Yes, Monty Hall again(adsbygoogle = window.adsbygoogle || []).push({});

I read many threads and explanations about it, but there is still something I'm not sure about.

I would just like to check if the approach I use is correct, so if anyone has enough patience to bear with me...

To simplify, let's assume it's 3 boxes, one with a white ball inside and two with a black ball. The prize is the white ball.

You pick one box at random.

The 'bank' (who knows where the white ball is, and that's the crucial detail) opens one box revealing a black ball.

Should you swap your box with the other one?

--> yes, because that doubles your chances of winning (in the long run, doing several repeats).

Classic explanation: as your initial chances of picking the white ball are 1/3, only 1/3 of the time the bank has the two black balls, whereas 2/3 of the time the bank has one white ball and one black ball. So by revealing the black ball the bank is left 2/3 of the time with the white ball, and you should prefer to have that than your own box.

Classic doubt people have: if you remove a black ball from the game, there are two boxes left, one black and one white, so both you and the bank have equal (1/2) chances of having the white ball.

The way I tried to rationalise this using formulae from conditional probability and Bayes is the following.

First, if the black ball were revealed by randomly opening one box, the chances of having the white ball are indeed equal between you and the bank at that point in the game.

Why? Because in that case revealing a black ball after you picked a black one, or revealing a black ball after you picked the white one, have the same chances of happening. Using conditional probability:

P(B_{1}∩ B_{2}) = 2/3 * 1/2 = 1/3

P(W_{1}∩ B_{2}) = 1/3 * 2/2 = 1/3

therefore:

P(B_{1}| B_{2}) = P(B_{1}∩ B_{2}) / P(B_{2}) = P(B_{1}∩ B_{2}) / [P(B_{1}∩ B_{2})+P(W_{1}∩ B_{2})] = 1/3 / [1/3 + 1/3] = 1/2

and symmetrically:

P(W_{1}| B_{2}) = P(W_{1}∩ B_{2}) / P(B_{2}) = P(W_{1}∩ B_{2}) / [P(B_{1}∩ B_{2})+P(W_{1}∩ B_{2})] = 1/3 / [1/3 + 1/3] = 1/2

So your chances of having the white ball given that the box that is opened has the black ball are 1/2 and equal to your chances of having the black ball. In this case, swapping the old box with the new one doesn't change your chances of winning.

What changes in this reasoning if the bank knows where the white ball is and makes sure that the box that is opened is the one with the black ball?

This is where I'm not sure.

My naive interpretation is that in that case you impose the constraint P(B_{2}) = 1.

So, recalculating the above:

P(B_{1}∩ B_{2}) = P(B_{1}) = 2/3

P(W_{1}∩ B_{2}) = P(W_{1}) = 1/3

P(B_{1}| B_{2}) = P(B_{1}∩ B_{2}) / P(B_{2}) = 2/3 / 1 = 2/3

P(W_{1}| B_{2}) = P(W_{1}∩ B_{2}) / P(B_{2}) = 1/3 / 1 = 1/3

proving that your chances of having the white ball when the black one is revealed are half your chances of having the black one, so you should swap boxes.

I applied this method to an extended case where there are initially 3 black balls and 1 white ball, and 2 black balls are revealed.

When I leave everything random, again I get P(B_{1}| (B_{2}∩ B_{3})) = P(W_{1}| (B_{2}∩ B_{3})) = 1/2.

When I impose the condition that the bank knows which boxes to open, P(B_{1}| (B_{2}∩ B_{3})) = 3/4, P(W_{1}| (B_{2}∩ B_{3})) = 1/4, so in that case apparently you would triple your chances of winning by swapping.

Do you think this approach is formally correct?

And can we confirm that in the 'deal or no deal' games, where (supposedly) the boxes are opened at random, there is no advantage in swapping?

Thank you!

L

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# Not Monty Hall again...

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