General Betting
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'--- and then the odds of there being 2 sons would be 2.0 ( or evens as you are now claiming).--- "
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$1/50c? I claim is not a valid mathematical question. It is the division of a material object by another material object.
(Sorry I can't conform with a pound key. Haven't used one since 1966). If the answer is 2, what is $1 x 50c? Is $1/$1=1? Is $1x$1=$1 squared or what? You can't divide a milk bottle by a milk bottle, or multiply them. I say the question must be set as a mathematical problem such as; The boy went to the shop with $1 to buy lollies. Lollies were 50c each. How many lollies could the boy buy? The calculation is then made abstractly. The answer to the mathematical problem is 2. The answer to $1/50c is not 2. The only person ever to agree with me was a female from University of NSW. Is there anyone else who agrees with me? |
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The issue I think he is getting at is what does x/yz mean?
I would read it as x/(yz) but strictly speaking it is (x/y)z |
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There are clearly some extremely tormented minds on here.
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Nobody disputes there are only 2 responses.
But they do, they expect it to be 2 boys, a boy and a girl or a girl and a boy, which is ridiculous unless you were asked about ages or birth order. The whole option of a third response is not required. Surely the point in question is " What are the odds that the response will be 2 sons ?" And the answer is 3.0. ( 1 in 3) Answers can be 2 sons, or a son and a daughter. There is no other answer. If you're going to use they could respond this way: two sons, a son and a daughter, or a daughter and a son. We weren't asked about the number of combinations or responses, we were asked the chances of the other child being a son. The easiest way to look at all these things is to to assess what the information on hand is telling you. Info on hand 1) Neigbour has two children. Hence, in order of birth, there is either BG,GB,GG,BB. Why do people insist on including hypothetical information, when the answer does not require any further information. We only need to know BB or BG, we do not need to know the order of GB or BG, it's totally irrelevant to the odds. You are taking the possiblity that one has to come first or for some strange reason we need to find out the order. 2) Neigbour has at least one boy. Hence there cannot be GG. It is eliminated completely. Therefore there are 3 options remaining, one of which produces 2 sons. If you had been given one further piece of info, in that the son you've seen is the eldest child, then you would have been able to eliminate BG, and then the odds of there being 2 sons would be 2 to 1 ( as you are now claiming). But these odds can only be arrived if you have this third bit of info. Why do we need to assess the chances of which one comes first, it's irrelevant unless asked what are the chances they have and older son!!! Then your answer is correct. I've had two bets at even money today. One of the bets won. What are the odds the other bet won? Now you get my drift. |
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Well I suppose at least one major question has been answered by all this.
We now all now know how you beat the overround. |
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Well we certainly now know why I'm able to.
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You're wrong though, bto. Think of it this way. As a percentage of all the people who have two children, 25% have two boys, 25% have two girls. The remaining 50% have one of each, that's why the bg/gb is important in this problem. By knowing a couple have one son eliminates only 25% of the cases leaving 75%. The likelihood of it being two sons is one third of the remaining 75%, ie one in three.
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It's amazing that this old chestnut is still doing the rounds on the forum - and that people are still getting it wrong.
Next it'll be the Monty Hall problem again. |
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Or the Ferrari and the goats, then we really will be in trouble
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Note to self: google anything you haven't heard of before posting
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 the mathematically challenged. Very large and important group in the ecosystem. |
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BTO
Perhaps you should have said " Well we certainly now know why I think I'm able to " |
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BTO, you're trying to portray this problem as a theoretical one that doesn't exist in the real world, this isn't true.
If you went to a Parent/Teacher meeting for parents of a single year of a boys' school and asked all the sets of parents to stand forward who had 2 kids, then you will find that around two-thirds of the sets of parents will have a boy and a girl (regardless of order) and around one-third will have two boys. It is very real. |
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Incidentally BTO, if I asked you whether Federer was going to win or lose a certain tennis match , there would only be 2 responses, yes or no.
Would you thus make the odds then 2.00 that he's going to win, irregardless of who he is playing. If so, please be my bookie for a few days. |
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Incidentally BTO, if I asked you whether Federer was going to win or lose a certain tennis match , there would only be 2 responses, yes or no.
Would you thus make the odds then 2.00 that he's going to win, irregardless of who he is playing. If so, please be my bookie for a few days. Now now, I'd like to firstly point out that being a reptile, frog's don't have hair  No because his chance of winning is independant of his previous result. It's likely that his odds will be shorter than they would be if he lost his previous game, but I would assess his odds based on a combination of career performance, recent performance and opposition. I.e. he will be shorter if he is "in form". But there are variables, a coin toss or black or red have no other variables (other than a big fat green 0). |
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How about my example BTO, do you believe that the real world example above would show different results than stated (i.e. 50/50) or that there is still a problem with the question?
