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Beat the overround.
If I flip 2 coins until at least one of them is heads, the probability that the other is heads is 1/3. I'd give you just over evens all day long. Get out some coins and try it. Flip two coins at the same time, if it comes up TT then discard it, then record how many heads and tails you get. There's no difference between the coin flipping and the children, and there's no difference between 'probability' and 'combinations' (sic). Those combinations are part of sets, and probability theory and set theory are linked, you can't pretend combinations and probability are different things. Each of those combinations are equally likely, and there are 3 equally likely combinations. Let's do this in events. We flip two coins. We have four events. A: TT B: HT C: TH D: HH P(A)=P(B)=P(C)=P(D) = 1/4 But now whenever D occurs we ignore the result and re-flip the coins. But because we only ever discard D, it must be true that P(A)=P(B)=P(C) still, the relative probabilities of A,B, and C cannot change because we simply reflip the coins when D occurs. So P(A)=P(B)=P(C) = 1/3. Which means if I tell you that I've flipped two coins, and you ask if I've got a heads, and I say yes, then you can say my other is more likely to be a tails. This is mathematics, and it's counter-intutive, but it's wonderful. |
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I am still offering odds of 1.25 (as per my original answer) but it does not look like I am going to get matched - too many sharp minds on here imo.
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It's not helping that some fecker keeps offering £1.99 @ 1.26
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Dunlaying: The handshakes problem isn't a probability question, because it's deterministic not random.
People aren't randomly shaking hands until all people have shook hands, they are simply finding people they've not shook yet and shaking them. The solution as was said earlier is very easily provable to be 21. (7x6/2), but I won't prove it here. An interesting problem is the happy meal problem. Suppose there are 6 different happy meal toys, and each happy meal has a random toy in it, with the probability of any given toy the same*. On average** how many happy meals do you need to buy before you have the set? Hint: It's not six! To do it in 6 you'd need to get a different toy each time, which is unlikely (1/720) and it's impossible to do it in less than six, so cannot possibly be the average/expected value. *Important to state this, as people mistake "random" to mean "equal probability" which is nonsense.) ** What is the expectation |
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Two more interesting ones:
1) How many randomly selected people do you need in a room before it becomes odds-on (less than evens) that two of them share the same birthday? 2) If one in every six children born into the world is Chinese, how many children would you need to have in order to guarantee that one of them was Chinese? |
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If I flip 2 coins until at least one of them is heads, the probability that the other is heads is 1/3. I'd give you just over evens all day long.
No, see this is wrong based on the puzzle. Parents can have a son and a daughter, two sons or two daughters, there is no other combination they can have. A daughter and a son is not relevant as they are the same thing, UNLESS we were asked about permutations or possiblities. We weren't, we were asked about probabilities. We know one is a son, the other has to be a son or a daughter, there is no other possibility. It's an even money shot, no matter how the quirky mathematicians want to try and trick you or make it something it isn't. How do two coins come into it? We already know the outcome of one of the coins! We are not flipping the chances of a son, it's already a known result in the first instance from the information provided. The only thing we need to know is the chances of the second coin being heads or tails. There are only two choices, unless it lands on it's edge. One of the coins is a known result, there is no further information required for that instance. Different story if we didn't know that they already have a son, they are independant events, they are not a set. Which means if I tell you that I've flipped two coins, and you ask if I've got a heads, and I say yes, then you can say my other is more likely to be a tails. You can't be serious  |
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Google the "Monty Hall Problem" or there are other variants (Bertrands Box, Three Prisoners etc).
All based on the same logic. |
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BTO:
I have a degree in statistics, and I'm deadly serious that you're wrong. "A son and a daughter" and "A daughter and a son" are the same thing, there is no order to the children. You've not asked "is your eldest a boy", you've asked "do you have a boy". You are right that one (of them) is a son. This leaves three not two possibilities. Youngest Son, Eldest Daughter (They were talking about their youngest when asked if they had a son) Youngest Daughter, Eldest Son (They were talking about their eldest when asked if they had a son) Both sons (They were talking about both when asked if they had a son). I think you're getting confused because you think "Is one a son" is an ordering. It's not, you're not asking of "Child number one" is a son, you're asking "is one of them a son", i.e. Do you have a son at all. This simply excludes the Girl,Girl combination leaving 3 equally likely combinations. It's also true for the coins. Note, it's not true if I flip one, then answer your question then flip the other. Crucially I flip them both at the same time, so there is no ordering to the coins (except in our analysis, but even that's not needed, so you could simply state P(a heads and a tails) = 0.5 if you don't care about order. |
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It's not the logic that's at fault, it's the interpretation of the question. We need look no deeper than the actual wording of it.
