The Monty Hall Problem Revisited (1 Viewer)

[dickey moe]

So tell me again why this needed to be revisited??

Because Vox said so.

Well when do you plan on revisiting it? Do you disagree with the solution? So far all you've done is post a problem that has already been solved and had a giant thread devoted to it.

I'm interested in opinions in here on this. Feel free to put me on ignore or hang out in one of the werewolf threads.

 

[CBusAlex]

I think we need to revisit -5^2, guys. Players' Tribune magazine said so.

 

[TheIronSheik]

So tell me again why this needed to be revisited??

Because Vox said so.

Well when do you plan on revisiting it? Do you disagree with the solution? So far all you've done is post a problem that has already been solved and had a giant thread devoted to it.

I'm interested in opinions in here on this. Feel free to put me on ignore or hang out in one of the werewolf threads.

God forbid you want to have a fun, thoughtful conversation. HE'S A WITCH!!!! BURN HIM!!!

 

[jon_mx]

The car looked like it was out of the70's. The goat might be a better value.

 

[dickey moe]

I think we need to revisit -5^2, guys. Players' Tribune magazine said so.

Playas gonna play.

 

[Juxtatarot]

Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

 

[SWC]

Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

i do not give a crap give me a goat brohan

 

[jon_mx]

Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

1/3

 

[CletiusMaximus]

So tell me again why this needed to be revisited??

I'd rather revisit the suicidal dotted monks.

 

[Officer Pete Malloy]

I think we need to revisit -5^2, guys. Players' Tribune magazine said so.

I once bookended -5^2 on the 4th row of a 38-man onslaught. My opponent didn't know whether it was positive or negative and forfeited. Tishilkov rewrote the rulebook before the next ConSlaught to disallow this technique as an illegal variant of the Euler Attack.

Tishilkov pretty much burned every bridge with that move. There's a reason why he's banned from both IOF and NAOAC. Of course, if go by the spirit of the law, your move was technically illegal. Yet at the time it did not violate the letter of the law. IMO, it was a well played shift. A great move against someone who isn't versed in Asgard or 70-82 G-Fogs or Zulu/Pulaski deformations.

 

[Mr. Pickles]

So tell me again why this needed to be revisited??

I'd rather revisit the suicidal dotted monks.

 

[tommyboy]

I take the goat

 

[Dinsy Ejotuz]

Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

BB

BG

GB

GG

1/3?

 

[Mr. Pickles]

I take the goat

Between grief and nothing... I'll take grief.

 

[Officer Pete Malloy]

I take the goat

Between grief and nothing... I'll take grief.

http://i.imgur.com/8dwSeS1.jpg

 

[jhib]

So tell me again why this needed to be revisited??

Because Vox said so.

Well when do you plan on revisiting it? Do you disagree with the solution? So far all you've done is post a problem that has already been solved and had a giant thread devoted to it.

I'm interested in opinions in here on this. Feel free to put me on ignore or hang out in one of the werewolf threads.

God forbid you want to have a fun, thoughtful conversation. HE'S A WITCH!!!! BURN HIM!!!

Nah, that was another Monty's problem.

 

[Das Boot]

http://www.vox.com/2015/12/1/9821256/monty-hall-problem-mansplainers

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say Door #1, and the host, who knows what's behind the doors, opens another door, say Door #3, which has a goat. He then says to you,"Do you want to pick Door #2?"

Is it to your advantage to switch your choice?

Goats?

You mean like Michael Jordan or Jack Nicholas?

 

[shuke]

Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

1/3

There is a 1 in 3 chance that the other child is a boy.

But the probability that both children are boys is 1 in 4.

These are two different questions.

 

[jon_mx]

Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

1/3

There is a 1 in 3 chance that the other child is a boy.

But the probability that both children are boys is 1 in 4.

These are two different questions.

Wrong on both counts. If you learn the sex of a particular child, the odds of the sex of the other child is 50-50 between either being a boy or girl. This would be a situation such there are two cards on the table representing the sex of each child and you pick one and reveal the sex of one of the children. If you learn from someone who knows the sex of both child that at least one is a boy, then it is 1/3 chance that both are boys.