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To be fair it is counter-intuitive which is why this and other similar problems generate such fierce debate. The trick is to work out the proportion of the original set of possibilities that has been eliminated by the piece of information provided.
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Absolutely cat.
I first heard this problem over 26 years ago and was struck by its counter-intuitive nature. I am hopefully trying to help BTO to see the problem for what it is with the real-world example above (which was also the way that the question was posed when I first heard it). |
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So you've essentially found the flaw in your own argument BTO in answering my ridiculous question on Federer.
The fact that there are only 2 responses, yes or no, has nothing to do with the odds associated with each response. Maybe it's time for you to start remembering the 1st rule of holes perhaps ? No shame in admitting you're wrong and moving on is there ?. After all nobody is saying that these sorts of " trick" questions have anything to do with gambling ability. I'm certainly not. |
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Bto, if you found everyone who had 2 bets at evens today, then took only those with at least 1 winning bet, then only one third of that group would have won both their bets.
It's real, but you refuse to believe it. Just do a naive simulation on Excel if you're still not convinced! |
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The above post assumes all peoples bets were at true odds and independent, I.e. they weren't betting on the same events.
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As already said a couple of times, BTO needs to look at this in terms of 100 random families moving in next door with 2 kids.
25 families have two girls 25 families have two boys 50 families have one of each "Do you have a son?" 75 families step forward. Of the 75 that step forward, "do you have another son?" is asked. 25 families say yes. Therefore 1 in 3 of those saying they have a son in the first question, end up having 2 sons. |
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Contrarian
Date Joined: 19 May 03 Add contact | Send message When: 29 Nov 11 14:56 Joined: Date Joined: 19 May 03 | Topic/replies: 844 | Blogger: Contrarian's blog It's amazing that this old chestnut is still doing the rounds on the forum - and that people are still getting it wrong. Next it'll be the Monty Hall problem again. There are two options left so it's 50/50 swithching doesn't matter.  |
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What's incredible about the Monty Hall problem is that so many clever people were getting it wrong, when this is also supremely simple to solve by simulation.
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To be fair investor, in those days simulation wasn't available to many, even academics.
These days you can bring up Google spreadsheet from the phone in yer pocket, no excuses! |
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Investor2, I reckon you actually want to win the goat ;)
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jabmast
29 Nov 11 15:17 Joined: 16 Jan 09 | Topic/replies: 204 | Blogger: jabmast's blog BTO, you're trying to portray this problem as a theoretical one that doesn't exist in the real world, this isn't true. If you went to a Parent/Teacher meeting for parents of a single year of a boys' school and asked all the sets of parents to stand forward who had 2 kids, then you will find that around two-thirds of the sets of parents will have a boy and a girl (regardless of order) and around one-third will have two boys. It is very real. Except this sample would be skewed by parents of two boys being significantly more likely to try for a third child! |
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^^ Nice appropriate user name there :)
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ror
If you have not seen this one it may amuse you for a few minutes,it amused me for a lot longer  .LOTTERY PROBLEM A straight forward selection of 6 from 49. Only one draw. P required is that the 6 drawn numbers contain 3 consecutive numbers and only 3 consecutive. EG 1 5 18 19 20 49 or 1 2 3 25 36 42 |
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Wonderful problem! I've been working on it all night, it lead me down a few false paths.
Is it 5/98? If not, don't tell me the answer, just that I'm wrong, I'd like to work it out! |
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I really wish I could be bothered with things like this
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ror,
I am loath to say that you are wrong but it is not the answer that has been agreed upon. I have had the trusty Casio whirring for 20 minutes and ,at long last,I have an answer that I find familiar. 0.0388308885 When I said it kept me amused for a long time I hope that did not give the impression that I had solved it.I did not,on top of which I have had a deal of help in understanding the answer. The answer is a masterpiece of lateral thinking. |
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BTO: I've got a biology one. When did a frogs become reptiles?
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I got 0.044649 but I'm allowing 1 2 3 7 8 9 which I presume I shouldn't.
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Bayes,
No there must only be the 3 consecutive. It is a sod of a question is it not? |
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ror,
I realise I have given you the final figure but the method is the real answer. |
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Technically frogs are reptiles, but amphibians.
knee deep, knee deep. |
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*not*
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I realised on my walk to work this morning that I was over estimating, because I was allowing 1 2 3 4 9 27 to count as a combination twice.
my initial thinking was that there are 47 sets of 3s, each of which then has 46C3 combinations for the other 3 numbers, so went for 47*46C3/49C6 Clearly as I said this counts 1234xx as a combination twice, so I'm going to have to figure out another method. Thanks for the problem it is great. |