Probability Conundrum

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• Oct 3rd 2010, 06:54 AM
undefined
Quote:

Originally Posted by Nath
I've tried flipping the two coins from the previous example and I get the result from the previous example :-)

Make sure you are counting thus:

Keep track of two counts, say, c1 and c2, both initially set at 0.

We are equating "girl" with "tails."

If you get HH increment neither c1 nor c2.
If you get HT increment c1 but not c2.
If you get TH increment c1 but not c2.
If you get TT increment both c1 and c2.

At the end, the probability is c2/c1.

If you are not counting thus, then you are probably solving a different problem from what you specified.
• Oct 3rd 2010, 07:19 AM
undefined
This may be my last valiant attempt at explaining the same thing over again:

The problem statement (1) "I have 2 children. One is female. What is the probability the other is also female?" is not very well stated.

We can restate as follows. (2) "I have 2 children. At least one is female. What is the probability that they are both female?"

For this problem, the probability is 1/3.

If, however, we were to interpret it as: (3) "I have 2 children. One of them is standing in front of me right now, and she is female. What is the probability that the other one is female?" then in that case the probability is 1/2.

Edit: Yet another way to restate, which is equivalent to (2), is: (4) "I have 2 children. At least one is female. What is the probability that, if I pick a child and that child turns out to be female, then the other child is also female?" It's less concise but includes the phrase "the other."
• Oct 3rd 2010, 08:36 AM
CaptainBlack
Quote:

Originally Posted by Nath
I've tried flipping the two coins from the previous example and I get the result from the previous example :-)

Post the results of your coin flip experiments.

CB
• Oct 3rd 2010, 09:13 AM
Quote:

Originally Posted by Nath
With both the Fortran simulation example and the Bayes Theorem quotation, the event space is skewed because again, it assumes that BG & GB are separate events, but that GG is only 1 event, even though the 2 Gs are separate entities (i.e. G1G2 & G2G1 are separate events in the event space).

************************
If \$\displaystyle G_1G_2\$ and \$\displaystyle G_2G_1\$ are seperate,
then you could model the situation as follows.

Of the 4 children Helen, Mary, Roger and Sean,
in how many ways can you form 2-child families with these children ?

This will be the "Sample Space"

The first child will be the younger one

Helen Mary
Mary Helen
Helen
Roger
Roger Helen
Helen Sean
Sean Helen
Mary Roger
Roger Mary
Mary Sean
Sean Mary
Roger Sean
Sean Roger

If Helen is the younger child, there is a 1/3 chance she is the younger child of 2 sisters.
If Mary is the younger child, there is a 1/3 chance she is the younger child of 2 children.

If the younger child is a girl, there is a 2/6 chance she is the younger of 2 sisters.

If Helen is the older child, there is a 1/3 chance she is the older child of 2 sisters

etc etc.
************************

Therefore I stick with my original proof:

I define 2 sets c1 & c2, each containing the set of possible states each of the coins can take after each is thrown:

c1 = {H1, T1}
c2 = {H2, T2}

Throwing both coins at the same time gives the following set of possibilities:

{H1, H2}
{H1, T2}
{H2, T1}
{T1, T2}

So I throw the 2 coins so that you can see them when they land, but I cannot. You tell me that one of the coins is in a state that is a member of the set {T1, T2}. You then ask me the probability that the other coin is in a state that is also a member of this same set. The set {H1, H2} is therefore eliminated, leaving the following possibilities:

{H1, T2}
{H2, T1}
{T1, T2}

3 equally likely scenarios with a 1/3 chance of TT

But if a throw of c2 gives T2, then {H2, T1} is eliminated, so there remains a 50% chance that c1 = H1 and a 50% chance that c1 = T1

One way we can model this with children is to tie a girl and boy back to back using rope.
With both boys tied to both girls facing in opposite directions, we can turn them around and around
and stop them in an unbiased way.
Now, suppose Helen is tied to Roger and you are told that after spinning, Helen is facing the front.