 

[jfranco77]

Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

50-50. The one you don't know is either girl or boy. In this particular scenario, whether that child is a boy or girl is not influenced by the sex of the other child.

Flip a coin and get heads. What are the odds *now* that you get a second head? The first one doesn't matter. 50-50.

 

[FUBAR]

Easiest way for me to think about it is this:

• If you play this game 99 times, you're going to pick the "car" door on your first guess ~33 times.
• This means that you're going to pick a "goat" door ~66 times
• If you never switch, you will win the car ~33 times
• If you switch every time, you will win the car ~66 times

thank you. that's probably the best explanation I've read.

 

[Politician Spock]

Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

1/3

There is a 1 in 3 chance that the other child is a boy.

But the probability that both children are boys is 1 in 4.

These are two different questions.

Wrong on both counts. If you learn the sex of a particular child, the odds of the sex of the other child is 50-50 between either being a boy or girl. This would be a situation such there are two cards on the table representing the sex of each child and you pick one and reveal the sex of one of the children.If you learn from someone who knows the sex of both child that at least one is a boy, then it is 1/3 chance that both are boys.

You're missing the point of variable change, which is what the Monty Hall example is about.

The probability of you picking the right door at the beginning is 33%, because each of the three doors has an equal probability of being right. However, the set of doors you didn't pick has a combined 67% chance of being right. The variable change occurs when Monty reveals one of the doors in the set you didn't pick. When this occurs no probability has changed. The probability of your door being the correct one is still 33%, and the combined probability of the set of doors you didn't pick is still 67%. In this case, you can benefit from changing your decision because of the variable change. You switch your answer because the variable change puts all the combined probability of the set of doors you didn't pick on one remaining door in the set.

Likewise, the probability of both children being boys is 25% and 75% probabile they are not both boys (which is combined set of: the first child being a boy is 25%; the second child being a boy is 25%; and neither children being boys is 25%). Revealing that at least one of them is a boy is like Monty Hall revealing a door that has a goat. When he does that variable change is introduced, but when variable change occurs probability does not change. There is still a 25% probability that both are boys and a 75% chance that they are not both boys. In this case the introduction does not provide you any advantage because you haven't made any decision where you can benefit from changing your decision because of the variable change that has been introduced.

In these examples, variable change helps you make better decisions. It doesn't change probabilities. For the probabilities to change, the data has to change, not just be revealed.

 

[B-Deep]

So tell me again why this needed to be revisited??

Because Vox said so.

Well when do you plan on revisiting it? Do you disagree with the solution? So far all you've done is post a problem that has already been solved and had a giant thread devoted to it.

I'm interested in opinions in here on this. Feel free to put me on ignore or hang out in one of the werewolf threads.

back it up there mr sassy pants, don't go dissing werwolf!

 

[sartre]

You encounter a man on the street who bets you $20 on a shell game. It is made clear to you that there will be one and only one round. He shuffles the 3 shells and you completely lose track of which has the pea.

You randomly pick a shell. He responds by turning a different one over, showing it is empty. He offers you the chance to switch your pick if you want.

Do you switch?

 

[jon_mx]

Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

1/3

There is a 1 in 3 chance that the other child is a boy.

But the probability that both children are boys is 1 in 4.

These are two different questions.

Wrong on both counts. If you learn the sex of a particular child, the odds of the sex of the other child is 50-50 between either being a boy or girl. This would be a situation such there are two cards on the table representing the sex of each child and you pick one and reveal the sex of one of the children.If you learn from someone who knows the sex of both child that at least one is a boy, then it is 1/3 chance that both are boys.

You're missing the point of variable change, which is what the Monty Hall example is about.

