Sunday Answers
You've suffered enough, here are the answers to Friday's puzzles.
The best way to work out why, is to change the problem. Imagine instead of 3 doors there are 1000 doors. Again the host knows which 1 door has the gold. When you've chosen a door, he opens 998 doors to show they're empty, and offer for you to swap the door you've chosen with the 1 door he didn't open. In this case it's almost certain that you didn't choose the right door first off, and therefore almost certain that the one door he didn't open holds the gold. The 3 door version is just a scaled down version of that.
Another way to think about it is the total odds involved. The odds always have to add up to 1. When you choose one door, you've got a 1/3rd chance of being right, and each of the other 2 doors have 1/3rd each, adding up to a total of 1. When the host opens one of the other doors to show it's empty, he removes the odds of it being right, making it zero. Your door hasn't changed, so it still only has a 1/3rd chance of holding the gold. That means to get a total of 1, the other door now has a 2/3rds chance of holding the gold.
Now the other one:
At first glances you think that's what being asked here. You're told that one child is a girl, the other child has to be fifty-fifty girl or boy, right?
Wrong, actually the other child is twice as likely to be a boy than a girl. Or in other words, 2/3rds of the time it will be a boy.
The reason is subtle. The question doesn't say, for example, "the oldest child is a girl", it says one of the children is a girl. What this question is really asking is: what are the odds that a family with two children has a mixture of one boy and one girl?
It helps to think of the children as X and Y. Child X has a (roughly) 50:50 chance of being a boy or a girl. So has child Y. Knowing this, we can make a table of all the possible combinations of children.
So there are four different options, each equally likely. We know that one of the children is a girl, so the first option is out. That leaves three other options, two of which have a boy/girl mix. Therefore in 2 occurrences out of 3, the other child will be a boy.
Brain hurt now?
You are on a game show and the host shows you three doors. The host tells you that behind one of the doors is a pot of gold, and the other two are empty. The host knows which door holds the gold. He gets you to choose one, then he chooses another one and opens it to show it's empty. He then offers you to keep the door you chose, or swap to the third, unopened door. Is it better to stay with your door or swap?The answer is it's better to swap, because the third door is twice as likely to hold the gold.
The best way to work out why, is to change the problem. Imagine instead of 3 doors there are 1000 doors. Again the host knows which 1 door has the gold. When you've chosen a door, he opens 998 doors to show they're empty, and offer for you to swap the door you've chosen with the 1 door he didn't open. In this case it's almost certain that you didn't choose the right door first off, and therefore almost certain that the one door he didn't open holds the gold. The 3 door version is just a scaled down version of that.
Another way to think about it is the total odds involved. The odds always have to add up to 1. When you choose one door, you've got a 1/3rd chance of being right, and each of the other 2 doors have 1/3rd each, adding up to a total of 1. When the host opens one of the other doors to show it's empty, he removes the odds of it being right, making it zero. Your door hasn't changed, so it still only has a 1/3rd chance of holding the gold. That means to get a total of 1, the other door now has a 2/3rds chance of holding the gold.
Now the other one:
A married couple have two children. You know that one of the children is a girl, what are the rough odds that the other one is a boy?It took me an age to get my head around this one. When we do basic probability at school we're taught that probabilities have no memory. Tossing a coin and it coming up head doesn't make it any more or less likely for it to come up heads the next time you toss it.
At first glances you think that's what being asked here. You're told that one child is a girl, the other child has to be fifty-fifty girl or boy, right?
Wrong, actually the other child is twice as likely to be a boy than a girl. Or in other words, 2/3rds of the time it will be a boy.
The reason is subtle. The question doesn't say, for example, "the oldest child is a girl", it says one of the children is a girl. What this question is really asking is: what are the odds that a family with two children has a mixture of one boy and one girl?
It helps to think of the children as X and Y. Child X has a (roughly) 50:50 chance of being a boy or a girl. So has child Y. Knowing this, we can make a table of all the possible combinations of children.
| Child X | Child Y |
| boy | boy |
| boy | girl |
| girl | boy |
| girl | girl |
So there are four different options, each equally likely. We know that one of the children is a girl, so the first option is out. That leaves three other options, two of which have a boy/girl mix. Therefore in 2 occurrences out of 3, the other child will be a boy.
Brain hurt now?
See the boy/girl one I still dont get. It is not exactly 50:50 boy/girl, but its near enough. It's either xx or xy. Thats 50:50
If you toss a coin and get 19 heads, it is still 50:50 that the next one will be a head, surely?
Unless its a double headed coin?
So it doesn't matter at all what their other child is. This has no memory either, its exactly the same as the coin. Unless you are bringing in some area of biology I don't know about.
Which is why I am a biologist, and not a mathematician, I guess...
Try it yourself, toss a pair of coins few hundred times (or mock it up in a spreadsheet package), then delete all the instances when there are two heads. The mixed tails-heads combos will be twice as common as the tails-tails combination.
Stuff like this expands the mind in a good way, it's worth trying to get familiar with it. I look forward to some biology puzzles on your blog. ;-)
'the other child is twice as likely to be a boy than a girl'
Remove option one and you have options totalling 4 girls and 2 boys, which I see as twice as many chances of a girl than a boy, not vice versa.
You can't just count the numbers. There are three options left. In two of the three options there is a boy in the set. In one there isn't. Therefore 2 out three.