We were not asked for the order they came in, we were not asked the chances of the older sibling being a daughter. We were only asked the chances of the "other" one. Youngest Son, Eldest Daughter (They were talking about their youngest when asked if they had a son) Youngest Daughter, Eldest Son (They were talking about their eldest when asked if they had a son) Both sons (They were talking about both when asked if they had a son). Based on the question, you've read too much into it, we were not asked about age or older or younger we were asked the chances the other one. Based on your interpretation, you've left out twins. Boy twins. Boy Girl twins. So whatever way you interpret it isn't correct. As someone who has a degree in statistics, you've been trained to look at all possibilities, even the possibilities that aren't listed in the question. I compile odds in the real world and medical statistics for the government. If I back two favourites in the first two races, and one of them wins, I still have an even money chance of the other winning, REGARDLESS of the order the bets were made, or the order the results are known. The possible combinations are: Win Win Lose Lose Win Lose Lose win is irrelevant to the overall odds, because it makes no difference to the end result. Lose Win and Win Lose are the same, there are no odds that will change that end result. If you ask me the chances of the first bet being a winner, that's a different story, but you asked me the chances of the next bet being a winner when the result is known for one of them. |
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But we weren't asked about order, we were told one is a son.
The other has to be a son or a daughter. Please tell me why we need to include age when age was not part of it and if you want to go down that route, what if they're twins? |
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No, you're getting it wrong, I've not asked you the chance of the NEXT bet being a winner, I've asked the chance of the OTHER bet being a winner.
That allows for the possibility that your winner was the second bet! And you're right, the questions asked don't include ordering, that's the point! There is no ordering involved, it's precisely because there is no order that allows for EITHER to be the boy. Let's go through this again, there are four equally likely outcomes of people having two children. GG BB GB BG You ask me if I have a boy. I say I do. This means that I don't have GG. Therefore I can have either: BB GB BG I don't understand why you think GB isn't possible, especially since it's identical to BG, given that ordering isn't important! I could equally say there are 3 possible outcomes with different probabilities: A: Both girls (prob=0.25) B: One of each (Prob=0.5) C: Both Boys (prob=0.25) Now, we have a random variable X that takes one of those 3 outcomes wiht those probabilities. What we want to know is that given that at least one is a boy, what is the probability that both are boys. P(C|X=¬A) (Probability that X is C, given that X is not A) For the other to be a boy, you need both to be boys, so require outcome C. We are given information that X is not A. For this we use conditional probability. Generally written this is P(A|B) = P(A and B)/P(B) In this specific case, we can therefore say that P(C|A) = P(C and not A)/P(not A) So, what is P(C and not A), well C and A are mutally exclusive, therefore this is just P(C) = 0.25 What is P(not A)? Well P(A) = 0.25 therefore P(not A) = 0.75 Therefore P(C and not A)/P(not A) = 0.25/0.75 = 1/3 QED Mathematical proof. There are lots of ways of looking at this, unfortunately you're getting confused between this and the "past flips don't affect future flips" which is true, but this isn't what's going on here. This is more subtle and is about how information can change probabilities in counter-intuitive ways. |
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ror,I try not to read them all
.Your question is just a "coupon collector" job is it not? ...................................................... Here is a question that anyone can solve if they think about it,it requires no mathematical knowledge. There are n people arranged randomly in a circle and two of them are called Able and Zeb and all persons have different names . Probability required, that there are exactly k people between Able and Zeb. |
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Yes my question is about coupon collecting
Ooh that's a good problem dun, I've not come across it before. Clearly it can't be bigger than n/2-1, the question is whether the symmetry holds or not, and I suspect there is a different formula for whether n is even or odd. (After-all, if there are 3 people then expectation is is zero, four people it's 1/3 (I Think? that's a half-guess based on fixing Abel and moving Zeb), Five it's 1/2?) Ooh this is lovely, I'll go away and get Prob in terms of k for you I hope! I might try to get the expected (minimum) distance too. |
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No,it is a surprising answer ror.The answer given in the book was even more surprising,it was wrong.
