Monty Hall [edited title]

Cecil's Columns/Staff Reports
1

Just a quick comment regarding the three doors, goat and car explanation. I suppose there is no right answer as if Monty knows what is behind each door it is up to him if he bluffs you or not. I presume if u picked a door with a goat he would open that door straight away and hence you lose. so if u pick the prize he would try hedge his bets and offer you another chance. however if this was a gamshow then this trick would soon lose out as everytime he offered you another door u would know u were right. so i am also presuming he would sometimes bluff to keep up suspense. however il shut up on that one.

on the children question if a couple have two kids and you know one is a girl then what is the sex of the another. in my opinion this must be 1 and 2. as you pointed out there are four alternatives

  1. child a = girl and child b = boy

  2. child a = girl and child b = girl

  3. child a = boy and child b = girl

  4. child a = boy and child b = boy

as you point out number 4 is out as there is a girl but you have not assigned whether the girl you know about is child a or b as she cannot be both. hence number 1 and 3 are the same! im not very good at challenges so im probably wrong but glad if you can explain.

2

The assumption is that the observer doesn’t know if the girl is the older or younger child. If the observer did know, the odds of her sibling being a boy would be 50/50. Consider the table of possibilities:
[list=#][li]Girl, boy[/li][li]Girl, girl[/li][li]Boy, girl[/li][li]Boy, boy[/list][/li]
Say a girl walks in. Her parent says “This is one of my two children.” Based on this information, #4 gets eliminated. In 2 of the 3 remaining possibilities, her sibling is a boy.

In contrast, say a girl walks in and her parent says “This is the older of my two children.” Based on this greater information, you can eliminate option #4 and option #3, leaving only option #1 (the sibling is a boy) and option #2 (the sibling is a girl), thus 50/50.

Option 1 and 3 are not the same. They describe different outcomes; girl first, then boy; or boy first, then girl.

3

Posted by Bryan Ekers

This is not a good example. If a girl walks in and her parents say “this is one of my two children,” then the family that has two girls is twice as likely to make this statement than the two families that have only one daughter. The two families that have both male and female children are equally likely to display their son as they are their daughter.

4

No, your example is wrong. If you see the kid, then you know that the one you are seeing is a girl, and the other has a 50/50 chance of being a boy. There is no requirement that you know which is older.

The difference between your example and the one in Cecil’s statement is that the parent is effectively being asked “is either of your children a girl?” instead of a particular one.

5

Welcome to the SDMB, jiffy1234.

A link to the column you’re commenting on is appreciated. Providing one can be as simple as pasting the URL into your post, making sure to leave a blank space on either side of it. Like so: www.straightdope.com/classics/a3_189.html

6

Also appreciated are descriptive thread titles. I have changed the title of this thread accordingly.

bibliophage
moderator CCC

7

That is a good point. Monty Hall doesn’t act randomly, and he doesn’t always have to give you the option of going to another curtain. Obviously (and it’s important to the understanding of the ‘classic’ answer), he’s never going to show you the million-dollar curtain.

But you’re right - by the time you’ve gotten into this position there are things affecting outcome you don’t know.

It may be that producers of the show realize that if you switch, you have a 2/3 chance of being right, so they told Monty ‘always give them the option to switch if they already picked the right one, but only give them the option one half the time if they’ve picked the wrong one.’ Then the probability would be 1/2 again.

8

Well, I sincerely hope so.

This is where there’s a misunderstanding. If the question was phrased “I have a child who is not in the room at the moment. Is my child a boy or a girl?” then it’d be a 50/50. The question being asked is actually “I have two children, one of whom is in the room at the moment and is obviously a girl. What is the gender of the other one?” For a family with two children, the odds are 50% that they are of the same gender and 50% they are of different gender. If you learn one of the children is a girl, half of the first option evaporates (it is been shown impossible for the family to have two boys) and the 50/50 becomes a 25/50, i.e. 33/67, or two-thirds likely the unseen child is a boy.

I don’t know how you came to that interpretation:

Whether you’re told they have a female child or you see the female child makes no difference, unless you want to throw in the odds of your friends lying to you.

Using Cecil’s playing-approach (and Vos Savant’s million-doors approach), consider a family of six children, five of whom show up and are all girls. What are the odds on the gender of the unseen child? The table is as follows:


      Child   1  2  3  4  5  6
Gender        M  M  M  M  M  M
              M  M  M  M  M  F
              M  M  M  M  F  M

... all the way to line 64...