In this way, we can get Roger out of the equation.

After spinning the other pair of children around, we can say that Mary and Sean
have equal likelihood of facing the front also.
Then the probability of Mary facing the front is unrelated to the probability of Helen facing front and is 0.5

Conversely, if a throw of c1 gives T1, then {H1, T2} is eliminated, so there remains a 50% chance that c2 = H2 and 50% chance that c2 = T2.

You are generally falling short with the Sample Space.
• Oct 3rd 2010, 02:29 PM
Nath
It seems we're now in agreement then because undefined's (3) interpretation above is the most logical way to understand the problem, and undefined has admitted this leads to a probability of 1/2.

Also Archie Meade's clever example of tying them back to back also leads to the same probability of 0.5: "Then the probability of Mary facing the front is unrelated to the probability of Helen facing front and is 0.5".

So it seems we now agree.

Anyway, Captain Black and undefined asked me to run simulations. Here are the results:

---
#include <iostream>
#include <math.h>
#include <time.h>

int main(void)
{
int event_one_TAIL = 0;
int events_two_TAILs = 0;
int coin1;
int coin2;

enum coin_type
{
TAILS = 2
} coin_throw;

srand ( time(NULL) );

for (int loop =0; loop < 10000; loop++)
{
coin1 = rand() % 2 + 1;
coin2 = rand() % 2 + 1;

if ((coin1 ==HEADS) && (coin2 ==TAILS))
{
event_one_TAIL++; // coin2 specifically is TAILS
}
else if ((coin1 == TAILS) && (coin2 == HEADS))
{
event_one_TAIL++; // coin1 specifically is TAILS
}
else if ((coin1 == TAILS) && (coin2 == TAILS))
{
events_two_TAILs++; // both coins, specifically 2 coins, are tails
}
}
std::cout << std::endl << "event_one_TAIL: " << event_one_TAIL << std::endl;
std::cout << "events_two_TAILs: " << events_two_TAILs << std::endl;
std::cout << "P(2 TAILS)= events_two_TAILs/event_one_TAIL = "
<< static_cast<float>(events_two_TAILs) /
(static_cast<float>(event_one_TAIL+events_two_TAIL s)) << std::endl;
}
---
This leads to undefined's probability of 1/3, if compile it with the GNU g++ and run it.
---
But it still doesn't wash. If you know one *specific* coin has landed on TAILs, then at that precise point in time, this will then *exclude* one of the events in the event space, in which case one, and exactly one, of the statement blocks below can be commented out, either here:

#include <iostream>
#include <math.h>
#include <time.h>

int main(void)
{
int event_one_TAIL = 0;
int events_two_TAILs = 0;
int coin1;
int coin2;

enum coin_type
{
TAILS = 2
} coin_throw;

srand ( time(NULL) );

for (int loop =0; loop < 10000; loop++)
{
coin1 = rand() % 2 + 1;
coin2 = rand() % 2 + 1;

/* commented out because the one coin that landed on TAILS was NOT coin2
if ((coin1 ==HEADS) && (coin2 ==TAILS))
{
event_one_TAIL++; // coin2 specifically is TAILS
}
*/
if ((coin1 == TAILS) && (coin2 == HEADS))
{
event_one_TAIL++; // coin1 specifically is TAILS
}
else if ((coin1 == TAILS) && (coin2 == TAILS))
{
events_two_TAILs++; // both coins, specifically 2 coins, are tails
}
}
std::cout << std::endl << "event_one_TAIL: " << event_one_TAIL << std::endl;
std::cout << "events_two_TAILs: " << events_two_TAILs << std::endl;
std::cout << "P(2 TAILS)= events_two_TAILs/event_one_TAIL = "
<< static_cast<float>(events_two_TAILs) /
(static_cast<float>(event_one_TAIL+events_two_TAIL s)) << std::endl;
}

or here:

#include <iostream>
#include <math.h>
#include <time.h>

int main(void)
{
int event_one_TAIL = 0;
int events_two_TAILs = 0;
int coin1;
int coin2;

enum coin_type
{
TAILS = 2
} coin_throw;

srand ( time(NULL) );

for (int loop =0; loop < 10000; loop++)
{
coin1 = rand() % 2 + 1;
coin2 = rand() % 2 + 1;

if ((coin1 ==HEADS) && (coin2 ==TAILS))
{
event_one_TAIL++; // coin2 specifically is TAILS
}
/* commented out because the one coin that landed on TAILS was NOT coin1
if ((coin1 == TAILS) && (coin2 == HEADS))
{
event_one_TAIL++; // coin1 specifically is TAILS
}
*/
else if ((coin1 == TAILS) && (coin2 == TAILS))
{
events_two_TAILs++; // both coins, specifically 2 coins, are tails
}
}
std::cout << std::endl << "event_one_TAIL: " << event_one_TAIL << std::endl;
std::cout << "events_two_TAILs: " << events_two_TAILs << std::endl;
std::cout << "P(2 TAILS)= events_two_TAILs/event_one_TAIL = "
<< static_cast<float>(events_two_TAILs) /
(static_cast<float>(event_one_TAIL+events_two_TAIL s)) << std::endl;
}
---
In which case both lead to the value of 0.5
• Oct 3rd 2010, 02:47 PM
My back-to-back example was chosen to illustrate the similarity to the two coins.
In real life, we don't do that to children. Hence the 1/2 achieved there doesn't fit.

Undefined may have meant something else because..

"I have 2 children, one of them is standing in front of me.
She is female"....

there is twice the chance of her having a brother than a sister, so it's still 1/3.

You will only get the p=1/2 case if you analyse "families that have 2 children".
you get the 1/3 case for a single 2-child family.
• Oct 3rd 2010, 04:39 PM
undefined
Quote:

Undefined may have meant something else because..

"I have 2 children, one of them is standing in front of me.
She is female"....

there is twice the chance of her having a brother than a sister, so it's still 1/3.

Suppose I have 2 children named Alex and Chris. (Both unisex names.) Given that Alex is female, what is the probability that Chris is also female? You're really going to argue that this is 1/3? Alex being female and Chris being female are independent events. So for this interpretation the solution is 1/2.

Quote:

Originally Posted by Nath
It seems we're now in agreement then because undefined's (3) interpretation above is the most logical way to understand the problem, and undefined has admitted this leads to a probability of 1/2.

Also Archie Meade's clever example of tying them back to back also leads to the same probability of 0.5: "Then the probability of Mary facing the front is unrelated to the probability of Helen facing front and is 0.5".

So it seems we now agree.

I like agreement! Until recent replies, I didn't realize interpretation was still a problem, I thought it was just the maths that was called into question.

For legibility, please enclose your source code in either [code][/code] or [php][/php] tags. (I realize C++ is not PHP, but the PHP tags give syntax highlighting whereas code tags do not.)

I thought that the following passage you wrote in post #8 supported statement (2) in post #17.

Quote:

Originally Posted by Nath
I define 2 sets c1 & c2, each containing the set of possible states each of the coins can take after each is thrown:

c1 = {H1, T1}
c2 = {H2, T2}

Throwing both coins at the same time gives the following set of possibilities:

{H1, H2}
{H1, T2}
{H2, T1}
{T1, T2}

So I throw the 2 coins so that you can see them when they land, but I cannot. You tell me that one of the coins is in a state that is a member of the set {T1, T2}. You then ask me the probability that the other coin is in a state that is also a member of this same set. The set {H1, H2} is therefore eliminated, leaving the following possibilities:

{H1, T2}
{H2, T1}
{T1, T2}

I took the sentence "You tell me that one of the coins is in a state that is a member of the set {T1, T2}" to mean "You tell me that (at least) one of the coins is in a state that is a member of the set {T1, T2}, and you don't specify a particular coin."

