Lotto Probability

[Just thinking]

... But if you know one is a boy, then it simply becomes P(B) = 1/2, since the two events are independent.

The other way is enumerating the options: BB, GG, BG, GB. In knowing that one of them is a boy, we have to assign one child that role. Most people, expecting a trick, would say "Now our options are reduced to BB, BG, GB, so the probability is 1/3" but really, GB isn't an option.

I'm not so sure what you are saying is correct. Please explain to me how GB is not a possible outcome knowing only that one is a boy.

 

[illuminatedwax]

I'm not so sure what you are saying is correct. Please explain to me how GB is not a possible outcome knowing only that one is a boy.

Oh, sorry. I'm saying that deciding one is a boy also forces you to decide where in the order that boy is. So if you decide it is the first, GB is not an option. If you decide it is the second, BG is not an option. Either way, you're eliminating two of your original four possibilites, not just one.

 

[Just thinking]

Oh, sorry. I'm saying that deciding one is a boy also forces you to decide where in the order that boy is.

I had a feeling that's what you meant ... but then usually one would say "the older one is a boy" ... or the "younger" ... or "shorter". Just saying one is a boy doesn't really force one to pick one and stick to it -- it really leaves either position open to being a boy. (At least that's how I read it -- and I think most others would too.) That's why I said in that case P = 0.33; but once you specify it becomes 0.50.

 

[ynot]

Are you saying, in a toss of two coins, that a combination of heads and tails will come up more often than either two heads or two tails?

 

[Just thinking]

Are you saying, in a toss of two coins, that a combination of heads and tails will come up more often than either two heads or two tails?

Yes ... you are more likely to get two different results than exactly two heads or exactly two tails. But you are just as likely to get two mis-matched outcomes (HT or TH) as you are to get two matched outcomes (HH or TT).

 

[ynot]

Yes ... you are more likely to get two different results than exactly two heads or exactly two tails.

This is not an answer to my question.

But you are just as likely to get two mis-matched outcomes (HT or TH) as you are to get two matched outcomes (HH or TT).

This is an answer to my question - Thanks.

So we agree that a combination of heads and tails (a mis-match) is equally as likely as two heads or two tails. It follows therefore that the odds of any one of the combinations described above occurring is 1/3. Replace heads and tails with boys and girls

A woman has two children. She either has two boys, two girls or a combination of boy and girl. The odds that she has two boys is 1/3.

A woman has two children. You are told that one is a boy. The two girl option has been removed. She either has two boys or a combination of boy and girl. The odds that she has one or the other is 50/50.

 

[illuminatedwax]

This is not an answer to my question.


This is an answer to my question - Thanks.

So we agree that a combination of heads and tails is equally as likely as two heads or two tails. It follows therefore that the odds of any one of the combinations described above occurring is 1/3. Replace heads and tails with boys and girls

A woman has two children. She either has two boys, two girls or a combination of boy and girl. The odds that she has two boys is 1/3.

A woman has two children. You are told that one is a boy. The two girl option has been removed. She either has two boys or a combination of boy and girl. The odds that she has either one is 50/50.

Ah, but there are two ways that a mother could have mismatched children. There are also two ways that you can have one Heads and one Tails. You forgot to count both of them.

Looking at it from a combinatoric standpoint, [latex]$P(E) = \frac{|S|}{|\Omega|}$[/latex] where [latex]$S\subseteq\Omega$[/latex] is the set of outcomes that satisfy E (an event), and [latex]$\Omega$[/latex] is the set of all possible outcomes. What is the size of Omega? Well, first we flip the first coin. There are 2 possible outcomes. Then we flip the second. There are 2 possible outcomes for that, so that means there are 2*2 or 4 possible outcomes. Only one of those outcomes is "both heads."

Just Thinking didn't say that P(HH) = P(TT) = P(one of each). Just Thinking said P(HH or TT) = P(one of each).

You can also use the definition of independent events, as I did earlier to prove that the chances are 1/4.

 

[ynot]

Ah, but there are two ways that a mother could have mismatched children. There are also two ways that you can have one Heads and one Tails. You forgot to count both of them.

Looking at it from a combinatoric standpoint,

where

is the set of outcomes that satisfy E (an event), and

is the set of all possible outcomes. What is the size of Omega? Well, first we flip the first coin. There are 2 possible outcomes. Then we flip the second. There are 2 possible outcomes for that, so that means there are 2*2 or 4 possible outcomes. Only one of those outcomes is "both heads."

