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# One Girl - One Boy

Best Answer Riddari, 13 July 2007 - 07:38 PM

Spoiler for solution
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348 replies to this topic

### #41 Martini

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Posted 28 July 2007 - 01:03 AM

There are two aspects.

1] We are confusing sequential events with simultaneous events.

If we toss a coin, the probability of getting a head is 1/2. If we toss two coins together, the probability of getting 2 heads is 1/3 (3 possibilities, head-head, tail-tail and head-tail)...

No, you're the one who's confused. When you toss two coins, one at a time or together, the probability of getting one H and one T is not 1/3. You are assuming that HH, TT, and HT (as combinations, as you put it) are equally likely. They are not. HT/TH is twice as likely as the others. If you don't believe it, just toss two coins a hundred times, sequentially or together. I guess you didn't bother with the coin experiment I posted for you earlier?

Your whole post is infused with silliness about the magical relevance of simultaneity. (Do you think it makes a difference if events are almost, but not quite, at the exact same time; say, within 5 ms of each other? At what point do things switch to the simultaneous distribution? Does this happen discontinuously?)
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### #42 Riddari

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Posted 28 July 2007 - 01:09 AM

I will use your coin flip example just because I am sick of regering to children and want some variety. ;p

Anyway, let say you flip two coins, consecutively or sequentially as you like, and record the results. You might get something like this:

TH,HH,HT,TT,TT,TT,TH,HH,TH,HT (This is actual results that I recieved.)

Two of those ten combinations are HH, three of them are TT and five of them are mixed. This actually falls very close to what one would expect.

Now, let say we throw out all of the TT combinations because we only want instances that include a head. So, HH constitutes two of the seven remaining instances, or `28.6%. Inversely, lets say we throw out the HH combinations because we only want instances with a tail. The TT groups would constitute three of eight instances, or 37.5%. Remarkably enough, the average of those two results is ~33.0%

Mock up the numbers as you like and you will end up with similar results. The worst you can really do is force all instances to be the same and that does not support a 25% answer any more than a 33% answer. If you go unreasonable to one side, it will balance on the other.

HH,HT,HH,HT,HH,HT.HH,HT,HH,HT would give you 50% for HH or 0% for TT, but it is a ridiculous distortion of expectations.
HH,HT,HH,HT,HH,TT.HH,HT,HH,HT gets 55% for HH and 20% for TT, averaging to ~37.8%.

As with all the previous compelling arguments for the 33% answer, I am feeling pretty solid with on the logic behind this.

We are not talking about the possible gender of a single child, but rather the possible genders of a set of two children with the specific criteria that at least one of them is female.
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### #43 dsu

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Posted 28 July 2007 - 03:19 AM

We are not talking about the possible gender of a single child, but rather the possible genders of a set of two children with the specific criteria that at least one of them is female

Ok, I understand your point of view, but you do not note the fallacy that creeps in when you combine both permutation and combination! When you create your set, you are using permutation. When you are pointing out the girl, you are using combination. Let me ellaborate the fallacy:
What are the chance the other coin is H if the 1st coin is H? 50% or 1/2 (out of HT and HH).
What are the chance the other coin is H if the 2nd coin is H? 50% or 1/2 (out of HH and TH).
Combining the above two,
What are the chance the other coin is H if the 1st coin or the 2nd coin is H? How can the answer be less than either of the above two, as it includes both the cases??? That is the hidden fallacy of combining permutation and combination.

What are the chances of the other child being a girl if the 1st child is a girl? 50% or 1/2 (out of GG and GB)
What are the chances of the other child being a girl if the 2nd child is a girl? 50% or 1/2 (out of GG and BG)

What are the chances of the other child being a girl if the 1st or the 2nd child is a girl? How can the chance be less than the above two, especially as both the above cases are included??? The fallacy disappears if we say:
What are the chances of the other child being a girl if one of the child is a girl?
(in combination, we have these members in the set: GG, GB, BB. So the answer needs to be 1/2).

