Since I have already made my choice and making the assumption that the door has been opened and I know the result (Lamborghini or bicycle) and knowing that a bicycle was behind the door that the host opened, I would choose to stay with my choice if there was a Lamborghini behind my already open door but switch to the unopened door if I had a bicycle behind my previously opened door.

If the above assumption is wrong and my door choice hasn't been opened I would flip a coin to chose which of the two unopened doors to chose since there is a 50% chance of winning either the Lamborghini or the bicycle. The coin flip would add additional value to me for by adding a small element of excitement to the experience.

The above assumes that the Lamborghini has more value to me.

A fun exercise and it is easy to let your biases run you down wrong corridors (not saying that I haven't!).

Should switch.

Car is behind only one door so initial probability of winning the car = 1/3. That means, 2 out of 3 times switching would win the car. Only time switching doesn't win the car is when initially you had picked the door which had the card behind, ie 1 out of 3 times. Bottomline, host opening an unpicked door to reveal a bicycle doesn't alter the initial probability.

This excellent puzzle seems to be the one that has been around in mathematically similar forms since 1889 when it appeared as Bertrand's Box paradox, popping up again as the 3 Prisoners problem and the 3 Shells problem in 1959 and the "Monty Hall" problem in 1975, probably raised the most ruckus in 1990 when answered by Marilyn vos Savant in a newspaper column. If so then switching doors increases the probability of winning to 2/3.

Of course you need to switch the door. If you understand stuff, you would be able to explain it easily with no complex math behind. Here is the way, I would explain why. Imagine that instead of 3 doors, you had 1000 doors to pick, and you choose one, the probability is very small, so most likely you picked the empty door. Now, the host opens all other doors but one, and you're asked to choose between the one you already picked and the one other that is left. Would you change your selection? I would, because most likely your big price is behind it (with the chance 1- 1/1000). The same reason, but not that emphasized works for 3 doors.

First lets eliminate the game where , after your choice, if the host shows you one of the remaining two doors and you lets you pick either remaining doors, you always win, the information matrix is singular, no game here.

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Now if the host shows you the compulsory alternative, then although all is revealed you do not have access to all outcomes.

Outcomes are bike one and two, B1, B2 and one car C1.

First stage : no info: flat probs at 1/3:

B1 -> fail (value = 0)

B2 -> fail (value = 0)

C1 -> hit (value = 1)

Introduce partial choice, the host selection switch alternative,and there is now 12 outcomes. If you impose that the car is the default selection, you get 8 outcomes with 4 winners. Probability of win is now 1/2 up from 1/3.

Hence the Bayesian conditional information value is not nil.

Q1: what is the Bayesian conditional ?

Q2: what can you say about the determinant ?

Q3: do we have an eigenvalue twist ?

Please tell me where I am wrong. Cheers. Stephen

List of outcomes: ex ante

B1 -> B2 (Shown by host) : switch : get B2 bike (value = 0)

B1 -> B2 (Shown by host) : don't switch : get B1 bike (value = 0)

B1 -> C1 (Shown by host) : switch : get C1 car (value = 1)

B1 -> C1 (Shown by host) : don't switch : get B1 bike (value = 0)

B2 -> B1 (Shown by host) : switch : get B1 bike (value = 0)

B2 -> B1 (Shown by host) : don't switch : get B2 bike (value = 0)

B2 -> C1 (Shown by host) : switch : get C1 car (value = 1)

B2 -> C1 (Shown by host) : don't switch : get B2 bike (value = 0)

C1 -> B1 (Shown by host) : switch : get B1 bike (value = 0)

C1 -> B1 (Shown by host) : don't switch : get C1 car (value = 1)

C1 -> B2 (Shown by host) : switch : get B2 bike (value = 0)

C1 -> B2 (Shown by host) : don't switch : get C1 car (value = 1)

Outcomes set by the car preference axiom:

B1 -> B2 (Shown by host) : switch : get B2 bike (value = 0)

B1 -> B2 (Shown by host) : don't switch : get B1 bike (value = 0)

B1 -> C1 (Shown by host) : switch : get C1 car (value = 1)

or 1/3 success and 2/3 failure

(as opposed to 100% failure with no choice)

B2 -> B1 (Shown by host) : switch : get B1 bike (value = 0)

B2 -> B1 (Shown by host) : don't switch : get B2 bike (value = 0)

B2 -> C1 (Shown by host) : switch : get C1 car (value = 1)

or 1/3 success and 2/3 failure

(as opposed to 100% failure with no choice)

C1 -> B1 (Shown by host) : don't switch : get C1 car (value = 1)

C1 -> B2 (Shown by host) : don't switch : get C1 car (value = 1)

or 3/3

Total outcomes 8 : total automatic desired outcomes 4

Probability of success is 1/2.

So winning outcomes that are automatic by the selection set axiom.

B1 -> C1 (Shown by host) : switch : get C1 car

B2 -> C1 (Shown by host) : switch : get C1 car

C1 -> B1 (Shown by host) : don't switch : get C1 car

C1 -> B2 (Shown by host) : don't switch : get C1 car

If you extend this path, is it ITO compatible ? If so what is the sigma, vega and volta ?

To win the car you have to win 2 coin flips consecutively. You pick heads. Its tails. Now if you stay with heads those of you saying switch would argue your chances are less with heads the second time? Such that what happened on the last coin flip changes the probabilities of the next one?

