Variants on Monty Hall...the Problem, that is

[nitrah55]

http://tierneylab.blogs.nytimes.com/200 ... ex.html?hp

I'll post my answers shortly. By the way, as fascinating as I find this, I'm lousy at coming up with the right answers. Kind of like my relations with women.

[nitrah55]

Spoiler

1. Odds are 50/50 that the other kid is a boy.

2. I think you should switch because the odds are 9 in 10 that the car is behind 5 or 8. They should call this the "Howie Mandel Problem."

3. Flipping the coin reduces the odds to 1 in 3.

I could really be wrong about this.

[MarleysGh0st]

Spoiler

1: 33.3%

2. Odds of car behind door #1 = 80%, 10% behind door #5 and 10% behind door #8. (This is just a general hunch, extrapolating from the classic three door problem.)

3. 50%

[TheCalvinator24]

Spoiler

1. 1/3

Assuming 50/50 sex distribution, there are three possible combos:

BB
BG
GB

Of the three, only one results in the "other" child also being a boy.

(Out of the three problems, this is the one I am least sure of)

2. Should switch. Should win 45% of the time as opposed to only winning 10% by sticking

3. 50% 1/2*1/3 + 1/2*2/3 = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2

[MarleysGh0st]

Spoiler

1: 33.3%

2. Odds of car behind door #1 = 80%, 10% behind door #5 and 10% behind door #8. (This is just a general hunch, extrapolating from the classic three door problem.)

3. 50%

Spoiler

Doh! That's what I get for rushing the answer.

Reading nitrah's answer, I see my odds are completely reversed. 10% behind door #1, 45% behind #5, 45% behind #8.

[slam]

Spoiler

1: 1/3 (approximately, because chances of a girl and a boy are not really equal).
His children could be (older/younger): B/B, B/G, G/B (we've eliminated G/G). Those three possibilities are approximately equal. So, it's only 1/3 that his other child will be a boy.

2: 90%: Chances of the door you originally picked having the car is 10%. Therefore, the car is behind the other doors 90% of the time. Since you have no way of distinguishing betweed doors 5 & 8, it's 45% for each of them.

3: 1/2

If you don't switch, you have a 1/3 chance of winning. If you do switch, you have a 2/3 chance of winning. You have a 1/2 chance of switching and a 1/2 chance of not switching. So, your total chances are:

1/2 = (1/2)(1/3) + (1/2)(2/3)

[slam]

Looks like Calvinator and I agree exactly down to which problem we are least sure about. I think there's a bit of a semantic problem in the way the first one is phrased, but I'll stick with my answers.

[silverscreenselect]

Here's a variant on the first extra problem. There are three coins in a bag, a two-headed coin, a two-tailed coin and a normal coin. You take one coin out of the bag at random and put it face up on the table. It is heads. What are the odds that the other side of the coin is also a head?

Spoiler

2/3.

Let's call the two heads of the two headed coin H1 and H2, the two tails of the two tailed coin T1 and T2 and the normal coin H and T.

Originally, there were six possibilities for what you could have placed face up on the table, H, H1, H2, T, T1, and T2. But since you know there's a head showing, that leaves three, equally probable possibilities : H, H1 and H2. Two of these possibilities are the two-headed coin, so the odds that the other side of the coin is also a head are 2/3.

You can try this by taking three pieces of paper and putting an X on one or both sides of the paper. 2/3 of the time, whatever you see on the first side will also be on the second side.

[MarleysGh0st]

Spoiler

2. I think you should switch because the odds are 9 in 10 that the car is behind 5 or 8. They should call this the "Howie Mandel Problem."

Spoiler

But this is not like DoND in that Howie isn't the one opening cases, carefully using some prior knowledge to not reveal the $1 million case until the end. The contestants do so, usually "spoiling" the game at random.

[Bob Juch]

Spoiler

1. 1 in 3
2. 9 in 10
3. 1 in 2

[ToLiveIsToFly]

Spoiler

1. Mister Smith has two children, at least one of whom is a boy. What is the probability that the other is a boy?
1/3 - four initial possibilities: boy + boy, boy + girl, girl + boy, girl + girl. Last possibility is eliminated, so one of the other 3 situations has a boy in it.

2. John Allen Paulos, a mathematician at Temple University, has proposed a 10-door version of the Monty Hall Problem. Once again, there’s a car behind one door and goats behind all the others. You pick one door, but before it’s opened Monty will always open seven other doors to reveal goats. Then he’ll give you a chance to stick with your original door or switch to either of the remaining two unopened doors. Suppose you originally chose Door 1, and Monty opens every other door except Door 5 and Door 8. Should you switch to 5 or 8? And what are the odds the goat is behind either 5 or 8?
You should switch. There's a 90% chance initially that the goat is behind a door that you didn't pick. So when he's done showing you goats, there's a 90% chance the car is behind 5 or 8. So 5 is 55% goat, 45% car, 8 is 55% goat, 45% car, 1 is 90% goat, 55% car. There's a 90% chance the goat is behind either 5 or 8.