If you can solve the fallacies pointed out above, I shall accept your point of view.

Your whole post is infused with silliness about the magical relevance of simultaneity

I understand what you mean - maybe I have used the wrong terms or expressions. What I meant to point out is probability is a tricky thing which may easily give wrong answers if used in the wrong context.
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### #44 Martini

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Posted 28 July 2007 - 04:31 AM

How can the answer be less than either of the above two, as it includes both the cases???

Because there isn't any inclusion going on here; you've re-conditionalized your probability. First, you were conditioning on the 1st coin being H. Then, you were conditioning on the 2nd coin being H. Finally, you conditioned on either one being H. You can't just add up conditional probabilities in the normal way, however, if they don't all have the same antecedent condition. Your last case doesn't "include" any of the previous cases, because they all have different conditions. What would be closer to the inclusion relationship you're going for is this:
What is the probability that the second coin is H and the first coin is H? 1/4
What is the probability that the first coin is H and the second coin is H? 1/4
What is the probability that both coins are H and either the first or second coin is H? 1/4
That is, we turn the conditionalizing into conjunctions instead. This makes the third case subsume the previous two, and thus the third case has probability >= each of the previous two. [As it happens, in this example, all three cases turn into the same thing when we do this. Ah well.].

I understand what you mean - maybe I have used the wrong terms or expressions. What I meant to point out is probability is a tricky thing which may easily give wrong answers if used in the wrong context.

Actually, what you claimed about the odds of coin flips landing on head or tails being dependent on whether or not they landed simultaneously had nothing to do with being wrong based on terms and expressions. The logic was nonsensical.
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### #45 dsu

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Posted 28 July 2007 - 06:21 AM

What is the probability that the second coin is H and the first coin is H? 1/4
What is the probability that the first coin is H and the second coin is H? 1/4
What is the probability that both coins are H and either the first or second coin is H? 1/4

It is remarkable you said 1/4, not 1/3 in each of the above!

But you say

the chance of there being two heads if the 1st coin or the 2nd coin is H is 1/3.

Ironic, isn't it? You talk about "re-conditionalized your probability", but aren't you doing just that?

In the first, TT is a member of the set, but in the second, it is (conveniently) not! But there is essential no difference between the two statements "What is the probability that both coins are H and either the first or second coin is H? 1/4" and "the chance of there being two heads if the 1st coin or the 2nd coin is H is 1/3"

You are repeatedly ignoring the fact I am trying to point out - you cannot use permutation and combination together - one in making the set, the other in setting up the conditions. Either use permutation, or combination, not both. If you use permutation to define the set, do not use the word "other". If you use the word "other", use combination in defining the sets.
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### #46 imtcb

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Posted 28 July 2007 - 06:31 PM

A lot of the initial confusion in this riddle is that it was worded differently the first few days. Initially, it was implied that a couple was going to have 2 children (future tense), and the question was, if the first child was a girl, what would the probability be for the second to also be a girl.

I do see with the wording now, the actual intent of the OP. It is indeed 1/3 the way it is worded now, just as it is 1/4 the way it was worded originally.

The problem with applying the gamblers fallacy to all probability is that gambling is set up to give better odds to the house, which means that the longer you gamble, the more you will loose. No casino will give straight odds on a 50/50 coin toss, because in the end they will only break even. This is why in roulette black and red are 2:1 payout, yet the odds are not 50/50 (2 greens).

Before the events happen, if the probability is 50/50, 2 consecutive events would yield a 25% probability for each of the 4 possible outcomes. BB / BG / GB / GG
Between events, the probability would be 50% for each of the 2 possible outcomes. GB / GG
After the events, when one of the two is given (but not the order), the probability would be 33 1/3%. BG / GB / GG
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### #47 skale

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Posted 28 July 2007 - 07:00 PM

A lot of the initial confusion in this riddle is that it was worded differently the first few days. Initially, it was implied that a couple was going to have 2 children (future tense), and the question was, if the first child was a girl, what would the probability be for the second to also be a girl.