The probability of you picking the right door at the beginning is 33%, because each of the three doors has an equal probability of being right. However, the set of doors you didn't pick has a combined 67% chance of being right. The variable change occurs when Monty reveals one of the doors in the set you didn't pick. When this occurs no probability has changed. The probability of your door being the correct one is still 33%, and the combined probability of the set of doors you didn't pick is still 67%. In this case, you can benefit from changing your decision because of the variable change. You switch your answer because the variable change puts all the combined probability of the set of doors you didn't pick on one remaining door in the set.

Likewise, the probability of both children being boys is 25% and 75% probabile they are not both boys (which is combined set of: the first child being a boy is 25%; the second child being a boy is 25%; and neither children being boys is 25%). Revealing that at least one of them is a boy is like Monty Hall revealing a door that has a goat. When he does that variable change is introduced, but when variable change occurs probability does not change. There is still a 25% probability that both are boys and a 75% chance that they are not both boys. In this case the introduction does not provide you any advantage because you haven't made any decision we re you can benefit from changing your decision because of the variable change that has been introduced.

In these examples, variable change helps you make better decisions. It doesn't change probabilities. For the probabilities to change, the data has to change, not just be revealed.

I fully understand the Monty Hall problem. But the boy problem is different. In the Monty Hall problem, the reveal is planned and has strict guidelines. In the Boy-Girl problem, it can be different depending on your assumptions of the reveal.

Lets say you have 4 sets of two cards representing all the boy-girl combinations, in an order which the top card is the youngest and the bottom is the oldest. I pick a set of cards, which if that set is boy-boy, I win. While you are not looking, I cheat and peek at the bottom card, revealing a boy. In the Monty Hall problem, the reveal teahes me nothing about the door I chosen, only about the door I have not chosen. In this case, I have learned about my selection, and that is that the bottom card is a boy (the oldest). In the beginning it was a 25% chance, but now that I know that the oldest is a boy, my chances just improved to 50-50.

 

[B-Deep]

I've got the knife, now turn on the *** **** lights!

 

[AcerFC]

Easiest way for me to think about it is this:

• If you play this game 99 times, you're going to pick the "car" door on your first guess ~33 times.
• This means that you're going to pick a "goat" door ~66 times
• If you never switch, you will win the car ~33 times
• If you switch every time, you will win the car ~66 times

Why does one door being opened change the percentage. I am a Math idiot so bear with my faulty logic here

If you were given two doors and you picked one, you would have a 50% chance of the car

So here, you are getting three and they will always open the bum one first.

You still have two doors with a goat in one and a car in the other. Seems like its still 50/50. The fact that you started with three shouldnt matter since the one they opened has nothing in it

 

[kentric]

One door is already out of the equation. Therefore, you odds drop to 50/50. Unless there is some research done which shows that whenever they offer to change the door there is a 57% chance that your original choice had the car, there is no benefit to change.

 

[TheIronSheik]

You encounter a man on the street who bets you $20 on a shell game. It is made clear to you that there will be one and only one round. He shuffles the 3 shells and you completely lose track of which has the pea.

You randomly pick a shell. He responds by turning a different one over, showing it is empty. He offers you the chance to switch your pick if you want.

Do you switch?

In this scenario, he wants you to lose. If I picked the wrong one, he would have no incentive to flip over a shell showing me it is empty and asking if I wanted to switch. So I would not switch.

 

[Politician Spock]

Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

1/3

There is a 1 in 3 chance that the other child is a boy.

But the probability that both children are boys is 1 in 4.

These are two different questions.

Wrong on both counts. If you learn the sex of a particular child, the odds of the sex of the other child is 50-50 between either being a boy or girl. This would be a situation such there are two cards on the table representing the sex of each child and you pick one and reveal the sex of one of the children.If you learn from someone who knows the sex of both child that at least one is a boy, then it is 1/3 chance that both are boys.

You're missing the point of variable change, which is what the Monty Hall example is about.

The probability of you picking the right door at the beginning is 33%, because each of the three doors has an equal probability of being right. However, the set of doors you didn't pick has a combined 67% chance of being right. The variable change occurs when Monty reveals one of the doors in the set you didn't pick. When this occurs no probability has changed. The probability of your door being the correct one is still 33%, and the combined probability of the set of doors you didn't pick is still 67%. In this case, you can benefit from changing your decision because of the variable change. You switch your answer because the variable change puts all the combined probability of the set of doors you didn't pick on one remaining door in the set.

