Posted by: truevoid | June 21, 2009

monty hall problem

i wanted to read this book ever since quiller has told me about Christopher and his liking towards prime numbers. i wanted to see his take on prime numbers and compare it with mine. as far as the book is considered i felt that the first hundred odd pages keep you gripped and is fast paced, however the author loses it in the second half. having said that i strongly recommend ‘the curious incident of the dog in the night-time’ – and i just noticed that none of the first letters of any word are in uppercase on the cover page – a style i often follow – right from contacts names in my phone book.

there is mention about the possibility that the stars which we see (their light reaching us) might be dead i.e. it takes so many light years to reach us that the star might have died already. when i read that sentence i felt i am cheated, i felt like i was living in a slow world like a ‘deferred live’ situation.

this post is not meant to review the book but discuss the ‘monty hall problem’ which in my opinion Christopher got it wrong or i am wrong somewhere in the probability theory. quiller argued that Christopher is right despite my logical and simple explanations. i did not read the book then. now that i have, i will explain it using the same block diagram that was used in the book.

the author says that if you change your choice after one of the door-not-having-car is revealed, the chances of winning are 2/3 as opposed to 1/3 if you insist on your initial choice. for this he has given the six outcomes (2 for each door, one – you if you stick to the same choice and two – if you change). the outcomes when you change your choice are highlighted in the picture. so there are two outcomes that will get you a car out of three when you changed your choice.

Monty Hall Problem

lets say i chose Door/Path X, and later changed my choice – the outcome says that i will get a car – because the author is eliminating the possibility of selecting the Door/Path Y because the host has revealed that Door/Path Y doesn’t have a car. if the host has not shown the Door Y, then even i change my choice i am not sure if i will get a car and the probability is ½. the author has eliminated either Door X or Door Y (because they have goat) to get the outcomes row (in the diagram).

so when you are finally calculating the probability shouldn’t you exclude one of the paths/doors? i will show how the diagram should look like in all the possible cases after making the second attempt.

i chose Door Path X and using the information that Path Y is no longer contender i have the outcome row. the final probability should not involve the path Y. now you see (Path X or Path Z, you changed your choice are not – the probability is ½)

Monty Hall Problem

i chose Door Path Y and using the information that Path X is no longer contender i have the outcome row. the final probability should not involve the path X. Now you see (Path Y or Path Z, you changed your choice are not – the probability is ½)

Monty Hall Problem

i chose Door Path Z and using the information that Path Y is no longer contender i have the outcome row. the final probability should not involve the path Y. Now you see (Path X or Path Z, you changed your choice are not – the probability is ½)

Monty Hall Problem

i chose Door Path Z and using the information that Path X is no longer contender i have the outcome row. the final probability should not involve the path X. Now you see (Path Y or Path Z, you changed your choice are not – the probability is ½)

Monty Hall Problem

there you go – the author is confusing the readers with successive probability and probability after the information has changed – the probability of choosing the correct door is ½ (if you think the problem is a fresh one after you got more information) .

there are so many ways to prove that the author is wrong. if we go back to basics of probability – you might have read simple problems like there are 10 green balls and 10 white balls, what is the probability of picking 2 green balls in successive attempts.
The answer is 10/20 * (10-1)/(20-1) = 10/20 * 9/19 because after you pick one green ball the information has changed, there are only 19 balls and out of which only 9 are green!

So if we go to our car-goat problem the probability is 1/2 as the information has changed before the second attempt and it is fresh problem altogether

i rest my case

p.s. if any of you find that i went wrong somewhere in the probability theory … do let me know … else i will stick with this logic and the probability of not changing my views is 1

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