Monty Hall Question

Cecil's Columns/Staff Reports
1

The current column contains a link to the earlier Monty Hall Question, which I have some concerns with. Two questions in particular seem to be twisted to support incorrect answers.

The way I see it is that the odds they also have a son are 1/2. It is a fact that 2 of the answers from the supplied list of 3 options will be correct if the other child is male, but this is not the question we are asked to answer! The chance of the other child being male or female are 50/50. The specific wording of the options provided does not change this, as options 1) and 3) in the list provided in the analysis are the same option with different wording, since the order of the children is irrelevant. Reword the options provided to read A) both female, B) both male, C) one male one female and the false logic behind the choices presented collapses.

Again, the answer of 2/3 is not the answer to the question asked. The odds that the original card was an ace are 1/2. The odds that you picked the original card are 1/2. The odds that the card that you picked was also an ace may well be 2/3, but this is not what the question asks.

Discuss and criticise (like I need to say the last :smiley: )

2

You are completely correct. The column actually makes this clear by claming that a paradoxical situation exists. A paradox, by definition, cannot exist! Therefore the column is wrong.

It is also possible to realize this not only by rewording the question, as you suggest, but by realize that with the original wording:

  1. Child A female, Child B male
  2. Child A female, Child B female
  3. Child A male, Child B female
  4. Child A male, Child B male

the knowledge that one child is female is also random. You do not know whether the child about which you’ve learned the identity is child A or child B, in fact, there is a 50/50 chance of each. The key here is that you have learned the identity of one of the children. It is exactly the same scenario as that in which a child comes to your door - but worded very poorly.

The ace problem is similar in that we know the probability that the first card was an ace (much like we know the probability that a child is male). If instead, we are not told what the probability of the identity of the initial card is, and follow through with the example, then we can claim that the probability that the first card was an ace was 2/3.

3

The problem with this logic is that A B and C are not equally probable. Given an infinite number of families with 2 children, 25% will be in category A (both female), 25% will be in category B (both male) and 50% will be in category C (one male, one female). Since we know that the family in question is not category B, there is a 2/3 chance that it will be C rather than A.

4

The solution to this problem is very similar to the one for the Girl-Boy problem, which FatBaldGuy captured well.

The original card is either a King or an Ace with equal probability. You toss in your own Ace and pick from the pair of cards with equal probability. There are then four possible scenarios that could happen — with equal probability:
[ol]
[li]The original card was an Ace; you drew your own Ace.[/li][li]The original card was an Ace; you drew that Ace.[/li][li]The original card was a King; you drew your own Ace.[/li][li]The original card was a King; you drew that King. (Rejected)[/li][/ol]
Case #4 is rejected by a stated condition of the problem: you’re given that you didn’t draw the King. Therefore you must be in one of the other three scenarios, which, remember, have equal probability. Therefore the probability that the original card was an Ace is 2/3, and that’s indeed what the problem asks for.

5

And, here’s another way of looking at the Girl-Boy problem (as I’ve been calling it)…

Suppose you are told that a particular couple has 10 children, and that 9 of them are girls. What’s the probability that the remaining child is a boy? It’s not 1/2.

There are 2[sup]10[/sup] or 1024 possible boy/girl sequences that a ten-child couple might have, all of which are equally likely. However, a condition of the problem — you’re told there are at least 9 girls — eliminates all but 11 of those sequences: one sequence has all girls, the other 10 have exactly one boy. Therefore the probability that the remaining, sex-unspecified child is a boy is 10/11.

Or to generalize a little: for N children, N-1 of which are known to be of one sex, the probability that the unknown child is of the opposite sex is N/(N+1).

6

Perhaps I don’t understand probabilities. The way I see it, at the time that child A was born, there was a 50/50 chance that it would be male. At the time that child B was born, there was equally a 50/50 chance that it was male. How does being told, at the present time, the gender of one of the children retroactively change the odds that, at the time the other child was born, it would be male?

But my issue here is that the question asked is what was the probability that the original card was an ace. This is 1/2. You had 2 cards to select from, ace and not-ace, and you drew one of them and put it into the hat. The fact that you subsequently added another ace and drew one of the two cards to find it was an ace is irrelevant to the question asked.

7

For a family with two children, all the following outcomes are equally likely:

  1. First child is a boy, second child is a boy.
  2. First child is a boy, second child is a girl.
  3. First child is a girl, second child is a boy.
  4. First child is a girl, second child is a girl.

How can ruling out one possibility make the others be not equally likely?

If you doubt this, take a quarter and a nickel, and flip them both about 100 times. Take the number of times when both coins are heads and divide it by the number of times that at least one coin is heads.