Likewise, the probability of both children being boys is 25% and 75% probabile they are not both boys (which is combined set of: the first child being a boy is 25%; the second child being a boy is 25%; and neither children being boys is 25%). Revealing that at least one of them is a boy is like Monty Hall revealing a door that has a goat. When he does that variable change is introduced, but when variable change occurs probability does not change. There is still a 25% probability that both are boys and a 75% chance that they are not both boys. In this case the introduction does not provide you any advantage because you haven't made any decision we re you can benefit from changing your decision because of the variable change that has been introduced.

In these examples, variable change helps you make better decisions. It doesn't change probabilities. For the probabilities to change, the data has to change, not just be revealed.

I fully understand the Monty Hall problem. But the boy problem is different. In the Monty Hall problem, the reveal is planned and has strict guidelines. In the Boy-Girl problem, it can be different depending on your assumptions of the reveal.

Lets say you have 4 sets of two cards representing all the boy-girl combinations, in an order which the top card is the youngest and the bottom is the oldest. I pick a set of cards, which if that set is boy-boy, I win. While you are not looking, I cheat and peek at the bottom card, revealing a boy. In the Monty Hall problem, the reveal teahes me nothing about the door I chosen, only about the door I have not chosen. In this case, I have learned about my selection, and that is that the bottom card is a boy (the oldest). In the beginning it was a 25% chance, but now that I know that the oldest is a boy, my chances just improved to 50-50.

No one asked you to pick a set. The question is what is the probability that both children are boys?

The probability of your decision and the probability of the set are not the same after variable change is introduced.

If you were asked to pick a set prior to you learning that at least one of them is a boy, each set has a 25% probability, so your decision has a chance of being right that equals the probability. When variable change occurs, your decision can now have a higher (or lower) chance of being right than the probability of each set. If you had picked the "both girls" set, as soon as you learned that at least on of them is a boy, your chances went from 25% to 0% as a result of the variable change. The other three options each inherited one third of the chances the set you chose just lost. The probabilities don't change due to variable chance, but your chances of your decision being right do. The Monty Hall scenario is a little more unique in that because Monty won't reveal to you that you picked a goat, all of the chances of the door he does reveal is inherited by the remaining door you didn't pick. The probabilities don't change at all, but the chances of your decision being right shift around because of the variable change. The original door you picked didn't inherit any more chances. The remaining door inherited twice as much. Change your decision. If you never made a decision, then there is no decision to change.

 

[jhib]

Easiest way for me to think about it is this:

• If you play this game 99 times, you're going to pick the "car" door on your first guess ~33 times.
• This means that you're going to pick a "goat" door ~66 times
• If you never switch, you will win the car ~33 times
• If you switch every time, you will win the car ~66 times

Why does one door being opened change the percentage. I am a Math idiot so bear with my faulty logic here

If you were given two doors and you picked one, you would have a 50% chance of the car

So here, you are getting three and they will always open the bum one first.

You still have two doors with a goat in one and a car in the other. Seems like its still 50/50. The fact that you started with three shouldnt matter since the one they opened has nothing in it

A big part of it is in how the revealed door is chosen and the timing of it. If there were the three doors and then they randomly revealed one of the goats BEFORE you picked a door, then yeah, it's 50-50 when you do pick.

But after you pick one of the three, they don't just randomly show you one of the goat doors; they randomly show you one of the goat doors that you didn't pick (they only have a choice of doors to show you if you picked the car door in the first place). Once you pick and know you have a 1/3 chance of being right, the reveal of the goat door doesn't tell you anything new about the one you picked, since you already knew that at least one of the other two had a goat in it. It instead concentrates the 2/3 chance that used to be spread out between two doors on the one door that wasn't opened.

