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And as others have said, the coins is a perfect analogy. It's the same question. You've got a pair of things that can be one thing or another and are asked to discount one permutation.
Take two coins and flip them repeatedly. Record the results of two heads, two tails, or one of each. Now, cross out any results for whom the answer to "is there at least one head?" is "no". Would you expect to see a similar tally of two-heads to one-of-each? If your answer is "yes" I strongly suggest that you actually try it as an exercise.
Coin tosses are not a valid analogy, we are looking at a single pair of dogs and we know that one of them is male.
Coin tosses are a valid analogy, we are looking at a single pair of coins and we know that one of them is heads.
Coin tosses are a valid analogy, we are looking at a single pair of coins and we know that one of them is heads.
And you are now suggesting that the probability of throwing another head is 1/3?
Okay here's my spreadsheet...
https://docs.google.com/spreadsheets/d/1T7BqW-IdB-5xZp3CE0cql6qaWHLtLGSFCriMCpoYRsQ/edit?usp=sharing
It generates 1000 scenarios of two dogs, where each dog has a random 50:50 chance of being male or female.
You can see for yourself that empirically, in actual results, the probability of both dogs being male consistently comes out at around 1/3
No mathematical trickery involved.
A man sees a sign in a window advertising two recently-tossed coins mounted in a display case. He goes in and tells the shopkeeper he will only take the coins if there’s at least one showing heads.
The shopkeeper phones his wife who is washing the case and asks her if there’s at least one facing heads-up. She says yes.
What is the chance there are two heads-up coins?
That’s not another way of wording it, it’s a different question entirely. You’re arbitrarily singling out a specific dog and then asking questions of the other one. What if it is, in fact, the other dog which has been identified as male?
if it was the other dog that had been identified as male then we'd have the other dog that would be male or female.
I know SP doesn’t like it but the coin toss analogy works well here. Toss two coins. HH 1/4, HT 1/2, TT 1/4. Now discount any results where you don’t have at least one head (ie one or more of them is a head, so all you can discount is TT) the results will now be HT 2/3, HH 1/3.
that just messes my head up trying to think that through, so im sticking to what we've been told about these 2 dogs. ones male, whats the chances of them both being male then. 50/50
And you are now suggesting that the probability of throwing another head is 1/3?
Why are you throwing one of them again? They're already tossed, you're then simply observing the results.
Observing the coins doesn't change the probability, putting one arbitrarily aside and re-tossing the second demonstrably does.
LOOK AT THE SPREADSHEET.
It will all become clear.
Hopefully.
Is it any wonder that bookies always win (sigh)
It generates 1000 scenarios of two dogs, where each dog has a random 50:50 chance of being male or female.
You can see for yourself that empirically, in actual results, the probability of both dogs being male consistently comes out at around 1/3
nice spreadsheet, but we've got a bit more information now, we're told that one of the dogs is male, thats a certainty. 100% male. so whats the chances of the second dog being male and hence having 2 male dogs? 50/50.
i can see what youre doing/saying, but all your maths is from a point where we know nowt about them poor wet pooches, we now know that at least one is male. you dont need a spreadsheet for that.
LOOK AT THE SPREADSHEET.
That's got my geek gene a purring.
we’ve got a bit more information now,
Yes but that information only lets you narrow down which scenario you are in. It doesn't change the original odds.
The original odds are that if you do this 1000 times you'll see "both boys" about 250 times and "mixed or both girls" about 750 times. Correct?
The information that at least one is a boy means you can ditch the "both girls" scenarios and you now know you are either in one of the 250 scenarios where they are both boys or one of the 500 scenarios where they are mixed.
Do the 50%ers still think the chance of all 3 being male is 50%?
Well I dont. But as I said my problem was with the wording, which was imo different from what he wrote in the explanation of why the answer wasnt 0.5.
It might make a little more sense if you think about the odds improving, not getting worse.
Before you have any information there is a 1/4 chance that they are both boys.