I think this thread is a little more interesting from a linguistics standpoint than a mathematics standpoint but all in all a bit of an exercise in frustration at that. Hopefully all is (almost entirely) cleared up now though.

I'm wondering if the essence of the ambiguity is that the phrase "one of them" can be interpreted as "a particular one of them" or "at least one of them" or "exactly one of them". Haha.

I'm inclined to agree that the sentence most reasonably should be interpreted to give answer 1/2, but based on the answer given by the question-makers, it is clear they meant the interpretation that gives answer 1/3. They just weren't careful with language.
• Oct 3rd 2010, 04:54 PM
Quote:

Originally Posted by undefined
Suppose I have 2 children named Alex and Chris. (Both unisex names.) Given that Alex is female, what is the probability that Chris is also female? You're really going to argue that this is 1/3? Alex being female and Chris being female are independent events. So for this interpretation the solution is 1/2.

I won't argue with anyone.

If they are independent events, you are basically saying

"what's the probability of a child being female"

Clearly a different question.
• Oct 3rd 2010, 05:14 PM
undefined
Quote:

I won't argue with anyone.

If they are independent events, you are basically saying

"what's the probability of a child being female"

Clearly a different question.

Of course we should assume they are independent events.

(3) "I have 2 children. One of them is standing in front of me right now, and she is female. What is the probability that the other one is female?"

There is nothing to suggest that the gender of one depends on the gender of the other.
• Oct 4th 2010, 03:45 AM
Quote:

Originally Posted by undefined
Of course we should assume they are independent events.

(3) "I have 2 children. One of them is standing in front of me right now, and she is female. What is the probability that the other one is female?"

There is nothing to suggest that the gender of one depends on the gender of the other.

The problem with that is that it is yet another fragmented sample space.

The sample space of possibilities is Af Cf, Af Cm, Am Cf, Am Cm.

The probability of Alex being female is 1/2
Then the probability of Alex standing in front of you and being female is 1/2

The probability of Alex being female and Chris being female is 0.5(0.5)=1/4
The probability of Alex being female and Chris being female is 0.5(0.5)=1/4

Those probabilities don't sum to 1.
• Oct 4th 2010, 04:27 AM
CaptainBlack
Quote:

Originally Posted by Nath
It seems we're now in agreement then because undefined's (3) interpretation above is the most logical way to understand the problem, and undefined has admitted this leads to a probability of 1/2.

Also Archie Meade's clever example of tying them back to back also leads to the same probability of 0.5: "Then the probability of Mary facing the front is unrelated to the probability of Helen facing front and is 0.5".

So it seems we now agree.

Anyway, Captain Black and undefined asked me to run simulations. Here are the results:

:
:
:

---
This leads to undefined's probability of 1/3, if compile it with the GNU g++ and run it.
---
But it still doesn't wash. If you know one *specific* coin has landed on TAILs, then at that precise point in time, this will then *exclude* one of the events in the event space, in which case one, and exactly one, of the statement blocks below can be commented out, either here:

You don't know that one specific coin has landed TAILs, all you know is that at least one of them has. The game is this: I toss two coins, if they land HH I toss again. For any other result I show you a coin that has landed T. Now we ask what is the probability that the other is T.

Consider this. We meet in the street and I volanteer that I have two chidren, then I mention that the elder is a girl. Then the probability that the younger is a girl and hence that both are girls is 1/2.

But that is not what I say, what I do is volanteer that I have two chidren, then I mention that at least of them is a girl.

This clearly contains less information so you should expect that in this case the probability that both are girls is < 1/2.

The above is the problem that is being discussed, there is no information about the gender of the other child, and no identification of which is the child I have in mind.

But no matter, I have closed the thread due to a non-meeting of minds.

CB
• Oct 4th 2010, 11:27 AM
mr fantastic
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Quote:

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