 

[sartre]

You encounter a man on the street who bets you $20 on a shell game. It is made clear to you that there will be one and only one round. He shuffles the 3 shells and you completely lose track of which has the pea.

You randomly pick a shell. He responds by turning a different one over, showing it is empty. He offers you the chance to switch your pick if you want.

Do you switch?

In this scenario, he wants you to lose. If I picked the wrong one, he would have no incentive to flip over a shell showing me it is empty and asking if I wanted to switch. So I would not switch.

Exactly. I was hoping to illustrate a shortcoming with the wording of the original problem.

It is implicit that regardless of your choice, the host will reveal a goat from the remaining two choices. Most are familiar with the show, so this isn't a big leap.

But people are inherently distrustful. If there is no guarantee that the host is neutral, if there is no guarantee that regardless of your selection, the host will reveal a goat from the remaining choices; well then you might reasonably conclude your odds are not improved by switching.

 

[Don Quixote]

Easiest way for me to think about it is this:

• If you play this game 99 times, you're going to pick the "car" door on your first guess ~33 times.
• This means that you're going to pick a "goat" door ~66 times
• If you never switch, you will win the car ~33 times
• If you switch every time, you will win the car ~66 times

Why does one door being opened change the percentage. I am a Math idiot so bear with my faulty logic hereIf you were given two doors and you picked one, you would have a 50% chance of the car

So here, you are getting three and they will always open the bum one first.

You still have two doors with a goat in one and a car in the other. Seems like its still 50/50. The fact that you started with three shouldnt matter since the one they opened has nothing in it

They control which one to open before the switch can be made. The other two had a combined probability of 67% before the bum door was opened, and that probability is transferred to the remaining door. It's basically, you can keep your one door, or you can take two.

 

[Officer Pete Malloy]

http://img.hsmagazine.net/2013/04/my-brain-hurts.png

 

[shuke]

Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

1/3

There is a 1 in 3 chance that the other child is a boy.

But the probability that both children are boys is 1 in 4.

These are two different questions.

Wrong on both counts. If you learn the sex of a particular child, the odds of the sex of the other child is 50-50 between either being a boy or girl. This would be a situation such there are two cards on the table representing the sex of each child and you pick one and reveal the sex of one of the children.If you learn from someone who knows the sex of both child that at least one is a boy, then it is 1/3 chance that both are boys.

Know what the sex of one child is does not change the overall probability that the pair are both boys.

 

[the moops]

One door is already out of the equation. Therefore, you odds drop to 50/50. Unless there is some research done which shows that whenever they offer to change the door there is a 57% chance that your original choice had the car, there is no benefit to change.

Let's play a game.

I have a tied up jon mx behind one door, a timoschet on speed who is reading his novel behind another, and a naked adriana lima behind the other.

Choose Door A, B, or C please.

 

[coyote5]

One door is already out of the equation. Therefore, you odds drop to 50/50. Unless there is some research done which shows that whenever they offer to change the door there is a 57% chance that your original choice had the car, there is no benefit to change.

Oh Oh here come the math wizards, now I see why this needed to be revisited!

my apologies to the OP

 

[the moops]

http://www.shodor.org/interactivate/activities/SimpleMontyHall/

Just played 20 times.

Switched every time. Won 65%

 

[jon_mx]

http://www.shodor.org/interactivate/activities/SimpleMontyHall/

Just played 20 times.

Switched every time. Won 65%

Which isn't 2/3rds! Need to revisit this again.

 

[coyote5]

We've done the "Two children, one is a boy" discussion to death. It relies on a very specific wording, and a specific interpretation of that wording, to justify either the "normal" or "trick" answer. It's not worth going over again here.

Agreed. Definitely needs it's own thread!

 

[CBusAlex]

Wrong on both counts. If you learn the sex of a particular child, the odds of the sex of the other child is 50-50 between either being a boy or girl. This would be a situation such there are two cards on the table representing the sex of each child and you pick one and reveal the sex of one of the children.

If you learn from someone who knows the sex of both child that at least one is a boy, then it is 1/3 chance that both are boys.