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O.K. ROR,
One last attempt at my explanation. Your calculations are not wrong. But what you've calculated based on the information provided and the question asked encompasses some other determination you've made from information not given. You've listed every possible permutation: BG GB BB GG These of course are the possiblities. We are not asked for all possible combinations, we are asked the chance or odds of having two sons, knowing already one of them is a son. The chance of having two sons, knowing that already there is one son, is still 50%. We are not asked about order, we are asked about the chance of two sons. So breaking it down: It could be BB, or BG regardless of the order. Because we are talking chance or odds BG = GB because B is the known. So let me compile it: We have BB BG GB B is the constant, there are three possible combinations, but the chance or odds for two of them are precisely the same because B is the constant. BG GB is the same thing because we weren't asked to assess the chance of order. |
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I agree with BTO. But it could be a trick question and the second child is a hermaphrodite. Sorry im just a very suspicious individual.
Im going to throw a very simple puzzle out there, apologies as im sure most have come across it; A bat and a ball cost £1.10 The bat cost £1 more than the ball. How much does the ball cost? |
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BTO, ror is right.
So breaking it down: It could be BB, or BG regardless of the order. Because we are talking chance or odds BG = GB because B is the known. So let me compile it: We have BB BG GB B is the constant, there are three possible combinations, but the chance or odds for two of them are precisely the same because B is the constant. BG GB is the same thing because we weren't asked to assess the chance of order. It's fine if you say BG and GB is the same thing, but, if you are told someone has 2 kids and you are only interested in how many are girls and how many are boys, the probabilities are: 2 boys = .25 2 girls = .25 1 boy 1 girl = .5 If you don't believe it, just do a simulation. you can solve it like this: 1) Do a simulation of 2 kids, so it looks something like this: 1. BB 2. GB 3. GG 4. GB 5. BG etc. 2) Remove all instances of GG, as you have been told there is at least 1 boy. Of the remaining sample, 1 boy 1 girl will occur about 2/3 of the time, and 2 boys 1/3 of the time. |
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1) Do a simulation of 2 kids, so it looks something like this: 1. BB 2. GB 3. GG 4. GB 5. BG etc.
2) Remove all instances of GG, as you have been told there is at least 1 boy. Of the remaining sample, 1 boy 1 girl will occur about 2/3 of the time, and 2 boys 1/3 of the time. OK we have left BB, GB, BG left. But BG and GB is the same. I have two children. One is a girl, one is a boy. I say I have a girl and a boy. I don't get an extra family allowance because I can also say I have a boy and a girl! It only matters if you want to consider possible combinations not odds. I have a daughter (really) I have neither a greater nor lesser chance of having a son or a daughter in future because of my daughter, or the order in which she was born. The odds of me having a son are still 50%. Possible combinations because you don't know the age, have nothing to do with the probability or odds of me having a son. And the original question was "what are the chances of the other child being a son", not "what are the chances of this combination....." |
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Its evens a boy and the ball cost 5p
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oh hang on its 2/1 a boy but the ball is deffo 5p
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methane.... is the answer to Q1 Abt 30 ?
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I ave a question for BTO.
If I lay 160 shorteners to a fixed backers stake,all between evens and 10/1 and make 56 points profit,whats the chances I've finally got it cracked ? |
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dunlaying:
I've worked out the circle problem, but it definitely depends on even or odd n. For odd n it's easy, it's just k=2/(n-1) for all k in [0,(n/2-1)]. For even n it's harder, because of the unique spot opposite, so it's 2/(n-1) for k |
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It is 2/(n-1).