Just Thinking didn't say that P(HH) = P(TT) = P(one of each). Just Thinking said P(HH or TT) = P(one of each).

You can also use the definition of independent events, as I did earlier to prove that the chances are 1/4.

“A woman has two children.” She’s not about to give birth to (or in the process of giving birth to) two children. She has children, not babies. The birth happened some time ago. It’s all over after the shouting (sorry ladies). What you’re doing is describing the combinations of odds relating to the order in which the children may have been born. The order in which she gave birth to the two children is totally irrelevant to the questions you posed. If you don’t agree with this (and I assume you don‘t), please explain exactly why.

Must admit tho - The math looks pretty

 

[ynot]

I say that the woman adopted the two children. The children had different mothers and they were born exactly at the same time. What say you now?

 

[illuminatedwax]

I say that the woman adopted the two children. The children had different mothers and they were born exactly at the same time. What say you now?

I say you don't know how to count.
Nonetheless: if you have an event E1 that has X possible outcomes, then you have X possible outcomes. If you consider two events, E1 and E2, where E2 has Y possible outcomes, then the number of total possible outcomes is X*Y. Example: you can choose to be a skeptic or a woo. This is E1. Your friend can also choose to be a skeptic or a woo. This is E2. Considering both your decisions, how many different outcomes can there be?

Answer: 4.

Let's put the child question a different way. Let's say you have a sibling, but for some unknown reason, you don't know if you are a boy or girl, and you don't know if your sibling is a boy or girl. How many possibilities are there?

Answer: 4.

You see, it has nothing to do with born first, same time, or any temporal stuff like that because the two events are INDEPENDENT, meaning that the probability of one does not depend upon the other. I guess "order" was misleading - the two children/coin tosses are distinct.

 

[BillyJoe]

About the New Information and the Monty Hall Puzzle.....

Okay, I agree.

The contestant is told right at the start that the host will
open an empty door. Therefore when the host actually does
open the empty door, there is no new information.

It all seems perfectly clear now

BJ

 

[BillyJoe]

About The Two Children Puzzle.....

The probabilities are not quite right.....

 

[ynot]

Before two coins are tossed there are a number of permutations that are possible as a result of the toss. The coins are tossed and the result is obviously one of those permutations. You are then asked what the odds are that the result is either a head plus a head or a tail plus a tail or a head and a tail. The answer is 1/3. You are not asked which if the possible permutations that were possible that the result represents. If the result happened to be a head and a tail, whether the head was coin A or coin B is irrelevant.

 

[Rasmus]

Before two coins are tossed there are a number of permutations that are possible as a result of the toss. The coins are tossed and the result is obviously one of those permutations.

Agreed.

You are then asked what the odds are the result is either a head plus a head or a tail plus a tail or a head and a tail. The answer is 1/3.

I would have said "1".

Thren I realised, you meant I would be asked one out of three possible questions.

The answer, then, would be diffierent depending on which question you'd ask:

a head plus a head: 1/4
a tail plus a tail: 1/4
a head and a tail: 1/2

You are not asked which if the possible permutations that were possible is the result. If the result happened to be a head and a tail, whether the head was coin A or coin B is irrelevant.

It might be irrelevant which coin was which, but that doesn't change the fact that there is only one way for two heads to come up,and two ways for a head and a tail to come up.

If you do not belive this, then please take two coins of the same value and try it out. Make a chart and count how many times out of 100 you get a tail and a head. (also, count the other results just for completeness)

Then take two coins of different value, repeat, this time also counting which coin was heads and which was tails if they show two different sides.

Then report back here with what you saw.

 

[drkitten]

So we agree that a combination of heads and tails (a mis-match) is equally as likely as two heads or two tails. It follows therefore that the odds of any one of the combinations described above occurring is 1/3. Replace heads and tails with boys and girls

That's (ahem) not right.

We agree that the odds of a match in general are the same as the odds of a mismatch in general.

It does not therefore follow that the chances of any one of the combinations occurring is 1/3 -- quite the opposite. The chance of a match happening is 1/2, of which there are two ways of getting it (HH or TT, both with probability 1/4). The chance of a mismatch happening is also 1/2, of which there are technically two different ways of getting it (HT and TH, probability 1/4) -- but you may or may not be able to distinguish these cases experimentally depending on how you throw the coins.

The chance of two heads is therefore 1/4; the chance of two tails is therefore 1/4, and the chance of one of each is therefore 1/2.

Doing the replacement:

A woman has two children. She either has two boys, two girls or a combination of boy and girl. The odds that she has two boys is 1/3.