Once you know that at least one is a boy then you know there is a 1/3 chance they are both boys.
Yes but that information only lets you narrow down which scenario you are in. It doesn’t change the original odds.
but why do you need to keep the original odds the same? the odds have now changed, as we now have more info.
The original odds are that if you do this 1000 times you’ll see “both boys” about 250 times and “mixed or both girls” about 750 times. Correct?
yes.
im sticking to what we’ve been told about these 2 dogs. ones male, whats the chances of them both being male then. 50/50
Riddle me this. I have a big bag of balls. 50% are red, 50% are blue. You close your eyes, pick out two at random and hold them in your closed fists. What are the odds of them both being red?
The ball in your left hand could be red or blue. 50% chance of either. The ball in your right hand could be red or blue. 50% chance of either. So our possibilities are,
You have a red ball in both hands.
You have a blue ball in both hands.
You have a blue ball in your left hand and a red ball in your right.
You have a red ball in your left hand and a blue ball in your right.
Each permutation equally possible, 25% chance of each. Yes?
I then look at them and tell you "at least one of them is red." What are the odds of them both being red now?
You now know that both cannot be blue. But you cannot rule out any of the other options. Of the three equally likely permutations, one states that they're both red. Observing them does not change the probability, they're balls (or dogs) not quantum particles.
but why do you need to keep the original odds the same? the odds have now changed, as we now have more info.
How does the new information change the 50:50 likelihood of any single dog being girl or boy? It doesn't.
It might make a little more sense if you think about the odds improving, not getting worse.
Before you have any information there is a 1/4 chance that they are both boys.
Once you know that at least one is a boy then you know there is a 1/3 chance they are both boys.
im not looking at it as improving or getting worse. and you may be right to the laws of mathematics, i dont know, im not good enough at maths to check that bad boy out. to me, the question doesnt mention 'using the laws of mathematics, what is the probablity......', it just gives us a scenario, to which the answer is 50/50 😀
No-one has said or even implied the first dog/coin has been randomly determined. It could have been selected precisely because of what it is. So you'd be better of saying, coin A has been PLACED heads up. Coin B is tossed. What is the probability of both coins being heads. Still think it's 1/3?
but why do you need to keep the original odds the same? the odds have now changed, as we now have more info.
The odds of there being two females or not being two females - the ratio between those two outcomes - has changed, yes.
However, the odds of their being either one male or two males - the ratio between those two outcomes - has not.
How does the new information change the 50:50 likelihood of any single dog being girl or boy? It doesn’t.
thank you. the likelihood of that dog being male is 50/50 as you correctly say. so as we know the first dog is male.....
Riddle me this. I have a big bag of balls. 50% are red, 50% are blue.
no no no no noooooo...... lets keep to the original question, i can understand that. why do we have to keep going to doors, coins, balls etc??
I'm making a list of who is in my gang.
@sadexpunk - the error you are making is that you assume the wife knows which one is male and which one she hasn't checked. That isn't the case with the wording of the question so MF & FM are different meaning that MM is one of three options.
So you’d be better of saying, coin A has been PLACED heads up. Coin B is tossed. What is the probability of both coins being heads. Still think it’s 1/3?
No, because again, that is a different question. Nowhere in the riddle does it say "you arbitrarily choose one dog and declare it to be male." You're changing the puzzle to fit your solution. Stop it.
@sadexpunk – the error you are making is that you assume the wife knows which one is male and which one she hasn’t checked.
we keep going back to the crux of the matter, which is where it will be won or lost.... it doesnt matter which she has checked, one or both, she may even know that there are two males, but she answers 'at least one is male'.
Nowhere in the riddle does it say “the sex of both dogs has been determined randomly” You’re changing the puzzle to fit your solution. Stop it.
so as we know the first dog is male…..
We. Do. Not. Know. This. You're adding information which isn't there. Where's this "first" dog come from?
lets keep to the original question, i can understand that. why do we have to keep going to doors, coins, balls etc??
