Game show conundrum - better to change choice or stick with original pick?

I first ran into this while working as a programmer, so I was surrounded by a large number of professionals with a mathematical education, most of whom didn't accept the answer. Several of them went back to their desks and wrote simulation programs to run through a large number of examples and produce a statistical evaluation. All of them then agreed the original answer, which they had disputed, was correct.

The problem (originally called the" Monty Hall problem" after the host of "Let's Make a Deal"--though I see people who know that don't know any more how to spell his name) is based on the way the show operated. Monty obviously did know which box had the prize, because in many, many cases like this, he never revealed the prize at the wrong time. (Of course, no show would want the suspense killed, so they made sure he didn't.) But if he simply revealed a random unchosen box, and it turned out not to have the prize, a contestant would be no better off switching. The instinctive response, which is that you don't want to switch, is probably based on the fact that if you DID switch and found you'd switched AWAY from the one with the prize, you'd feel a lot worse about losing.

Here's another try at convincing you of the correctness of the solution: Your initial choice has a 1/3 chance of being the winner. Because Monty Hall isn't allowed to reveal what's in your box prematurely, that's still your chance of being the winner after the revelation. Because he doesn't ever reveal the real winning box, the one he revealed has no chance of being the winner, and after the revelation, you know that. There are two choices left: yours, with a 1/3 chance of winning, and the one you could switch to. But it's a certainty that there's a prize somewhere (another rule of the game, as played by Monty), so the chance of winning by switching must be

1 - 1/3 = 2/3

and you've got twice as good a chance by switching.

In the many-boxes problem, your best strategy is to stick with your original choice, while Monty repeatedly reveals that other boxes aren't winners, until there are just two left--then switch! The reason is that the more boxes there are to choose from, the smaller the chance that the chosen one is the winner, so by sticking with one box until the last chance to change, you can keep that choice's chance of being the winner as small as possible. If there were 22 to start with, you'd be hanging onto that choice because its chance of winning is 1/22, and that means that when we get down to two boxes left, the other one would have a

1 - 1/22 = 21/22

chance of being the winner. Again, this depends on the host following Monty Hall's rules: on each step, always reveal a box that is NOT a winner, and then offer you another chance to change your mind.

If this still doesn't convince you, the simulations are easy to program, and they have always proved effective at convincing their programmers.

Rather than thinking about the odds of picking the right box, what are the odds of picking the wrong box? When the first choice of 3 is made (A), the likelihood is 66.7% to pick an empty box. So the player is more than likely sitting on an empty box. Now the host takes away an empty box (C) leaving the one previously chosen (most likely empty) and another one (B). Because the one the player initially chose is most likely empty, and the one already taken away is definitely empty, then the other one (B) is most likely to have the prize.

Next query re the host knowing or not knowing which one is empty - it doesn't matter because there are only three boxes. If the host reveals the prize then the game is over. If the host shows an empty box, then the same odds are still in place. Whether the host chooses by random or by prior knowledge, the only thing that is important is that the box that is taken away is empty. If it has the prize in it, then the player cannot win as the prize has gone and changing boxes makes no change to the outcome.

Last section in the case of deal or no deal. His likelihood of selecting the big prize is 1 in 22. Or the other way around it's 21 in 22 , very likely not to have selected the big prize. If the other boxes are being dwindled out by chance then the other last box is equally unlikely to have the big prize. But if the boxes are removed with knowledge to avoid the big prize, then swap as there is a much greater chance of the prize being in the other one than in the player's!

This is the monty hall problem.

When the contestant picks a box at random he has one chance in 3 of picking the right one.

If he does then the game show person opens a box at random and if the person switches he will always lose: Therefore he will lose 1/3 of the time if he switches.

The contestant has 2 chances in three of picking the wrong box.

If he does then the game show host MUST open the sole remaining wrong box leaving the jackpot box as the hidden one. The contestant therefore MUST win if he switches. Thus he will win 2/3 of the time if he switches.

With regards to 2) it depends on whether the choices are forced or not! If they are then the contestant has a 95% chance of winning if he switches (i.e. he only wins by sticking if his original choice was the winning box) If the choices are at random then it is 50-50.

You have practically answered your own question,it doesnt defy logic but it is a tricky concept to get a grip on.Look at your question and you will see why the principle is correct.

With 22 doors it is 1/22 p for the first choice.

With only 2 doors left ,ie your first choice and the remaining door left unopened by the wretched Monty Hall,the probability of the last door having the car behind it is 1/2.

I am going to have a beer ,I hate that man Hall..............

EDIT

You have 3 doors after you have chosen 1 there remains

goat car

car goat

goat goat

as possibilities for the remaining 2 doors.

Dear Monty is honour bound to show you a goat which leaves you with possibilities

car

car

goat

2/3.

No?

>> All explanations say that he is better to swap, although I don't accept the explanation! <<

The theory is that your first pick was most likely wrong, so you should change.

That said, I'm with you, as the above is faulty logic (in addition, the host ALWAYS knows which box to reveal as a wrong choice).

With 2 boxes remaining, the odds are 50/50, and WHEN you made a choice doesn't matter.

Let's suppose you flip a coin 3 times and got a heads all three times.

What are the odds your next flip will be heads.

Answer: 1/2. Future flips do NOT depend on flips in the past.

this is an old one!

the mc has a FORCED choice when one box has prize and other doesnt. This {after doing the appropriate math} gives a 2:1 favor for the switch

It makes no difference whether you accept the explanation or not. In the Monte Hall Problem, it's always better to switch your selection after Monte reveals an empty box.

Your acceptance or non-acceptance of that fact does not change reality. Google the Monte Hall Problem and study the published information. Repeating it here is a futile exercise.

1) Yes

2) Yes