Bayesian probability and monte carlo simulation

[gkathotia]

Has anyone got a clue to this question?

Monte Carlo Game and Bayesian Probability.


You must be aware of the Monte Carlo Game or its Hindi equivalent the ‘Khulja SimSim’. Briefly, a player faces three doors and behind any one of the doors is a treasure. The player has to choose a door and if the choice is correct the treasure is the player’s; if not player loses nothing.

The host knows exactly where the treasure is. The routine is that when the player chooses a door, the host either opens the chosen door or prolongs the game. If the host chooses to open the door then whatever is behind the door is the player’s gain. It may or may not have the treasure. The game stops.

But usually the host will prolong the game. The host will prolong so by opening any one of the other two (un-chosen) doors. The door opened by the host will always be the one that does not have the treasure. The host then will offer the player to shift the choice. For e.g. consider a player having chosen door A; the host may prolong the game by opening door B (that the host knows not to have the treasure) and the host will offer to the player the choice of shifting from A to C.

Please note that regardless of the fact that the first choice of the player is right or wrong (e.g. A in this case has or has not the treasure), the host may prolong the game by opening a door that does not have the treasure and offer the player a chance to shift.

The question is what should the player do? Should the player retain the original choice A or shift to C?

The answer of course is that the player should shift every time an offer is made. The reason lies in using Bayesian Probability Theorem.

Prove that the player should switch by using Bayesian Probability Theorem.


Waiting for answers....

Gk

[Psychodementia]

Can you rephrase that in English ? Greek and Latin were never my pet subjects

Jokes apart, another realization that I'm not your "quant" types, seriously looking forward to people contributing to this thread.

Arun

[bingo]

Hi gaurav kathotia
man!! where were u so long? i spoke to bharatram a coupla days back and enquired about you and sujan, howz u guys doing? Its been ages man!!!!!!!!!!
anyways, im no quant stud, so dont understand much of baysean and all, too bad at it.But ya, logically, here is the answer....
Suppose you chose A and C is the correct answer. Before I reveal that B is incorrect, you have a 33% chance of getting the right answer if you stuck to A. But the moment I tell you that B is incorrect, the probability changes. Now this is extremely counter-intuitive. Your mind is thinking along the following lines:

"How can the probability change because of what he is telling me? I chose A and that's 33% likely to be correct, and this was the case even before he told me that B is incorrect. So there is no change in status of my choice. So, I'll stick to my choice."

The problem with the above line of thinking is that it ignores what happens to the other choice - of switching to C. While it's true that not changing choice from A will not make a difference to its chances of being correct, its also true that changing the choice from A to C, after new information has been revealed, doubles the chances of switching being the right strategy.

In other words, you're focusing on your choice of A whose chances have not changed. But you're ignoring that the other choice of switching to C has become twice as likely to be correct as A.
The key thing to know is that while new information may not have any impact on your prior choice, it may make something else far more attractive, thus making your prior choice, relatively unattractive.

[D e e p t i]

this was my line of thought - the treasure is in one of A, B n C.

=> 33% of each being corrent

once its not there in B, its in A or C

=> 50% in each.

Whats wrong with this line of thought?

[poltergeist]

hi
this is a pretty old puzzle.
the actual version of the puzzle is called the Monty hall Puzzle....
i had the Monte Carlo Simulation codes for the puzzle in my old comp, but lost them when my hard drive crashed..
the monte carlo iteration worked pretty well provided the mathamtical modelling tool u use is pretty accurate.
the probability of winning increases to 2/3 if u switch choices and is only 1/3 if u stick to the old choice.

check this link out...
it has a Bayesian explanation....
http://exploringdata.cqu.edu.au/montyexp.htm

there is a research paper by MIT which gives the full explanation..will post it if i can get hold of it.

regards,
rakesh

[gkathotia]

Hey Bingo

Nice to hear from you . Though i have not been much active in this forum.,.i do visit sometimes. I have also been posting threads relevant to some XLRI issues.

Met Bharath a couple of days ago in the XL-IIMC meet and boy did he render " O Humdum suniyo re" in Gusto. Good voice! I won the Table tennis match but we lost the TT event. The best thing however was that we drubbed IIMC 13-8 altogether and the partying hasn't stopped yet.

Regards
Gaurav

[khanna_sumit]

Originally Posted by gkathotia

Has anyone got a clue to this question?

Monte Carlo Game and Bayesian Probability.


You must be aware of the Monte Carlo Game or its Hindi equivalent the ‘Khulja SimSim’. Briefly, a player faces three doors and behind any one of the doors is a treasure. The player has to choose a door and if the choice is correct the treasure is the player’s; if not player loses nothing.

The host knows exactly where the treasure is. The routine is that when the player chooses a door, the host either opens the chosen door or prolongs the game. If the host chooses to open the door then whatever is behind the door is the player’s gain. It may or may not have the treasure. The game stops.