The chance that she has two boys is 1/4.

A woman has two children. You are told that one is a boy. The two girl option has been removed. She either has two boys or a combination of boy and girl. The odds that she has one or the other is 50/50.

Again, no.

 

[illuminatedwax]

Before two coins are tossed there are a number of permutations that are possible as a result of the toss. The coins are tossed and the result is obviously one of those permutations. You are then asked what the odds are that the result is either a head plus a head or a tail plus a tail or a head and a tail. The answer is 1/3. You are not asked which if the possible permutations that were possible that the result represents. If the result happened to be a head and a tail, whether the head was coin A or coin B is irrelevant.

Ah, but it is a permutation! Therefore whether the head was coin A or coin B is relevant! Foiled by your own words!

Seriously, we've given like 5 proofs that the odds are 1/4. This is basic probability here. Are you just testing us??

 

[illuminatedwax]

About The Two Children Puzzle.....

The probabilities are not quite right.....

Which probabilities are not correct? The ones I stated or that 1/3 nonsense?

 

[Just thinking]

As in many (if not all) probability problems, the wording of said problem can make a big difference; if one does not agree with this, we can all argue a problem into the next millennium and get absolutely nowhere. So let's all please try our very best to be precise -- including me. (Example: If one means at least one is a boy, please say exactly that -- not just one is a boy.) That said, the probability of a given event is usually defined as the number of ways a favorable outcome can occur divided by the total number of possible outcomes. The rub here is in counting those ways or outcomes -- it's not always easy or obvious.

Making a table can often times help and clear up ambiguities. Let's look at the woman with two children and call them Child A and Child B ... what are the possible sexes for her children given no other information other than she has two children? Here's a table ...

Child A ----- Child B

Male -------- Male
Female ------ Female
Male -------- Female
Female ------ Male

We see 4 (four) possible outcomes; this will be the denominator. Now, how many result in two males? Just 1 (one); this is the numerator. Hence, the odds of having both males is 1/4 or 0.25.

What is the probability of having two of the same sex? Well, we have 1 of Female - Female and 1 of Male - Male ... thats 2 ways of having the same sex for both. So, in this case P = 2/4 or 1/2 or 0.50.

The probability for both to be of different sex also comes out to be 0.50 -- but I'll leave it up to the reader to see it for himself.

Finally, we see that the odds for having two children of the opposite sex is the same for having two of the same sex, since 1/2 = 1/2.

If one has a problem containing the phrase at least one is a boy, then we see that the total number of possible outcomes changes to just 3 (three), since one of the outcomes is no longer possible; the Female - Female option. This (3) becomes the new denominator -- it's no longer 4. Hence, the probability of having two males given at least one is a male is 1/3 or 0.33.

Hope this helps.

(A side note: The above is true given that all four outcomes are equally likely to occur -- when one gets into weighted events, things get even hairier.)

 

[ynot]

Here's another good pedagogical example of probability:

A woman has two children. What is the probability that both are boys?

Now you are informed that one of the children is a boy. Now what is the probability that both are boys?

Perhaps JT is correct and the disagreement is to do with the interpretation of the question.

I don’t think the question asks what the odds were during the process of determining the result. Rather that now that the result has been determined, what are the odds that the result is what it is.

The woman has a pair of children which could be any one of the following . . .

(1) A pair consisting of two males.
(2) A pair consisting of two females.
(3) A pair consisting of one of each gender.

Not . . .
(1) A pair consisting of two males.
(2) A pair consisting of two females.
(3) A pair consisting of one of each gender.
(4) A pair consisting of one of each gender.

I suppose it could be claimed the one or both of the children were hermaphrodites.

 

[Molinaro]

Molinaro,

Regarding the highlighted bit below....

Maybe you need to point this out, because it is not obvious to me what this new information is.
Before you even draw a card, you know that, if you draw a blue card, the probability that the other side is blue is 2/3. This does not change when you actually draw the card. Therefore, no new information.
This puzzle would seem to me to be quite different from the Monty Hall Puzzle where you are shown an empty door after your selection.

BillyJoe

The new info is right here in this line:

You pick 1 card at random, and look at the side that is face up. If it is blue, what is the probability that the other side of that card is also blue?

Before any card is chosen, you would expect the probability to be 1/2 -- that the other side of the card you choose, will be blue. However, here we are asking the question AFTER you pick a card and ONLY IF the face up side is blue. That essentialy changes the problem from one with a population of 4 events (4 different sides could be the 1 face up) into one with a population of 3 events (3 blue sides that could be the 1 face up).