Know what the sex of one child is does not change the overall probability that the pair are both boys.

One child is a girl. Still a 25% chance that both are boys?

 

[the moops]

Mr Smith has 100 children. At least 99 of them are boys. What is the probability that all 100 of them are boys?

 

[SWC]

mick at the bar has had two beers what are the proabilities that he is being a jagalope well brohans here is a hint it is one hondo percent because mick is a total laroosh take that to the bank

 

[GregR]

Easiest way for me to think about it is this:

• If you play this game 99 times, you're going to pick the "car" door on your first guess ~33 times.
• This means that you're going to pick a "goat" door ~66 times
• If you never switch, you will win the car ~33 times
• If you switch every time, you will win the car ~66 times

Why does one door being opened change the percentage. I am a Math idiot so bear with my faulty logic here

If you were given two doors and you picked one, you would have a 50% chance of the car

So here, you are getting three and they will always open the bum one first.

You still have two doors with a goat in one and a car in the other. Seems like its still 50/50. The fact that you started with three shouldnt matter since the one they opened has nothing in it

Walk through this set of variations of it and it might make more sense. First version, you made your choice. Monty says, "I will let you switch and take the best prize of the other two doors". Do you switch? Of course, with an obvious 2/3 chance of winning a car.

Now second version, same exact thing but Monty adds, "... and after you agree to change, I'll open first a goat door, before we see the prize you'll keep". Would that change anything? Of course not, you obviously still have a 2/3 chance of winning the car.

In the actual version asked, the only difference is he's moved up the reveal of the door you won't keep the prize from if you switch. You still are getting the better prize from the 2 doors just now he's telling you which of those 2 other doors has the better prize.

It isn't the same as just having a choice of your door or 1 door you switch to for a 1/2 chance. You have a choice of 1 door, or of the better of 2 doors. That someone with knowledge uses that knowledge to reveal the door you'll discard doesn't change that you get the benefit of the better of the 2 doors.

Now if he didn't have that knowledge and had just randomly chosen a door to open and it could have been the door with the car, then it would truly be the 50-50 shot for you to switch.

 

[Politician Spock]

Mr Smith has 100 children. At least 99 of them are boys. What is the probability that all 100 of them are boys?

The probability of all of Mr Smith's children being all boy is a 1 out of 2^100.

And the probability of exactly 99 of Mr Smith's children being boys is 100 out of 2^100.

Since all other probabilities have been eliminated, the probability of all of them being boys has a 1 out of 101 chance of being correct, as it is only one out of 101 possibilities that remain possible from the set of 2^100 possibilities that exist.

 

[Fred Lane]

Maybe this helps....

Let's say I have 1 million $1 lottery tickets, and I sell you one. Then I put aside 999,998 of them so that I have one and you have one, and I offer to trade. Do you trade? Do you think you have a 50/50 chance of holding the winning ticket?

Then we'll do it again. I put aside the 999,998 so that we both have one. Do you still have a 50/50 chance? You think you we're able to have a 50/50 chance on holding the winning ticket in two consecutive lotteries?

 

[CBusAlex]

mick at the bar has had two beers what are the proabilities that he is being a jagalope well brohans here is a hint it is one hondo percent because mick is a total laroosh take that to the bank

I object to the racial slurs and stereotypes against the good people of Ireland contained in this post, brohan.

 

[Dinsy Ejotuz]

Know what the sex of one child is does not change the overall probability that the pair are both boys.

It eliminates the possibility that there are two girls -- i.e. 25% of the cases.

So if you know at least one of the two children is a boy the denominator drops from 100% to 75%, and the odds that both are boys is now 25%/75% -- or 1/3.

 

[SWC]

mick at the bar has had two beers what are the proabilities that he is being a jagalope well brohans here is a hint it is one hondo percent because mick is a total laroosh take that to the bank

I object to the racial slurs and stereotypes against the good people of Ireland contained in this post, brohan.

no that is micks name and he is not irish he is just a laroosh take that to the bank all the same all apologies to any from the emerald isle down under who were offended