8

Just to keep things straight, in post #6 the second quote should be attributed to Bytegeist, not to me.

9

Yes, but by drawing a card you have more information than you had originally, allowing you rule out some possibilities.

As an analogy, assume that you have a coin that is either normal (heads/tails) or a fake (heads/heads). There was a one of each in a jar and someone selected one at random. So, there’s a 50% chance the coin is normal or fake.

Now start flipping the coin. If you get a tails, you know for certain that the coin wasn’t fake. With that new information you can now state that the probability of the coin being normal is 100%. On the other hand if you get 50 heads in a row, the probability of that coin being fake is higher, though not 100%. As you learn more information, you know more about the situation and can better refine the probabilities.

10

Well, the way you’ve phrased it, the chance would be 50/50. If you ask the question “the first child is a boy, what’s the probability that the second is a boy?”, then that’s 50/50.

But that’s not what the question is asking. You’re told that there is at least one boy, which means that the boy could be either the first or the second, without being specific. Note: I’m not saying that the order of which is first actually matters - what matters is that you’re talking about a specific kid in one case, and not being specific about whether there is a boy in the other.

When doing probabilities, you can list out all the equally likely possibilities and count them to get the answer. If they’re not equally likely, that technique doesn’t work. Again, consider looking at the question this way. Think of all the families with two kids. 50% will have a boy and a girl, 25% will have two boys, and 25% will have two girls. Do you agree so far?

You’re told that this particular family has at least one boy, which tells you that the family is in either group 1 or group 2, but can’t be in the two-girl group 3.

So now out of the reduced population that we’ve narrowed it down to, 2/3 of those have a boy and a girl, and 1/3 have two boys. Make sense?

11 


---+-- F --+-- F
   |       |
   |       +-- M
   |
   +-- M --+-- F
           |
           +-- M

This is supposed to be a little graph of the possible outcomes. So far we only know that there are two children, each equally likely to be male or female. Each outcome corresponds to a path from left to right. If a girl is born, you take the upper branch, if a boy is born, you take the lower one. There are four different paths through the graph (4 different outcomes.)

It is important to note “our” perspective. When a child is about to be born, we are at a junction and both branches are equally likely: Both possibilities are equally likely for this particular child.

However, if both children have been born already, one complete path has been taken, but we don’t know which one. As long as we have no additional information, all four are equally likely.
But in this case we have additional information. We know that there is at least one girl. With this knowledge, we can elimininate one path. The probabilities for the paths do not change retroactively.



---+-- F --+-- F
   |       |
   |       +-- M
   |
   +-- M --+-- F
           .
           ... .

If we eliminate one path, we know that one of the others has been taken. Two of those contain the birth of a boy, one doesn’t. Keep in mind that we don’t know which child is boy or a girl.

12

Hmmm…

The last time this came up on the boards was during the recidivist lurker phase of of my six-month non-paying self-imposed exile.

I hunted around and found this page, which recreates the problem.

Play once, and you’ll see the statistics showing a near-perfect 2-1 distribution of winners who switched choices to those who did not switch.

Funny thing though. The numbers have been reset since I first saw this page. When I found it months ago, the number of players was in the five figures, showing the ratio of winners among those who switched choices to those who didn’t to be a dead-even 1-1.

13

I know the OP focuses on the subordinate questions raised in the column, but he did say he had issues with the Monty Hall problem.

Upon further review, I now cave on the MH problem. Cecil’s revised answer is correct.

Here is the clearest explanation I have seen for why it works as described.

14

Well, before you add the second Ace, and draw, one would have to say that. In fact we’re given outright that this probability is 1/2. If nothing else happened, that’s what you’d have to say.

But the problem keeps on going…

Careful. Those events go the other way around: you toss a known Ace into the hat, then you draw randomly from the pair. You knew that though, I see.

Your draw turns out to be an Ace (we’re given), though you can’t tell which Ace it is. Nevertheless, at this point, you have new information that changes your computation of the probabilities. Given what has just happened, you now know that the probability the original card was an Ace is 2/3 instead of 1/2 — by the reasoning given earlier.

The problem asks its question at the end, after it has told you about the other events: the adding of an Ace, and the drawing of an Ace. You’re meant to take this information into account. In fact you have to, otherwise, in the case of this particular problem anyway, you’ll get the wrong answer.

15

I find the easiest way to approach these problems is to imagine actually performing the experiment multiple times. For example, take a look at the card question:

Imagine setting up this question. Here, I’ll pitch in: You clear out your living room, and I’ll bring over one thousand hats, OK? So imagine we set up all 1000 hats in your living room.