Let me assure you that the only time post was edited was to add the solution tag. But am glad you see the reasoning now.
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### #48 unreality

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Posted 28 July 2007 - 08:11 PM

there is a lot of confusion between the two sides of this problem, and i've read all ur arguements, and it isnt about conditionals or conjunctionals or anything

Ok, so Teanchi and Beanchi are a married couple (dont ask me whose he and whose she)!

They have two kids, one of them is a girl, what is the probability that the other kid is also a girl.

Assume safely that the porbability of each gender is 1/2.

this is the problem.

let me requote the important part (middle line)

"They have two kids, one of them is a girl, what is the probability that the other kid is also a girl."

"they have two kids." They already have them. They're not expecting kids... they already have both of them born. The kids could be 20 and 18 years old for all we care.

"One of them is a girl"
okay... so if we have two kids, already born, and we can rule out B/B:
G/B
B/G
G/G

"what is the probability that the other kid is also a girl."

not: "what is the probability that the other kid will be a girl."

In that case it would be 1/2, since the other child does not affect the probability. But it doesnt say that. It said "is also a girl". The child is already born.

it all comes down to two sides:
Side A (answer is 1/2): We're looking at the second child's birth as an independent event.
Side B (answer is 1/3): We're looking at both together.

and I am sorry to say Side B is right (as I was on Side A for quite some time)

Why?

Because the second child has already been born. We dont care what was the probability that it was boy/girl while it was in its mother's womb. We know thats 1/2. The fact is, the gender has already been determined. We're just trying to figure out the probability they are both girls.

Since the options are:

G/B
B/G
G/G

we can safely say, G/G has a 1/3 chance.

There is no fallacy. There is no bla-bla-bla. It all comes down to wording:

They have two kids, one of them is a girl, what is the probability that the other kid is also a girl.

the "other kid" is already born. The solution doesnt call for the probability that the other kid will be a girl. It calls for the probability that they are (as in right now) BOTH girls.

We're not looking at one girl's chances. In that case there would be two options:
B
G

but we're not. BECAUSE OF THE WORDING, we are looking at both children at once:
G/B
B/G
G/G

It is 1/3.

It pains me to say this (I defended 1/2 for a while until i realized i was wrong) but I'm not being bitter and I'm trying to prove what I now see is right.

its a fine line, based on skale's wording. but the answer is 1/3. I rest my case. I think this topic should be done. The discussion cant go much elsewhere.
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### #49 Martini

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Posted 28 July 2007 - 09:54 PM

Ironic, isn't it? You talk about "re-conditionalized your probability", but aren't you doing just that?

There's nothing wrong with re-conditionalizing of probabilities. You just have to keep in mind that that is in fact what you've done, and make sure not to add things up in the wrong way.

But there is essential no difference between the two statements "What is the probability that both coins are H and either the first or second coin is H? 1/4" and "the chance of there being two heads if the 1st coin or the 2nd coin is H is 1/3"

Those are both true statements. Do you mean to say there is no difference between "The probability that both coins are H and either first or second coin is H" and "The chance of there being two heads if the 1st coin or the 2nd coin is H"? Because those two statements have all the difference in the world; the first of those is a nonconditional probability, and the second is a conditional probability. They're as different as "What is the probability of rolling a 2 and rolling an even number?" (which would be 1/6 on a normal die) and "What is the probability of rolling a 2 given that you've rolled an even number?" (which would be 1/3).

Did you try any of the experiments that I and others have posted? If you haven't, please take the time to try them. You'll find that they won't give the results you're saying they will and I don't think you'll have a convincing argument that the coins are lying.
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### #50 kippi3000

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Posted 29 July 2007 - 08:02 AM

unreality: well put. Your reasoning is the correct one, and the explanation is most aptly put.

for those who can't agree that the answer is 1/3, READ unreality's post before u post another absurd argument that it is 1/2.

good play, unreality!
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