[xzibit2008]

Am having truble figuring this two questions out.Any help will be apreciated
Question 1
there are three doors from which to choose. Two have goats behind them and the remaining one a pile of $1,000,000. The host selects randomly and shows you one door. You can now change your mind, should you? If you do, what is the probability of picking the door with $1,000,000?

Question 2
poker game. Poker = 52 cards.
if cards are randomly selected, what is the probability of selecting 3 cards of 1 type and 2 cards of another type?
(all cards are randomly selected from the 52 cards)

[kangabitt]

There are two doors. What are your odds of picking the one you want?

[ebaines]

Question 1
there are three doors from which to choose. two have goats behind them and the remaining one a pile of $1,000,000. The host selects randomly and shows you one door. You can now change your mind, should you? If you do, what is the probability of picking the door with $1,000,000?

This seems to be a modification of the famous "Monty Hall Problem," but in the original version of that the host purposesly opens a door that he knows is a goat, and then you are asked if you want to change your mind. The answer to that one is yes - it always improves your odds of winning the money. But in this case you say the host randomly selects which door to show you. My question for you is: suppose the host opens the door that has the $1M behind it - by these rules are you allowed to change your mind to select that door, or are you restricted to only the doors still closed? Obviousy if you're allowed to select the door that you know has the $1M that would always be a winning strategy. But if he randomly opens a door and there's a goat behind it, then it doesn't matter whether you change your guess or not.

Question 2
poker game. poker = 52 cards.
if cards are randomly selected, what is the probability of selecting 3 cards of 1 type and 2 cards of another type?
(all cards are randomly selected from the 52 cards)

Define what you man by "type." Do you mean 3 cards of one suit and 2 of another suit, or do you mean 3 cards of one face value and 2 of another?

[morgaine300]

You're contradicting yourself. If Monty opens a door he knows is a goat, versus opening the door and it just happens to be a goat, doesn't make any difference. The point is, it's a goat. And the two doors left have one goat and one money. Since the open door cannot be "repeated," we're now down to two doors. 50/50 chance. Under the assumption that the open door isn't the one you picked, changing doors doesn't increase or decrease the chances that your picked door is the correct one. i.e. your last statement. How does Monty's knowledge change that?

I would never change doors. It was a silly tactic on the show's part.

[ebaines]

Sorry Morgaine - you are forgetting that in the classic Monty Hall problem the host knows which door has a goat - his opening a door to show a goat is not random. When you;'re ddown to 2 doors left it s not 50/50 chance, because Monty purposely selected which door to leave closed. You will agree I'm sure that your chance of picking the correct door at first is 1 out of 3. The fact that he shows you a door with a goat doesn't change that - he can always show you a door with a goat. But since after opening one door there are only 2 doors left, that means that you have a 1/3 chance of winning if you stay with your original guess and a 2/3 chance of wining if you change to the door that is still closed.

Think through the possibilities here. You have a 1/3 chance of the prize being behind door 1, door 2, or door 3. Suppose you originally pick door 1. If the prize is indeed behind door 1, then Monty will open up either door 2 or 3,(doesn't matter which) to reveal a goat, and if you follow my strategy and change guesses, you lose. That's a 1/3 chance of losing. But if the prize is behind door 2, Monty will open door 3 (the one that he knows has a goat), you change your guess to door 2, and you win. Same thing if it's behind door 3 - he opens door 2, you change your guess to door 3, and you win. Hence your chance of winning with this strategy is 2/3. If you stuck with your original guess your chance of winning is only 1/3.

A thought experiment to try that will really drive this home is the following: suppose there were 100 doors, 1 with the $1M prize and the other 99 all with goats. You have only a 1:100 chance of picking the correct door, correct? Now after you make your guess Monty opens up 98 of the remaining doors to reveal 98 goats, leaving just your original guess with its door closed plus one other. Rememnbr - he knew which 98 doors to open to reveal goats, so he will always leave you wiyth your pick plus one other door left unopened and the $1M behind one of them. Do you change your pick? Of course you do, because that way you'll win 99 out of 100 times.

[morgaine300]

Well, I didn't "forget" anything because I never learned the classic Monty Hall thing to begin with.

I'm following your logic of the 1/3 and the 2/3, but something in my brain is not comprehending why it should come out that way. Obviously I'm missing some important tidbit in there that is making his knowledge of the doors be different than someone else who doesn't know and just happens to pick those doors. That's one thing annoying about probabilities -- sometimes things that sound difficult are quite simple. And sometimes things that sound simple are more difficult than they seem, because there's some little piece of something in there making some kind of difference. And I think this is one of those cases.

But I will ponder on it. Otherwise it will drive me crazy. :-)