              F  F  F  F  F  F

From the available information (at least five of the children are female), the only possible lines from the table are:



M  F  F  F  F  F
F  M  F  F  F  F
F  F  M  F  F  F
F  F  F  M  F  F
F  F  F  F  M  F
F  F  F  F  F  M
F  F  F  F  F  F

Seven possibilities, in which six have the unseen child as male, for about an 86% likelihood. Frankly, I invite anyone who still has doubts to simulate it with playing cards. Shuffle together two Kings and two Queens and then draw two cards at random. Look at one of them. If its a King, throw back the two cards and reshuffle. If it’s a Queen, look at the second card and record it as a Q or K on a piece of paper. After thirty of so trials, you should see more Ks than Qs on your record sheet. Run enough trials, and the Ks will outnumber the Qs roughly two-to-one.

9

Err, Cecil’s playing-card approach, of course.

10

Bryan, there is a whole thread devoted to Monty Haul question and the daughter question, you should check it out. I debate Cecil’s( and Marilyn’s) answer in the other thread, but your example in this thread is far worst. The fact is that once a child becomes specific, like you see the daughter, then Cecil’s four families don’t work. Cecil’s families work when a random family is the focus but the child is not. But when a random child is focus, then the random families list just causes confusion. What you have in your example is specific to the daughter, which makes her random to the question. She is either Child A or Child B. If she is Child A then (from Cecil) families #1(F/M) and #2(F/F) are possible. If she is Child B the families #2(F/F) and #3(M/F) are possible. Each combination has a equal chance to have another sibling that is either a sister or a brother. In your example we have four combinations:
Known daughter and a younger sister
Known daughter and a younger brother
Known daughter and a older sister
Known daughter and a older brother

Your King/Queen card experiment creates a false interpretation of the problem. First if you throw the card(s) back if the first card you see a King, then half the time you will be throwing back a King and a Queen. Therefore the only possibility for your “mixed hand” is if you draw a queen first. You could claim that there are 3 other cards for other card can be, two Kings and one Queen, but that doesn’t work either. If you only use one pile and never return the original card you draw, then based on the 3 remaining cards, a mixed set of cards will be 66% likely to happen. After a hundred draws using one pile you get this percentage:
QQ…16.5%
QK(KQ)…66.6%
KK…16.5%

As we know that all things are suppose to be equal, we should get:
QQ…25%
QK…25%
KQ…25%
KK…25%

Now, if we put the original card back into the pile then we would get the correct distribution of outcomes. A better way to do this experiment is to make two piles each with a king and a queen. One pile represents Child A and the other pile Child B. You can draw from either pile. No matter which pile you draw from the other pile has exactly a king and a queen. Your chances are 50/50 for either.

To fit this experiment to your example, we are told not only of a daughter but we physically see her. We can equate from that we have already drawn a queen. You can now draw a single card from either pile. As each pile as a king and a queen then your chances are again 50/50.

11

P.S. Sorry for my poor English skills. Please forgiven me, I’m from the Alabama.

12

Let me see if I can straighten things out a little bit. I’ve read most of your posts, Wissdock, and I’ve learned a lot, even though you’re a little hard to decode.

Say I discover a town that has 100 families, and each family has 2 children. I decide I want to make a little bit of money, so every mother I run into I ask if she has a daughter. If she says yes, I bet her a $100 that I can guess her other kid’s gender. Of course I’m going to be right 2/3 of the time when I guess boy, because I’ve weeded out the moms who only have boys. (This is in line with Cecil’s answer). Over time, I begin to see a profit.

This is when I start getting greedy. I decide if I can rip off the parents, I can rip the kids off too. I know that if I see a girl on the street, she can’t belong to a family will all boys, so the odds of winning my bet should be the same. So I start to bet every girl I run into that I can guess whether she has a brother or a sister, and subsequently guess brother. To my dismay, I’m not making any money as it appears the odds are now 50/50.

I ask myself, “what’s changed?” I then read a math article that wissdock linked.
http://64.233.161.104/search?q=cache:b2FuesRPPL8J:mathforum.org/library/drmath/view/52186.html+as+1/3+for+it+being+a+boy+and+2/3+for+being+a+girl.+&hl=en&lr=lang_en&ie=UTF-8)

Essentially, the chance of the sister having another sister is doubled because I have an equal chance of picking either sister in the girl/girl family.

