2>\n
\"Monty<\/a>

Monty Hall & Bayes’ Theorem – by Roopam<\/p><\/div>\n

Monty Hall is back, for one last time, to host the famous show from the 1960s \u2018Let’s Make a Deal\u2019. You are his final guest and the prize is really desirable \u2013 a red Ferrari 458 Italia. Monty is no generous host; he could be over 90 years old, but he is going to make you work for your prize. So here is the deal, there are three closed doors (A, B, and C) and behind one of these doors is the Ferrari and the remaining doors have a goat each. You need to pick a door and can take home whatever is behind that door \u2013 you really hope it’s the Ferrari. \u00a0After doing your calculation you picked door A \u2013 you know that was a wild guess. Monty being a generous host tries to help you out by opening one of the remaining two doors i.e. door B and displays the beautiful goat behind the door. By the way, Monty knows the location of the car and will never open the door concealing it. He asked you for the last time if you would like to switch from door A to door C. This is where it gets really difficult – whether to stick or switch. The decision is particularly difficult when a huge audience (studio and television) is watching you live. Despite the pressure, you have decided to do the long calculation for this problem using the Bayes’ theorem. I think you have made a wise choice.<\/p>\n

Bayes’ Theorem to Solve Monty Hall Problem
\n<\/span><\/h2>\n

You are aware of the difficulty of this problem. The solution to this problem is completely counter-intuitive. Marilyn Vos Savant was asked to solve the same problem by a reader in her column \u2018Ask Marilyn\u2019 in Parade magazine. Marilyn, by the way, is listed as the person with \u2018Highest IQ\u2019 by the Guinness Book of World Records. She recommended switching as that will increase one’s chances of winning the car by a factor of two. Her answer created a major furor with one of the readers (with a Ph.D.) writing her the following:<\/p>\n\n\n\n
“[Marilyn] You blew it, and you blew it big! Since you seem to have difficulty grasping the basic principle at work here, I’ll explain. After the host reveals a goat, you now have a one-in-two chance of being correct. Whether you change your selection or not, the odds are the same. There is enough mathematical illiteracy in this country, and we don’t need the world’s highest IQ propagating more. Shame!” <\/em>– Scott Smith, Ph.D. University of Florida<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

 <\/p>\n

You know it is not going to be easy but Bayes’ Theorem will guide you well. We have discussed Bayes’ Theorem a couple of times in previous articles on YOU CANalytics, you may want to refer to those articles Bayesian Inference – Made Easy<\/a> and O.J. Simpson case<\/a>.<\/p>\n

You have scribbled down your solution to the problem on a piece of paper. The following is a version of your scribbling<\/p>\n

\"Monty<\/a><\/p>\n

It seems Marlyn was right or was she? You want to solve it one more time to be doubly sure. So you start all over again, to begin with, you need to find the prior probability i.e. the chances of having the car behind each door. Since there is no other information available the prior probabilities is the same for all the doors. This was fairly easy as shown in the formula below.<\/p>\n

\"P(Car@A)=P(Car@B)=P(Car@C)=\frac{1}{3}<\/pre>\n
There is not much to choose and hence you have chosen door A. Now comes the difficult part, Monty Hall has revealed a goat behind door B this has infused some information into the above problem. This needs us to create a second layer of probabilities post the event i.e. probability of Monty to open door B when you have chosen door A. Remember that Monty knows the location of the car and will never open the door concealing the car.<\/p>\n
This requires us to calculate the conditional probability or chances of the event given the position of the car. What are the chances for Monty to open door B if the car is behind door A?<\/p>\n
\"P(Open\\r\n\r\nMonty could have opened either 'door B' or 'door C'. As You have chosen 'door A' (He cannot open this door)<\/pre>\n
Second question: What are the chances for Monty to open door B if the car is behind door B?<\/p>\n
\"P(Open\ \r\n\r\nMonty will never open door B if it was concealing the car.<\/pre>\n
Final question: What are the chances for Monty to open door B if the car is behind door C?<\/p>\n
\"P(Open\\r\n\r\nMonty has to open door B, he has no other choice. You have already chosen 'door A' and door C is concealing the car.<\/pre>\n
Having this knowledge leaves you with one final task to calculate the posterior probabilities. One of the calculations is shown below.<\/p>\n
\"P(Car@A|Open\\r\n\r\nwhere,\r\n\"X=P(Open\\r\n\r\n\"Y=P(Open\\r\n\r\n\"Z=P(Open\<\/pre>\n
Let us revisit your scribbling from the previous calculation.<\/p>\n
\"Monty<\/a><\/p>\n
 <\/p>\n
The chances of winning the car are indeed 2 times higher i.e. 2\/3 when you switch than when you stick i.e. 1\/3. Marilyn was right! Going by your calculation you have decided to switch to door C.<\/p>\n
Sign-off Note<\/span><\/h4>\n
Monty opens the door you chose i.e. door C and reveals your prize – a goat! Hard luck. If Monty Hall plays this game with you 1000 times you are most likely to end up with 667 Ferraris – that’s how probability works. You have decided to donate the goat to a charity and feeling good about it (probably better than driving a Ferrari).<\/p>\n","protected":false},"excerpt":{"rendered":"
Monty Hall Problem Monty Hall is back, for one last time, to host the famous show from the 1960s \u2018Let’s Make a Deal\u2019. You are his final guest and the prize is really desirable \u2013 a red Ferrari 458 Italia. Monty is no generous host; he could be over 90 years old, but he is<\/p>\n
Read More…<\/a><\/p>\n","protected":false},"author":1,"featured_media":1834,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_jetpack_newsletter_access":"","_jetpack_newsletter_tier_id":0,"jetpack_publicize_message":"","jetpack_is_tweetstorm":false,"jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":true,"jetpack_social_options":{"image_generator_settings":{"template":"highway","enabled":false}}},"categories":[64,65],"tags":[33,7,6,10],"jetpack_publicize_connections":[],"yoast_head":"\nBayes' Theorem - Monty Hall Problem – YOU CANalytics |<\/title>\n<meta name=\"description\" content=\"The solution to the Monty Hall Problem using Bayes Theorem. 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