Next, you close your eyes, and I’ll bust into my stash of playing cards. Then I’ll randomly toss either an ace or king of spades into each hat. When I’m done, there will be something like 500 aces and 500 kings in the hats, right? Not exactly, but pretty close, probably.

OK, now you can open your eyes. I give you my stack of playing cards, and you walk aroung the room and drop an ace of spades into every single hat. So now there’s about 500 hats with two aces, and 500 hats with an ace and king, right? With me so far?

Now comes the experiment. You walk around the room, and I’ll follow you with a clipboard. Here’s what we do:

  1. YOU pick a card from a hat and tell me what it is.

  2. If the card you pick is a king, I go “BZZZZZTT!!!” as loud as I can, and we throw the hat out the window.

  3. If the card you pick is an ace, then you peek in the hat and tell me what the other card is, and I’ll record it on my clipboard.

So what happens? There’s about 500 hats with two aces in them. When you pick a card out of those hats, it’s an ace (duh), and the second card will be an ace. So on my clipboard, I’ll write “ace” about 500 times.

There’s also about 500 hats with an ace and a king. For about half of those hats, you’ll pick a king, and I’ll yell “BZZZZZTT!!!” as loud as I can, and we’ll throw the hat out the window. For the other half–250 hats or so–you’ll pick the ace first. Then you’ll peek into the hat, see the king, and tell me. I’ll write “king” on my clipboard about 250 times.

So, when all’s said and done, you’ve picked an ace out of the hat about 750 times. In 500 (2/3) of those cases, the original card was an ace, and in 250 (1/3) of those cases, the original card was a king.

16

To address a couple issues left hanging in the air…

It’s better to say that Mr. Ragozin (the letter writer quoted in Cecil’s column) shouldn’t have used the word “paradoxical” when he really meant “surprising” or “counter-intuitive”. At least I assume that’s what he meant. But in fact, there’s no true paradox in any of these problems we’re discussing.

Although it might seem like you’re given a fact about some specific if “random” child (“You have been told this family has a daughter”), all you’ve really been told is that this family has at least one daughter, possibly two. And all that this fact allows you to do is exclude the category of families with two boys.

Actually, if you don’t know the probability of the initial card being an Ace — as in, it could be anywhere from 0 to 1, not just 0.5 — then you can’t precisely compute the probability that it was an Ace, after the drawing.

That’s an interesting variation (to me) of the original problem, so let’s take a look at it:

Again, there are four possible scenarios you might find yourself in, though this time they’re not equally likely:
[ol]
[li]The original card was an Ace; you drew your own Ace. Frequency: P x 0.5.[/li][li]The original card was an Ace; you drew that Ace. Frequency: P x 0.5.[/li][li]The original card was a King; you drew your own Ace. Frequency: (1–P) x 0.5.[/li][li]The original card was a King; you drew that King. Frequency: (1–P) x 0.5. However, (BZZZZZTT!)[/li][/ol]
Case #4 is again excluded outright by the problem, leaving the other three scenarios as the only ones you could be in. The final answer then is the sum of the first two frequencies divided by the total of the three frequencies, which simplifies to 2P / (P+1).

In the particular case of P = 0.5 (the original version of the problem), this probability evaluates to 2/3, as we should expect. For P = 0, “initial card always a King”, the probability it was an Ace is 0, which certainly makes sense. And for P = 1, “always an Ace”, the probability is 1, which also make sense.

17

Take a shuffled deck of cards and deal out 26 two card piles. Get someone else to go through and delete all the piles that contain no red cards. Now how many piles contain two red cards?

On the original Monty Hall problem, one way to think of it is to notice that you win whenever your original guess was wrong and you lose whenever your original guess was right. There are 2 chances in 3 of the former and 1 in 3 of the latter.

And FWIW, Cecil’s answer about the plane is wrong. I am not about to join the cacophony of the responses to this week’s column that has produced five screens of responses so far, so I will put my comment here. First consider the case of a jet plane. The jet produces no significant airflow over the wings and the plane cannot possibly take off. A prop plane is a little different because the prop does push some air over the wings but not nearly enough to have it take off. I could conceive of a specially shaped airfoil that might catch enough of the blowby to take off, I concede, but the action of a normal prop is to provide forward speed from the reaction and it is the airflow that provides the lift. If you have a plane with a 100 kt takeoff speed and try to take off downwind into a 100 kt wind you will require a ground speed of 200 kt to take off. If you take off upwind you will need no ground speed at all. My brother who was in the Air Force told me of seeing a plane take off from a taxiway in such a gale by basically turning into the wind.

18

The point of that question is not that engines produce flow over the wings. It’s that the engines will propel the plane forward regardless of what the converyor belt is doing. The plane will move forward (relative to the ground) no matter how fast the belt spins, air will flow over the wings, the plane will take off.