Because you demonstrably don't understand the question, you just think you do. Go through the balls question, tell me where you think I've gone wrong.
So you’d be better of saying, coin A has been PLACED heads up. Coin B is tossed. What is the probability of both coins being heads. Still think it’s 1/3?
No, because again, that is a different question. Nowhere in the riddle does it say “you arbitrarily choose one dog and declare it to be male.” You’re changing the puzzle to fit your solution. Stop it.
but its the same as looking at both coins, seeing that one of them is a head, placing it down heads up, and then asking the probablity of them both being heads? 50/50.
Nowhere in the riddle does it say “the sex of both dogs has been determined randomly”
How else would you determine it? Is the shopkeeper a genetic scientist on the side? Good grief.
It's not a trick arcane "laws of mathematics" question. It is empirical reality as the spreadsheet shows.
Look you've agreed that if we do this 1000 times then at the start we have roughly 250 scenarios where both are boys and 750 scenarios where they are either both girls or mixed.
If you picked any one scenario at random from that pot then you'd have a 250 in 1000 chance (1/4) that it was a both boys scenario.
So once we have the new information that they are not both girls, we are left with a pot containing 250 scenarios where they are both boys and 500 scenarios where they are mixed gender.
Which scenario are we in? Well the chance that we are in a both boys scenario has improved to 250 in 750 (1/3) but it is still twice as likely that we are in a mixed gender scenario.
the probability of the second dog being male is 50% but that isn’t what is being asked. MF & FM are valid and separate probabilities so have to be taken in to account since we don’t know which of the two dogs was checked for gender.
It is whats being asked. One dog is male. the other dog is M or F the order is irrelevant.
We. Do. Not. Know. This. You’re adding information which isn’t there. Where’s this “first” dog come from?
ok, so it may not be the first dog, its a dog. but its male. so the odds of them both being male now we know this ones male is 50/50. which is where i can see your argument from. at first, before any info is known, youll be correct about the probability. but we're asked the probabliity after we know one is male.
but its the same as looking at both coins, seeing that one of them is a head, placing it down heads up, and then asking the probablity of them both being heads? 50/50.
Excellent, now we're getting somewhere. This is precisely where you're going wrong. Go get a couple of coins, try this, and report back.
If you toss two coins, see that one is a head and put it aside, the probability that both are heads is NOT 50:50. Either coin could be a head. You'll throw two heads, a head and a tail, or a tail and a head with equal chance.
Read the analysis of ambiguity in the wikipedia page about this problem.
The paradox occurs when it is not known how the statement "at least one is a boy" was generated. Either answer could be correct, based on what is assumed.
However, the "1/3" answer is obtained only by assuming P(ALOB|BG) = P(ALOB|GB) =1, which implies P(ALOG|BG) = P(ALOG|GB) = 0, that is, the other child's sex is never mentioned although it is present. As Marks and Smith say, "This extreme assumption is never included in the presentation of the two-child problem, however, and is surely not what people have in mind when they present it."
https://en.wikipedia.org/wiki/Boy_or_Girl_paradox
I'm sticking with 50/50. good luck chaps!
In your "fixed" coin tossing example you've made a mistake. (another one but it's different)
You've already tossed the first coin so the chances of it being heads was 50% you've then ignored that and stated that the probability of both coins being heads is purely down to the probability of the second coin being heads.
@ whitestone, yep Green is the reason the house always wins at roulette,
calling red or black is near enough a 50 50 chance, calling red twice in a row is 1/4 odds, calling red then black is 1/4 odds. Green just guarantees its not 50/50 at all
so the odds of them both being male now we know this ones male is 50/50
No it's not.
Pick up any pair of dogs and there are 4 possible scenarios:
1) both dogs are male
2) the one in your left hand is male and the one in your right is female
3) the one in your left hand is female and the one in your right is male
4) both dogs are female
If you are told that "at least one is male" then you are in scenario 1, 2,or 3.