3. One unconvinced Lab reader, Scott, insisted yesterday that the best way to play the classic three-door Monty Hall game was to simply flip a coin. (I told him I’d be glad to let him bring his coin, and some bills, so we could play the game for money — with him flipping a coin and me using the switching strategy.) He’s wrong about the best way to play the game, but his strategy does raise a question: Suppose, when you’re confronted with the final choice to stick or switch doors, you flipped a coin — heads, you stick with your original door; tails, you switch. What would be your odds of winning the car?
He's got a 2/3 chance of winning if he flips tails and a 1/3 chance of winning if he flips heads. So he's got a 2/3 * 1/2 + 1/3 * 1/2 = 50% chance of winning

[nitrah55]

Cal was the first one in with all 3 answers correct. Marley, haste makes waste.

These are the correct answers:

1. The odds that the "other" kid is a boy are 1 in 3.

2. You should switch, the odds that the other two doors hold the prize are 9 in 10.

3. Coin flip reduces the odds to 1 in 3.

Here is a link to an exhaustive explanation of why, using "common sense," we tend to get these things wrong.

http://tierneylab.blogs.nytimes.com/200 ... ex.html?hp

And, unless I appear on Lets Make a Deal sometime soon, this is the last of my Monty Hall posts for a while.

[MarleysGh0st]

Cal was the first one in with all 3 answers correct. Marley, haste makes waste.

Yes, it does.

[megaaddict]

Spoiler

1. Mr. Smith has two children, at least one of whom is a boy. What is the probability that the other is a boy?

The two children are one of the following: both boys, an older brother and a younger sister, or an older sister and a younger brother. The other sibling is a boy in only one of these three equally likely possibilities.

2. John Allen Paulos, a mathematician at Temple University, has proposed a 10-door version of the Monty Hall Problem. Once again, there’s a car behind one door and goats behind all the others. You pick one door, but before it’s opened Monty will always open seven other doors to reveal goats. Then he’ll give you a chance to stick with your original door or switch to either of the remaining two unopened doors. Suppose you originally chose Door 1, and Monty opens every other door except Door 5 and Door 8. Should you switch to 5 or 8? And what are the odds the car is behind either 5 or 8? [Update: I originally wrote “the goat” instead of “the car” in that previous sentence, but I’ve corrected it. Thanks to Harris for noting the mistake.]

Only a 10% chance that the car is behind door 1. So there's a 90% chance it's behind either 5 or 8. With no other information, it is equally likely to be in one or the other, so both 5 and 8 have a 45% chance of revealing the car.

3. One unconvinced Lab reader, Scott, insisted yesterday that the best way to play the classic three-door Monty Hall game was to simply flip a coin. (I told him I’d be glad to let him bring his coin, and some bills, so we could play the game for money — with him flipping a coin and me using the switching strategy.) He’s wrong about the best way to play the game, but his strategy does raise a question: Suppose, when you’re confronted with the final choice to stick or switch doors, you flipped a coin — heads, you stick with your original door; tails, you switch. What would be your odds of winning the car?

50/50. Each heads gives you a 1/3 chance of winning. Each tails gives a 2/3 chance. Overall chance is the average of these, 1/2.

[slam]

Cal was the first one in with all 3 answers correct. Marley, haste makes waste.

These are the correct answers:

1. The odds that the "other" kid is a boy are 1 in 3.

2. You should switch, the odds that the other two doors hold the prize are 9 in 10.

3. Coin flip reduces the odds to 1 in 3.

Here is a link to an exhaustive explanation of why, using "common sense," we tend to get these things wrong.

http://tierneylab.blogs.nytimes.com/200 ... ex.html?hp

And, unless I appear on Lets Make a Deal sometime soon, this is the last of my Monty Hall posts for a while.

For #3, I think you meant to say that the coin flip reduces chances to 1 in 2 from 2 in 3.

[slam]

This thread also brings up the fact that people are generally very sloppy about the distinction between odds and probability.

Let's say an event happens 1 time out of 4. The odds that this event will happen is 3 to 1 (also written as 3:1). The probabilty (or similarly, chances) that this even will happen are 1 out of 4 or 25%.

[megaaddict]

Here's a variant on the first extra problem. There are three coins in a bag, a two-headed coin, a two-tailed coin and a normal coin. You take one coin out of the bag at random and put it face up on the table. It is heads. What are the odds that the other side of the coin is also a head?

Spoiler

2/3.

Let's call the two heads of the two headed coin H1 and H2, the two tails of the two tailed coin T1 and T2 and the normal coin H and T.

Originally, there were six possibilities for what you could have placed face up on the table, H, H1, H2, T, T1, and T2. But since you know there's a head showing, that leaves three, equally probable possibilities : H, H1 and H2. Two of these possibilities are the two-headed coin, so the odds that the other side of the coin is also a head are 2/3.

You can try this by taking three pieces of paper and putting an X on one or both sides of the paper. 2/3 of the time, whatever you see on the first side will also be on the second side.

Just a rec because I nailed the original Monty Hall variations but self-dursted royally on this one.

[nitrah55]

Cal was the first one in with all 3 answers correct. Marley, haste makes waste.

These are the correct answers:

1. The odds that the "other" kid is a boy are 1 in 3.

2. You should switch, the odds that the other two doors hold the prize are 9 in 10.

3. Coin flip reduces the odds to 1 in 3.

Here is a link to an exhaustive explanation of why, using "common sense," we tend to get these things wrong.

http://tierneylab.blogs.nytimes.com/200 ... ex.html?hp

And, unless I appear on Lets Make a Deal sometime soon, this is the last of my Monty Hall posts for a while.

For #3, I think you meant to say that the coin flip reduces chances to 1 in 2 from 2 in 3.

Thanks for being diplomatic as well as correct.