I was looking around on the internet, and I found something about Marilyn vos Savant. When I was reading about it I saw an interesting math problem. This was taken directly from Wikipedia, so I'm not certain about its accuracy in wording:

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

Please answer the poll before reading on.

I think Vos Savant said that it was a 1/3 chance, although the Wikipedia article wasn't very clear about it. However, I though it was interesting, and since this is seems to be a smart forum, I thought I'd ask it here.

50% surely...?

Unless I'm missing something.

No offfence but this thread is kinda stupid, and further more it's also pointless.

But whatever. I'm guessing 50%

well, if their is only 2 beagles, we know one is a male, then thier is a 50/50 chance of the other being male or female....

Does it matter what gender the other dog is?

The chance that both dogs are male is 25%. Flip two coins, whats the chance of both being heads or both being tales? 25% with a 50% that one will be heads one will be tails.

The same thing applies here. I think. I might have read the question wrong.

Okay. I feel stupid, but the 75% was jumping out at me.

Just a little flag going up in my brain somewhere. Either that, or early onset dementia.

basically you're saying forget the dog we know is male, what's the other one? thusly it's 50/50 it can be male or it can be female.

Should we factor in the chance that it could be a hermaphrodite?

It's a pretty basic math problem unless you're supposed to factor in what GhostOfSin suggested. Then it's just trying to be tricky.

75% works because we know one is male and the chance the other is male is half of that (1\2 of 25% + 50% = 75%)

I thought about this way too much.

ObnoxiousTwat:
75% works because we know one is male and the chance the other is male is half of that (1\2 of 25% + 50% = 75%)

I thought about this way too much.

No, it would be 50% because it's a separate beagle, the other one being male would have no effect on it.

ObnoxiousTwat:
75% works because we know one is male and the chance the other is male is half of that (1\2 of 25% + 50% = 75%)

I thought about this way too much.

Thats not the way probability works.

To find the probability of two things you must multiply them. The chance of one dog being male is 50% or 1/2.

1/2 is 0.5. 0.5 times 0.5 is 0.25 or 25%

The probability of both dogs being males is 25%. However looking at the question it appears to be asking what the chance of one dog being male is, which is 50%. It's a strange question that can interpreted differently.

TheGhostOfSin:
Should we factor in the chance that it could be a hermaphrodite?

You just made my day, GhostOfSin- heck, you just made my week.

50%.

Okay, factor in that it's a 100% chance that one of two is that it's female, and that the question asks what the other one will be, not regarding the male one.

It has to be 50%, because no matter what, there's an even chance of being male or female.

BASIC MATHS, PEOPLE: Read the question. They throw in Red Herrings on purpose.

Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

Of these three possible outcomes, only one of them results in the second beagle being male.
Thats where the 33% comes from...
That's the logic behind the problem, however there's also the idea of independent events... the mind boggles.

TheGhostOfSin:
Should we factor in the chance that it could be a hermaphrodite?

That's exactly what I was thinking after I finished reading the question...

But it's 50%, we know one dog's a male and they're asking what's the probability of the other being male. It's not asking for the possibility of a pair or not, just if the other is male or not.

50%

I just know there's something sinister behind this.

Damn you math!! Even outside of school you make me look stupid.

dirtface:
Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

Of these three possible outcomes, only one of them results in the second beagle being male.
Thats where the 33% comes from...
That's the logic behind the problem, however there's also the idea of independent events... the mind boggles.

Never thought to use the Monty Hall problem for this. Nice.

50%: It's either a male or its something else.

Thing is, the gender of the second has no apparent relation to the gender of the first unless you factor in a whole shit load of biological variables which I don't really care to. As such, the two are as mutually exclusive as one coin-flip from the next. (Keep in mind that -any- series of coin results is as unlikely as any other.)

So I'll say 50%.

Fire Daemon:

The probability of both dogs being males is 25%. However looking at the question it appears to be asking what the chance of one dog being male is, which is 50%. It's a strange question that can interpreted differently.

Not really.

dirtface:

Male Female
Female Male

This is the same thing

dirtface:
Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

Of these three possible outcomes, only one of them results in the second beagle being male.
Thats where the 33% comes from...
That's the logic behind the problem, however there's also the idea of independent events... the mind boggles.

This.

Then again, I don't remember very much about probability and the last time I studied that subject was 5 years ago.

Yay Probability!

If you flip a coin and it comes up Heads 7/20 tosses, what are the chances the 21st toss will be Heads again?

-deleted (not worth it)

dirtface:
Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

Of these three possible outcomes, only one of them results in the second beagle being male.
Thats where the 33% comes from...
That's the logic behind the problem, however there's also the idea of independent events... the mind boggles.

So is this thing right or were the rest of us right?

based on the question stated in the pole it's 50% the other one is male

simply because you can disregard anything about the other dog, the lady is unsure if it's male or female but one is definitely male. so this means the other dog has a 50% chance of being male or female irregardless of the sex of the other dog

dirtface:
Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

Of these three possible outcomes, only one of them results in the second beagle being male.
Thats where the 33% comes from...
That's the logic behind the problem, however there's also the idea of independent events... the mind boggles.

Give this person a banana. The two beagles are already selected as part of a set; thus the 33%. If you selected one beagle, discovered it was male, then selected another beagle, the chance would be 50%. But if you select them two at a time and identify one as male, you remove the chance that both are female before the second chance.

dirtface:
Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

Of these three possible outcomes, only one of them results in the second beagle being male.
Thats where the 33% comes from...
That's the logic behind the problem, however there's also the idea of independent events... the mind boggles.

Well, this is actually why I posted this topic. I think that this is the mathematically proven solution, atleast according to Wikipedia. However, if you think about the problem as the person randomly grabbed one and it happened to be male, instead of looking at both and then replying, I think the answer is actually 50%. This answer seems right to me because if they just randomly grab one, there's a 50% chance that he would have grabbed a male in one of the two m/f pairs, and a 100% chance in the m/m pair. So, at least to my tired brain, there's a 50% chance that the male is from the m/m pair and a 50% from the m/f pair, therefore a 50% of the other being male. That seemed to be vaguely like the Bertrand's box paradox.

dirtface:
Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

fixed: that says the same as the one above it so it's irrelevant
we know that the first is male so it would look like this
male/ male
male/female

werepossum:

dirtface:
Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

Of these three possible outcomes, only one of them results in the second beagle being male.
Thats where the 33% comes from...
That's the logic behind the problem, however there's also the idea of independent events... the mind boggles.

Give this person a banana. The two beagles are already selected as part of a set; thus the 33%. If you selected one beagle, discovered it was male, then selected another beagle, the chance would be 50%. But if you select them two at a time and identify one as male, you remove the chance that both are female before the second chance.

But the gender of the second one isn't dependent on the first one. It's still 50% right?

(ZHU) Michael:

dirtface:
Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

fixed: that says the same as the one above it so it's irrelevant

Ah, but there's a reason it was like that- it was the order.

Shivari:

dirtface:
Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

Of these three possible outcomes, only one of them results in the second beagle being male.
Thats where the 33% comes from...
That's the logic behind the problem, however there's also the idea of independent events... the mind boggles.

So is this thing right or were the rest of us right?

... The wikipedia article states its 1/3, i was just explaining it.
For the other guy who said that male female was the same as female male - you're sortof correct, but not really. they are different cases that can be grouped...

jamanticus:

(ZHU) Michael:

dirtface:
Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

fixed: that says the same as the one above it so it's irrelevant

Ah, but there's a reason it was like that- it was the order.

I came back for three mins. after leaving to have fun so im not getting into this cirle logic again