All are equally likely so it is a 1/3 chance you are in scenario 1.
That is very different from being told that "the dog in your left hand is male".
Then you could only be in scenario 1 or 2 so then and only then it would be 50:50
the odds of them both being male now we know this ones male is 50/50.
Again, you're singling out a specific dog and ignoring the fact that the male dog could be the other one. We do not know that "this" dog is male, at no point has the shopkeeper's wife identified a specific animal.
Let's try it this way.
A man sees a sign in a window advertising a kitten and a puppy for sale. He goes in and tells the shopkeeper he will only take the animals if there’s at least one boy.
The shopkeeper phones his wife who is bathing the animals and asks her if there’s at least one boy. She says yes.
What is the chance there are two boys?
Your argument is "we know the puppy is male so the kitten must be male or female." But we don't know that, we only know that one of them is. We could have a male puppy and a female kitten, a male kitten and a female puppy, or both males. Three possibilities, each equally likely.
You have a litter of dogs, the sex of which is determined (by observation, gene splicers not required 🙄 ) You pick one which you know to be male, and one other. Good grief, it ain't rocket surgery 😂How else would you determine it? Is the shopkeeper a genetic scientist on the side? Good grief.
You pick one which you know to be male, and one other.
Nope.
That is a specific dog. You pick any pair of dogs randomly from the litter then, as above, you are in one of four scenarios..
1) both dogs are male
2) the one in your left hand is male and the one in your right is female
3) the one in your left hand is female and the one in your right is male
4) both dogs are female
..etc
Then you are told that "at least one is male" so now you are in one of three equally likely scenarios:
1) both dogs are male
2) the one in your left hand is male and the one in your right is female
3) the one in your left hand is female and the one in your right is male
If nothing else this thread shows why people who understand probability and can derive the correct model based on a situation, even if that model isn't the intuitive obvious one, can MAKE MONEY APPLYING IT!
there's a healthy return to be made of y'all think the return of a situation is 50%, but really it only comes up 33% of the time.
not that the international mathematicians' puppy trading market is huge & lucrative, but its possible these misunderstandings and failures of gut feelings happen elsewhere too...
For those saying it doesn't matter which one was checked:
1 A - M B - M
2 A - M B - F
3 A - F B - M
4 A - F B - F
There are two possible assumptions here that affect the outcome of the problem. The first, as per Monty Hall, assumes the wife has knowledge of the dogs genders before checking and so checking is unnecessary. If at least one dog is male then line 4 is automatically discounted. That leaves three possibilities as to the assignment of gender for each dog. 1/3 (just got that there writing it out).
The second is that she has no knowledge of the genders so has to check the dogs individually, the first check is 50/50 as to whether the dog will be male or female and determines the outcome of the overall problem. Assume the first dog (A) she checks is male, options 3 and 4 are removed leaving only options 1 and 2, the chances then of the second dog being male being 1/2. Of course dog A could have been female, options 1 and 2 are removed leaving only options 3 and 4. From the answer given, had this been the case, then the chances of the other dog being male is zero since we first found a female and we know at least one is a male. The problem now is that we have no idea whether or not she found a female first or not.
Given the problem presented I would say it's not possible to give an answer without knowing the facts first since it could be 0, 1/2 or 1/3 depending on which assumptions you make with the limited (in that we are never told if she knows the genders beforehand) information given.
Which is also a fantastic analogy for Brexit.
IMO
The problem now is that we have no idea whether or not she found a female first or not.
Yep, so 3 equally possible scenarios where only one scenario is both dogs are male.
It's definitely 1/3
Just to add, this is NOT the same as Monty Hall as the problem is utterly reliant on the host knowing what is behind each door for the justification to work. Since we are never told this then the stated problem has multiple answers. Much the same as asking what the square root of nine is without stating what number systems are being used if we are to arrive at a single answer.