But usually the host will prolong the game. The host will prolong so by opening any one of the other two (un-chosen) doors. The door opened by the host will always be the one that does not have the treasure. The host then will offer the player to shift the choice. For e.g. consider a player having chosen door A; the host may prolong the game by opening door B (that the host knows not to have the treasure) and the host will offer to the player the choice of shifting from A to C.

Please note that regardless of the fact that the first choice of the player is right or wrong (e.g. A in this case has or has not the treasure), the host may prolong the game by opening a door that does not have the treasure and offer the player a chance to shift.

The question is what should the player do? Should the player retain the original choice A or shift to C?

The answer of course is that the player should shift every time an offer is made. The reason lies in using Bayesian Probability Theorem.

Prove that the player should switch by using Bayesian Probability Theorem.


Waiting for answers....

Gk

hi,

i was just checking the threads and found this question....i dont like baysian thm.....but i know one simple tool for finding the probability......p=fav. outcomes/total outcomes.....where all outcomes are equilikely to happen......
assume there are 3 doors named as A,B and C......treasure(lets say car) lies behind one of these doors.....i may choose one out of the three doors......
for one particular case where car is kept in a specified door......lets say A.....i may choose A,B or C....these three cases will be equilikely...lets say i'll always switch...for case 1 when i choose A, host can open any one of the other two doors for me, which are both empty.....switching door means i get and empty door i.e. i loose......in second case if i choose B, since car is behind A, host can open only door C which is empty, i'l switch and choose A, which contains car and i win....third case is same when i choose door C, in this case again only door B can be opened, i will switch and choose A and win again......so in these 3 cases if i switch always then i will win 2 times out of three times.....similarly if car is behing B or C, there will be three cases in each of these selections from which i'll win in two cases and loose in one....so i'll win 6 times out of 9 times if i decide to switch always.......so prob of winning if i switch is 6/9=2/3

[Eccentric]

Originally Posted by gkathotia

Has anyone got a clue to this question?

Prove that the player should switch by using Bayesian Probability Theorem.

Waiting for answers....

Gk

No need you thms..

Say I have 100 boxes. Only one of them has gift. I offer you a choice that either you are "First" means you come to me pick a box and Go to himalayas and live happily thereafter !
Or, you can be "2nd" when you come to pick one of remaining 99 boxes, I will open in fornt of you 98 empty boxes, you will take the 99th box and go to Niagara Falls.

which one you want to be . Obviously "2nd" ... that has 99/100 probability of picking the gift !


In the original case we are dealing with 3 boxes instead of 100 and switch / "2nd" has 2/3 probability for gift. what say 100 is too easy to visualize isn't it.

[meurly]

Originally Posted by gkathotia

Has anyone got a clue to this question?

Monte Carlo Game and Bayesian Probability.


You must be aware of the Monte Carlo Game or its Hindi equivalent the ‘Khulja SimSim’. Briefly, a player faces three doors and behind any one of the doors is a treasure. The player has to choose a door and if the choice is correct the treasure is the player’s; if not player loses nothing.

The host knows exactly where the treasure is. The routine is that when the player chooses a door, the host either opens the chosen door or prolongs the game. If the host chooses to open the door then whatever is behind the door is the player’s gain. It may or may not have the treasure. The game stops.

But usually the host will prolong the game. The host will prolong so by opening any one of the other two (un-chosen) doors. The door opened by the host will always be the one that does not have the treasure. The host then will offer the player to shift the choice. For e.g. consider a player having chosen door A; the host may prolong the game by opening door B (that the host knows not to have the treasure) and the host will offer to the player the choice of shifting from A to C.

Please note that regardless of the fact that the first choice of the player is right or wrong (e.g. A in this case has or has not the treasure), the host may prolong the game by opening a door that does not have the treasure and offer the player a chance to shift.

The question is what should the player do? Should the player retain the original choice A or shift to C?

The answer of course is that the player should shift every time an offer is made. The reason lies in using Bayesian Probability Theorem.

Prove that the player should switch by using Bayesian Probability Theorem.


Waiting for answers....

Gk

If you want a very entertaining solution to this problem go read MARK HADDON's "THE CURIOUS INCIDENT OF A DOG AT NIGHT TIME"

You will get solution to this and many other interesting problems. And also its a great book

[the_Game]

Just a thought:
Suppose we face a similar situation in the CAT (the gameshow ), where:
Out of the given four choices for a question, we (the host and the player) have to (God forbid) "guess" ,

We've made a guess, and then eliminated one choice after that.
It would **not** make a difference whether we switch our choice - reason being, ofcourse, the probability of choosing any of the other 2 options remains the same, because we **dont know** the correct answer at the time of choosing