Entropy

Without doubt, you should always switch...A simple explanation is to divide the choices into 2 groups...The door you choose is group #1 and has a 1/3 probability of having the car. The other doors combined have a 2/3 probability of having the car. In what circumstance would you ever not want to switch between your choice at 1/3 probability and the other one with 2/3 probability?

This is not the same as flipping a coin, as the probability of choosing the car in this case is no longer a 50-50 choice. Remember that two choices are at 50-50 when you know nothing about them. In this case, however, the host is filtering out the bad choice from the original doors. This gives you extra information that makes switching more attractive and worth more than 50% despite it being 2 choices.

3 reduced to 2 and now the probababity is 1/2 and so it doesnt matter staying with the same choice or switch as bcoz selecting a car is 50% and loosing the car is 50% irrespective of staying with the same choice or switch..But I prefer to stay with the same choice even if we loose the mental impact is less than changing choice and loose still..Risk is all about impact

Funny problem, but which way does it have any relationship to risk management in trading?

I'll give an example to prove why you should switch. Say instead you were told to pick a card from the deck but you will only win the car if you pick the ace of spades. You have a 1/52 chance of winning. Now the host gets rid of every other card until he has one left. He then tells you that one of the cards is the ace of spades. (this is basically what he is doing by eliminating the bike). Given that you had a 1/52 chance of winning before the host eliminated the cards, you should always switch as the host won't elininate the ace of spades. This means switching gives you a 51/52 chance of winning. The same is said for the above problem. Switching gives you a 2/3 chance of winning.

Alex Krishtop It's related to risk management because one must realize that probability risk changes even if you choose not to change your decision. Essentially, knowing when to change strategies based on changing conditions.

After the host has opened the door with the bike, we are like in the situation of a new game with only 2 doors. Therefore 50/50. But I will say the first impression is always the best one, so don' t change. You make assumptions based on your experience but don't combine the probability for a future event with the probability of a past event.

No, because it won’t increase my likelihood of winning the Lamborghini.

Using Bayes, P(A|B) = P(B|A) x P(A) / P(B):

P(A) = probability Lamborghini is behind door i, the door I chose = 1/3

P(B) = probability a bicycle is behind door j (before it was opened) = 2/3.

P(B|A) = probability a bicycle is behind door j given Lamborghini is behind door i = 1

Thus, P(A|B) = probability Lamborghini is behind door I, given a Bicycle is behind door j = 1 x (1/3) / (2/3) = 1/2

Joseph Lee nailed this Riddle perfectly. As Joseph describes the probability is not 50/50 as many assume. Excellent response and I really enjoyed the comments. I post these Risk Riddles periodically to show that risk is not as intuitive as we believe. I hope you enjoyed the Riddle and will continue to play!

That's "The Monte Hall" problem, named after a game show host of the same name. Always switch.

@Alex: I understand your reservations, but unlike you wasn't brave enough to raise the issue in public :)

Instead I spent some time reviewing the role of probability in my algorithm development process (since after all these years the process is more cut-and-paste from a toolbox than the skunk works it once was), to what extent I accept Bayesian logic, the extent to which I exploit it and whether opportunities have been missed.

While as Joseph Lee says and James Bone agrees the moral of the famous puzzle so far may be "knowing when to change strategies based on changing conditions", as a bot trader with some insight into the line between a bot decision process and our own and the fact that profitability depends on the bot relentlessly making decisions that we may or may not be equipped to make (hence may not know to code), is anyone willing to offer explained non-backtest examples of profit making that employs the logic :)

The problem being discussed here is a single event problem. Trading, at least systematical, is about series. In terms of this riddle, the question is not whether you switch or not, but whether you're able to guess the right door more frequently in a series.

Switch. The probability was two thirds that the Lamborghini was in the two doors you didn't choose versus one third it was in the door you chose.

Hi Alex,

I recognize a cold call when I see it--the only kind of reaction I respect and the one I hate most to receive.

The only recourse to this this thread that I can accept is all reconvene at a later date to show new evidence (from participants--not the time worn Interwebz variety) that this particular application of Bayesian stats work in trading.

Switching gives 2/3 of chance of winning. Use bayesian inference!

You should change the choice because your probability of winning increases at 50% from 33%!!

If the odds are increased by switching then they are measurable ... so what are the probabilities of winning the car with switching and without switching please? I see some people saying odds are 2 in 3 when switching. I will run all day doing the following game with anyone. We each get a pick. Every time we survive round 1 I will keep my pick, you switch your pick. Evey time I win round 2 you pay me $2. Every time you win round 2 I will pay you $1. If you think you would come out even since you have a 2 in 3 chance of winning then lets play :)

@ Joe Duffy This is the classic Monty hall problem and your probability of winning is increased to 2/3 when you switch. Also, to answer your question this is measurable and easy to prove. Given enough trials, your winning percentage will settle around the 33% mark. You can check the following link that explains it very well and gives you an interactive game application so you can see how it works.

http://betterexplained.com/articles/understanding-the-monty-hall-problem/

If the host looks behind all 3 doors and then opens one to reveal a bicycle, that is an entirely different situation then any one of the 3 doors opened without looking behind them. The question did not indicate that the decision to open one of the doors was made with any insight as to what was behind the doors first. If one of the 3 doors is opened randomly first and its not a car the chances are 50/50 on the other 2.

The law of large numbers concludes that the one who consistently chooses the option with 2/3 probability will perform better than the one who chooses with 1/3 probability. Thus, casinos are very profitable. In fact, this forum is probably geared toward those who use that theorem to isolate anomalies.

The Montey Hall Problem... Under the standard assumptions, contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their choice have only a 1/3 chance.