After I realize my mistake I decide to go back to only making bets with the mothers.

On one occasion, before I have the chance to ask the mother whether she has a girl, her daughter walks into the room. What’s just happened? I know that the mother has a daughter now, so if I guess her other kid is the opposite I should have 2/3 chance of winning. But if I bet the daughter she has a brother I should only have 50/50 chance of winning. Is this some kind of paradox?

This leads me to conclude that probability means NOTHING in isolated cases. Probability only means ANYTHING when the experiment is repeatable. Saying I have a 2/3 chance or a 50/50 chance of being right only means something in the context of how I repeat the experiment.

In other words: *ANY PROBABILITY VALUE I ASSIGN TO MY CHANCES OF BEING RIGHT IS ARBITRARY, THEREFORE THE ORIGINAL QUESTION IS MEANINGLESS WITHOUT MORE INFORMATION. *

Let me ask a question: how would you repeat the original experiment?

Let me ask another question: After you met the girl in the last meeting of my story, if the mother said “This is my oldest and tallest girl, and the one I love the most, and I think she will be President someday,” would that change your odds of winning the bet with the mother? (OF COURSE NOT). I only mention that because some people seem to think that once you know the girl as a specific, it changes the odds. That doesn’t even make sense to me. The odds will only change in relation to who you’re betting against.

Someone earlier, (in the other thread) stated that how the information is obtained should not affect how the probability is calculated. I could not disagree more.

It took me hours to think of this, and I’m no genius or mathematican, so if anyone would like to rebut, I’d love to hear why I’m wrong.

13

I mused it over and found what I think is the ambiguity:

Process 1:
Pick a random two-child family.
Determine if said family has at least one daughter.
Bet that the other child is a son.
Profit.
Process 2:
Pick a random two-child family that has at least one daughter.
Bet that the other child is a son.
Break even.
When the two conditions are separate, it’s a good bet. When they are compounded, it’s an even bet.

14

For what it’s worth, I tried simulating this in Matlab. With 1,00,000 trials, I got a girl 15478 times, and a boy 15719 times, or about 50/50. The code is below.



function [ng, nb] = mh(ntrials)

nb = 0;
ng = 0;
for i = 1:ntrials
  s = (rand(1,6) > 0.5);
  if all(s(1:5) == 0)
    if (s(6) == 0)
      ng = ng + 1;
    else
      nb = nb + 1;
    end
  end
end
15

The approach doesn’t work because the test is comparable to flipping a coin six times, and trying to predict the sixth outcome in cases where the first five are heads. What you’d need is a simulation that randomly builds a family, picks out ones where there are no sons or only one son:


t = (s(1) + s(2) + s(3) + s(4) + s(5) + s(6));
if (t == 0) or (t == 1)   // There are no sons or only one son
 then...

…then picks out five of the kids at random. If you get five girls, the odds are significantly higher than 50% that the sixth will be a boy.

16

Right. Your simulation specified which child had to be the boy (the sixth).

Imagine if you rolled dice. Everyone’s done dice probabilities, right?

Odds of two sixes = 1/36. Odds of a 5 and a 6 = 2/36.

That’s why 11 is more common than 12.

17

You have a clear way of explaining things that I like.

You don’t have a paradox, however, nor should you reach the conclusion that the question is meaningless without more info.

In the case you describe, make the bet. Here’s why.

There are 100 girls in your hypothetical town - 50 families have one girl, 25 families have 2 girls.

When the girl walks into the room, you have whittled out problems - the 50 girls that will make you lose are squished into 25 families. Even is she is one of the 50 girls, she’s in one of the 25 families, and you have a better chance of being in the 50 families that are boy/girl split.

So what did change? You have prevented those girls from “spreading out.” If the girls and boys in town each wore a number indicating their address, then you could still make the bet with the kids. You see a girl from address 30, and make the bet. If you lose, you won’t make the bet with the other girl wearing address 30.

Back to your hypothetical. By being in the house, you have prevented the 25 families splitting into 50 chances that are out in the crazy outdoors. If they are one of the 50 that make you lose, at least they’re both confined to one house.

18

I think I see what you’re saying, but hear me out: My problem is that in this isolated case that I’ve created there is no way to figure out my probability of being right. My odds are simultaneously 50/50 (50%) and 2/3 (66%) depending on who I bet with (the mother or the daughter).