Sorry.. I don't see a difference in whether its their oldest or not. Maybe its the wine.
If I toss a coin and its a Head. What is the chance of the next one being a head? It's still 50:50 as far as I can see. And will be everytime I toss the coin.
I know this is the Alan Davies response.. ;)
If I toss a coin 100 times, and its been heads all the way. The coin doesn't know that. So if I ask you what will it be next - its just as likely to be a head or a tail.. I've just been very very "lucky" in getting 99 heads so far, but its not impossible.
Hmm.. I'll sleep on it ;)
Perfectly correct, but the word you used is next. If you had said one of the coin tosses is tails the answer maybe different. Like Becky said get 2 coins and toss them, record each coin for each toss. if you toss them enough you will see that combination of coins that show at least one head is 3 times more likely than no heads, and the same for tails. Obviously then this can't be 50/50.
Biology is much harder than maths anyway!
Cathii
Yes, it is. The NEXT coin will be 50:50 heads or tails. You're a hundred percent right that one coin toss has absolutely no influence on the NEXT coin toss or any other coin toss. But that's not the question. The equivalent question in terms of coins is "I tossed two coins. One landed on heads, what are the odds the other one landed on tails?"
As I was writing this, Cathii made the same point better... well done Cath. ;-)
It's fascinating in that we get the "one dice roll doesn't influence the next" thing drummed into us when learning about probabilities, so much so that it misleads us when it comes to questions like this.
One great way to prove this would be to have a competition. :-)
We both bring £100 in pound coins. We get an independent person to toss two coins at a time in secret, then to tell us the status of either coin at random. (E.g. "One coin is showing a head"). We then both guess what the other coin is showing. If we're right, we get to keep both coins, if we're wrong, the other person gets them. If we're both right, one coin each.
I confidently predict if I was playing based on my theory, and you were playing on the "totally random" theory, I'd walk away with more money than you. In fact, I'd give you an extra £100 if I lost. ;-)
More seriously, getting back to my comment about juries in my original post, this is what I find interesting:
You're a very educated person and intelligent person, and you're having trouble understanding this fairly simple piece of probability (as I did at first).
Imagine then the average juror expected to make decisions about someones guilt based on things like, for example, "the likelihood that two babies in the same family would die of cot death". Probability issues like this are not necessarily identical to the puzzle above, but they still require more than a layman's understanding of how probability works. I wonder how many times a the application of "common sense" probability has led to miscarriages of justice.
I'm off to make babies in Excel....
I need to see it in black and white with the model to get a handle on it. And once I did it in Excel I see what you mean.
(53 BB, 102 GB, 45 GG)
Now, about these lottery numbers......
I needed to see it in black and white too, just to be sure. :-)
Here's the Excel spreadsheet I made.
I'm not convinced. The problem with your lovely spreadsheet is the assumption that each of the three possibilites - BG, GB and GG - are equally likely, given we know there are two children and at least one is a girl. In fact, the GG option is twice as likely as each of the other options. (Because there are two girls, instead of one.) To get the correct answer from your spreadsheet, you need to find the string "girl", then check whether the other sibling is "boy" or "girl". (Because "girl" appears twice in the GG families, you need to count each "girl".) You will find it is 50/50 after all.
Another way of looking at it is to consider the elder, younger situation. We don't know whether the girl is older or younger, but we know she is one or the other. This means one of the BG/GB possibilities can be ruled out. We don't know which one, but we know one of them can be.
Love,
Genette
And if you don't believe Dr. Math, perhaps you'd like to take me up on my pound coin challenge? ;-)
I'm sure I won't convince you, but maybe you could tell me which part of this is wrong:
The girl we know about could be
a) the elder child in a girl-boy family
b) the younger child in a boy-girl family
c) the elder child in a girl-girl family
d) the younger child in a girl-girl family
Each of these is equally likely.
The fact that c and d look the same, doesn't mean we can count them as being the same. To put it in terminology like Dr Math's (and to show why he is wrong!) it's not a case of counting the paths, but counting how often the path is travelled!
As for the money challenge, you amuse me far to much to upset you by taking money off you!
Love,
Genette
> the GG option is twice as likely as each of the other options
This is wrong. The GG option is exactly the same likeliness as the other two. All three options have a 1/3 probability.
However, because BG and GB both result in you having one boy and one girl, the probability of those combined is 2/3. Hence you're twice as likely to have a boy and a girl, than two girls.
Doctor Maths is my dad. True story
YES, I'M TALKING OF TRANSGENDER.
lol :P
(and what about sexual position which are suppossed to improve the odds of having a boy or a girl? Or what about the genetic history of the parents? or what if they artificially alter the baby? etc.)
Genette: I admit you make a convincing argument, and my lack of maths knowledge makes it hard to show exactly where you're wrong, but it's something like this.
You're assuming that a girl-girl mix is twice as likely as a boy-girl mix. This is wrong. Boy-girl mixes turn up twice as often as girl-girl mixes. There's a simple way to prove this. Just toss a pair of coins and count the ratio of heads-heads that appear compared to mixed heads-tails. Mixed heads-tails will turn up twice as much, in the long game, as heads-heads.
Oh, and when you've done that, and you're STILL not convinced, you might want to re-write the Wikipedia Article which also must be wrong. ;-)
Nobody's wrong. Comments closed. Moving on! :-)
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