Monty Hall Problem

[spiderweb]

Stick with the door you chose. Your odds just increased that you initially picked the right door as the host has eliminated one of the other possibilities. At this point it’s still a 50% chance.

Yep, after my initial response I had more time to think about it and switch sides. It is beneficial to switch doors. The first time you are 66% likely to have picked the wrong one. Eliminating one door doesn’t change that initial probability chance for it being the wrong door, but it does make the other door more likely.

Good puzzle.

 

[Xystus]

But yes, you are right - the goat is preferable anyway. No more mowing.

maybe nitpicking but goats are browsers, not grazers. you would more likely have your bushes/shrubs/ trees eaten than the grass.

 

[Tarsier]

Here’s another fun one, though it is not really probability, but just logic:

On an island, there are 96 people who have blue eyes, and the rest of the people have green eyes (though none of them knows there are 96, as nobody knows the color of his own eyes). At the start of the puzzle, no one on the island ever knows his own eye color. By rule, if a person on the island ever discovers he has blue eyes, that person must leave the island at dawn; anyone not making such a discovery always sleeps until after dawn.

On the island, each person knows every other person’s eye color, there are no reflective surfaces, and there is no discussion of eye color. At some point, an outsider comes to the island, calls together all the people on the island, and makes the following public announcement: “At least one of you has blue eyes”. The outsider, furthermore, is known by all to be truthful, and all know that all know this, and so on: it is common knowledge that he is truthful, and thus it becomes common knowledge that there is at least one islander who has blue eyes. The problem: assuming all persons on the island are completely logical and that this too is common knowledge, what is the eventual outcome?

 

[Tarsier]

No takers, which is fine. It’s a tough one. It helps to reduce the puzzle and start with a smaller population of blue-eyes, probably beginning with one.

Anyway, the answer is that all 96 of the blue-eyed islanders leave at dawn on the 96th day.

 

[Theo520]

Once the host opens either door B or C, the entire 66.7% probability transfers to the other door, so you are left with a 33.3% probability if you stick with A or a 66.7% probability if you switch to whatever door is left unopened.

No, you are making a new choice, and the odds are 50/50 each way.
Choosing to stay with your first choice is still a choice, with equal odds.

 

[Tarsier]

No, you are making a new choice, and the odds are 50/50 each way.

Choosing to stay with your first choice is still a choice, with equal odds.

That’s not how probability works. It is never as simple as saying that, with two choices, there is an equal chance of each.

Sometimes one choice outweighs the other in a probability that precedes your choice and continues in spite of it (and, to be clear, one’s “choice” never influences the statistical probability of the same action it is choosing, only the psychological probability).

Anyway, as pointed out above, every single simulation (computer or otherwise), run a sufficient number of times, will show that the switch gives you 66.6% probability.

I suppose it is the way that the proposition is presented that causes the problems, which should be a good lesson for those of us who evangelize - be careful of how one presents truth!

Let’s suppose the problem is presented this way:

You are given a choice of three doors, and you choose A.

Host says, “Hey, you can stick with A, or I’ll let you have BOTH B and C.”

You: “That’s great. My odds are double with both B and C. I’ll take that deal.”

Host: “Fantastic. By the way, the rules are that you can only open one door, and you only get the prize from the door you open.”

You: “Oh …”

Host: “Tell you what, I know what is behind both those doors. We both know at least one of them has a goat, right? Would you like me to open that one for you so you can open the other?”

Same scenario. Worded differently.

 

[Theo520]

Anyway, as pointed out above, every single simulation (computer or otherwise), run a sufficient number of times, will show that the switch gives you 66.6% probability.

lol, that all depends on the programming, doesn’t it.

What I said was true, for the second choice, the odds of that choice are 50/50. I was referring to that single choice while you are referring to cumulative odds.

 

[Tarsier]

lol, that all depends on the programming, doesn’t it.

What I said was true, for the second choice, the odds of that choice are 50/50. I was referring to that single choice while you are referring to cumulative odds.

No, it doesn’t depend on programming. That’s why I said “every single” simulation of sufficient size, including non-computer, inviting you to run your own. Create three doors. Have a friend put paper goats and cars behind them. Have that same friend, aware of which prize is behind which door, eliminate one of the doors you’ve not chosen. Do that 100 times.

You are correct that I’m not referring to a single choice, absent of the first choice, because I’m referring to the actual wording of the puzzle we are discussing. The puzzle makes reference to the fact that this is a two-part event. You were only 33.3% likely to have guessed correctly in the first place. The car is most likely behind one of the other two. The host, who knows what hides behind them, is just eliminating a definite wrong choice.

Here are the nine possible outcomes, 2/3 of which favor switching:

You pick door A - Car is behind Door A - Host reveals Goat behind Door B or C - You should STAY for the car.
You pick door A - Car is behind Door B - Host reveals Goat behind Door C - You should SWITCH for the car.
You pick door A - Car is behind Door C - Host reveals Goat behind Door B - You should SWITCH for the car.
You pick door B - Car is behind Door A - Host reveals Goat behind Door C - You should SWITCH for the car.
You pick door B - Car is behind Door B - Host reveals Goat behind Door A or C - You should STAY for the car.
You pick door B - Car is behind Door C - Host reveals Goat behind Door A - You should SWITCH for the car.
You pick door C - Car is behind Door A - Host reveals Goat behind Door B - You should SWITCH for the car.
You pick door C - Car is behind Door B - Host reveals Goat behind Door A - You should SWITCH for the car.
You pick door C - Car is behind Door C - Host reveals Goat behind Door A or B - You should STAY for the car.