I could go back out into the world after the bet and meet the aforementioned girl’s sister and bet with her and lose (consistent with 50/50 odds), or I could keep meeting mothers and keep my odds at 2/3. (By and by, some people implied that once you actually meet the girl as an actual person the odds change to 50% which is definitely wrong).

In my isolated case, my two ways of making a bet converged. This implies that in any given situation, if you ask a person, “what is your chance of being right?” that person has to ask “well how am I going to do it next time?” Otherwise any number value he assigns to his chances are meaningless.

My main point was that if the original question doesn’t imply how you would repeat the experiment, then probability is meaningless. Isolated cases can’t have probabilities associated with them. I might as well say I’ve got a 100% or 0% chance of being right, because in reality the kid is either 100% boy or 0% boy, not some uncertainty in a box.

If you were to repeat the original experiment, and the next family you met with had two boys, what would you do? Discard that family like I’ve been doing to win money? Somehow magically make it impossible to run into a family with two boys? Automatically call it a loss if they have two boys? Automatically call it a win? Or maybe you’ll avoid that situation by meeting girls on the street and then go meet their families. If you do that and you end up meeting a family you’ve already met, do you still make the bet? Or do you get to meet the other girl in the family after you make the bet so that you don’t make the same mistake twice? I don’t believe the original question tells you how to deal with this situation, so no meaningful odds can be calculated without more information.

19

Bryan Ekers wrote:

There’s nothing wrong with the way I simulated it the first time, but I’ll do it again like you describe above. See the code below. With 1,000,000 trials, I get a girl 15268 times and a boy 15807 times, still about 50/50. The total number of families with 5 or more girls was 108819. In 77744 of those cases, the one boy was one of the five people you met.

The reason the results are near 50/50 even though there are six times as many families with 5 girls and one boy as families with six boys, is that when you select five of the six kids randomly, only 1/6th of the families with five girls will have all five girls in the five you meet.

bup wrote:

Since the six are randomly assigned boy or girl, picking one at random, instead of last won’t make any difference. But I pick the child randomly below.



function [ng, nb, ng5] = mh(ntrials)

nb = 0;
ng = 0;
ng5 = 0;
for i = 1:ntrials
%
%   Randomly build a family.
%
  s = (rand(1,6) > 0.5);
  if (sum(s) <= 1)
%
%   Zero or one sons.  Now pick five at random.  p is between 1 and 6, 
%  and the five selected are those that aren't p.  I'll also keep track
%  of how many families have 5 or more girls.
%
    ng5 = ng5 + 1;
    p = ceil(rand(1,1) * 6);
    ii = find((1:6) ~= p);
%
%   ii contains 5 of the numbers 1 to 6 (the five not equal to p).
%
    if (all(s(ii) == 0))
%
%  The five are all girls.  Now check the sixth one.
%
      if (s(p) == 0)
%
%   So is the sixth.
%
        ng = ng + 1;
      else if (s(p) == 1)
%
%   Sixth was a boy.
%
        nb = nb + 1;
      else
%
%   Should never be here.
%
        disp('Error')
      end
    end
  end
end
20

And now here is the clincher because this is actually what the puzzle seems to be implying: Say I walk up to a hundred families and they each randomly tell me the gender of one of their kids. If I guess the opposite each time, how much money will I win?

Does anyone doubt that I’ll break even?

I think it should be obvious that I would (because the odds are now 50/50), but if you want a real world application: Have a friend randomly pick two cards from a deck and then have him randomly tell you the color of one of the cards (red or black). Guess that the remaining card is the opposite color of what he told you. Repeat 100 times. I bet you’ll only win about 50 times.

And this is why the puzzle is so deceptive, and FLAWED. The only way you’re going to bump up your odds is by ignoring the families with no girls, by following my above (earlier post) procedure. And do you think my procedure is IMPLIED by the original question?

If you just walk up to random family on the street and they tell you “we have a girl,” you stand NO BETTER CHANCE at winning a bet with them by guessing their other kid is a boy. The original question and the subsequent answer of 2/3 implies that you DO have a better chance.

This just goes back to my original assertion that odds mean NOTHING unless a way to repeat the procedure is implied.

Can you see the difference between the original question, and the Monty Hall question? The Monty Hall question has a very specific, repeatable procedure laid out. Does the Girl/Boy question have the same? Of course not.