19

There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal? (Answer: 2/3.)

Assume those are the only two possibilities [ no twins or mutants or w/e ]
Assume there is no trickery or mental games involved, no unaccounted for dead babies etc, this is being done purely to figure out the odds, noone wins or loses, noone has incentive to change the odds
in favor of one result over an other, and the only random factor is a 50% chance that occurs every time a child is born.
==“assuming the biological odds of having a male or female child are equal”:
Assume chance of generating a male is 50% and chance of female is 50%
==“There is a family with two children”:
only 2 children, a child born after the other [Child A], and a child born before the other **
==“You have been told this family has a daughter”:
one child, is a female, unknown whether its A or B
It does NOT say the female is the child born before [ B ]. If this were the case the answer would be 50%. This is one of the two things that throws off people

The question is:
“What are the odds they also have a son,”
keyword: also
ASSUME: its asking the chance of having…
==a combination that includes a female child and a male child, out of all potential combinations for a family having two children==
-This entire explanation only holds true under this assumption- any other interpretations of this question can lead to other possible answers
Assuming this…
It is NOT asking the chance of a child being male or female when generated, we are already told that the answer to that is assumed to be 50%, rather we are asked the chance of having a female and male child combo, assuming the only choices are female and male and each sex has an equal chance of appearing.
This is the other thing that throws off people

The only combinations possible at this point:
A = Male B = Female
A = Female B = Male
A = Female B = Female
these are absolutely the only combinations possible, and ALL the combinations possible, any more and any less is incorrect.

Its tempting to say that since theres only 3 possibilities, and two of them have a male, that the chance is 2/3 66%, however noone ever said THESE outcomes have equal chances. only that on generation of a child, theres a 50% chance of female and 50% chancec of male. In otherwords, its not 33% 33% and 33% with two having male for 66%, thats incorrect

because the chance of a child when it was being born was 50% male and 50% female,

Assuming we didnt know a child was F
1st child 50% M 2nd child 50% M
2nd child 50% F
1M 2M = 50%50% = 0.50.5 = .25 = 25%
1M FM = 50%50% = 0.50.5 = .25 = 25%

1st child 50% F 2nd child 50% F
2nd child 50% M
1F 2F = 50%50% = 0.50.5 = .25 = 25%
1F FM = 50%50% = 0.50.5 = .25 = 25%

However we know one child is F. This doesnt change the fact that the first child can be M or F nor the fact that the first child born into the family had a 50% chance of being M or F. This is undeniable since the first child was the first child, before a combination could occur. Building up from the first child, regardless of what the first child was, the 2nd child is a 50%M and 50%F again, REMEMBER: the question is asking what the chance of a combination of M and F is, not the chance of the 2nd child being M regardless of the first child. Again, we ONLY know that one child is an F, not whether it was the first born or 2nd born.
1st child 50% M 2nd child 0% M [impossible due to one child being F no matter what]
2nd child 100% F [since M is 0% in this branch, F must be 100%]
1M 2M = 50%0% = 0.50.0 = 0.0 = 0%
1M FM = 50%100% = 0.51.0 = 0.5 = 50%
If the first child born is a male, the other absolutely must be female 100%, agreed?

1st child 50% F 2nd child 50% F
2nd child 50% M
1F 2F = 50%50% = 0.50.5 = .25 = 25%
1F FM = 50%50% = 0.50.5 = .25 = 25%
If the first child is born a female, there is an equal chance that the 2nd child is male or female, since we assume when a child is born it has 50% of being either or.

Now the question asks for the chance of a male and female combination, there is a 50% chance of a male 1st and female 2nd combination
25% chance of a female 1st and female 2nd combination
25% chance of a female 1st and male 2nd combination
a female female combination doesnt meet the requirements, however a male female or female male combination does. Adding 50% and 25% gives 75%.
Answer = 75% chance that a family will have a male and female child combination out of all the child gender combinations possible in a family with 2 children.

The key is, regardless of the fact that we know one child is a female, the first born has a 50% chance of being M or F. We didn’t know that one out of two children would be female at the time of birth of the first child.
At this point the only thing left to argue is interpretation of the problem, which i dont really care about

20

Long and involved, but incorrect. An easy way to see this is to note that the original problem has no information about “first” or “second” children. Thus, I could retype your entire post, substituting “first” for “second” (and vice versa) and have the argument be just as valid. However, I’d some up with the probability that 2M 1F = 50%100% = 0.51.0 = 0.5 = 50%, which is clearly at odds with your post, which give the probability as 25%. Since these are conflicting, the underlying logic must be wrong.