I am not a mathematician but it works like this. Denominator: If the n were in a straight line there would be n! ways to line them up.But in a circle we have n possible starting points ,hence,(n-1)! Numerator: If we remove A and Z from the n people we have n-2. We can start with A but it makes no difference which one of A or Z we commence with. So we start ,then line up our (n-2) people "behind" our starter.We do this in (n-2)! ways and then we have 2 places where the other of A /Z can be situated to satify the conditions of the problem. Which gives us 2(n-2)!/(n-1)! which,in turn ,gives us 2/(n-1). The author gives the answer as 2/(n-1)! which I know is a misprint.I hope it is anyway as he is a world renowned expert. If I have time tomorrow I will tap in a real goody for you. |
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That agrees with me, but assumes n is odd.
e.g. How can that be the case with n=6. If it's 2/(n-1) then that is 2/5. So k=0, 2/5. k=1 2/5 But this is 4/5. Which leaves an option that must be 1/5. (Which is k=2). Probability of all events must add up to 1. This only happens where n is odd. Ew, for some reason half my previous response was cut off, I think I used angle brackets or something. What I meant to say that for n even it was 2/(n-1) for k less than (n/2-1), but 1/(n-1) for k=n/2-1. |
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I have a daughter (really)
I have neither a greater nor lesser chance of having a son or a daughter in future because of my daughter, or the order in which she was born. The odds of me having a son are still 50%. BTO, That's obviously correct, no-one is arguing with that. There are two different situations: 1) Someone has tossed a fair coin twice. You are told the first toss was heads, what is the chance the second toss was also heads? (answer 50%) 2) someone has tossed a fair coin twice. You are told one of the tosses was heads. What is the chance the other toss was also heads? (answer 33.33%) These are two different questions, with different answers. If you don't believe it, do a simulation. A final problem: You are told someone has 10 children, and at least 5 of them are boys. Do you think the probability of the other 5 children of unknown gender all being girls is equal to the chance of them all being boys? |
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Ror,
Yes I do believe you are right as in that case backwards=forwards.Very good!  |
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Here's a problem from a programming challenge site, but it's one with a neat mathematical formula solution so it doesn't require programming knowledge.
Rather than steal it I'll link it: www.spoj.pl/problems/EBOXES/ N large empty boxes (assume they are of type:1) are initially placed on a table. An unknown number of boxes (type:1) are selected and in each of them K smaller boxes (type:2) are placed. Again an unknown number of type:2 boxes are selected and K boxes of type:3 are placed inside. This process is repeated T times. Now a box is assumed to be empty when it has no smaller boxes inside it. Finally after all the processes are complete let there be F empty boxes in total. Given N,K,T,F, how many boxes are there in total on the table? e.g. N=11 K=8 T=2 F=102 Then the answer is 115 boxes. What is the general solution for with respect to N,K,T,F? |
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I think i understand the boy girl problem now, and want to retract my agreement with BTO, and switch sides
. Sorry BTO. |
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colonelll there are only 3 scenarios, as TT is no longer possible.
I can ask the same question in a different way: I tossed a fair coin twice, the outcome was not TT. What is the chance of HH Given that information? If you think it is unchanged, and the probability is still 0.25, then what are the probabilities of the only other two possibilities TH and HT? Remember they have to add up to 1. |
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oh! investor, if that is what u are asking then it is true it will be 33%, didnt see it that way. however i dont think anybody with the slightest knowledge of probality will find that proabablity dificult to calculate, why are u complicating it by suggesting simulation
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That is a completely different question.
You are also asking for the order of outcome. 1) Someone has tossed a fair coin twice. You are told the first toss was heads, what is the chance the second toss was also heads? (answer 50%) Correct, we agree. 2) someone has tossed a fair coin twice. You are told one of the tosses was heads. What is the chance the other toss was also heads? (answer 33.33%) Only if you are asking for the order, or age, or anything else the question didn't ask for. I know that what you are saying is the accepted mathematical answer. I am saying that if I were to calculate odds like that, I'd go broke. The mathematicians, like to twist and hide angles to demonstrate that "hey you missed this". But it was never there in the information provided, nor asked for in the question. The question was not, what are the odds the FIRST child is a son, the question was what chance the other one. Son is the constant. I know you can say I'm looking at it wrong etc. I'm saying the answer is flawed because it's not the product from the infomation provided. Bit like saying How much did it cost to get to the moon. You add up everything NASA expended. "OH you forgot about Neil Armstrong's taxi fare to NASA" We never had that information, it was never asked in the question. |
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I ave a question for BTO.
If I lay 160 shorteners to a fixed backers stake,all between evens and 10/1 and make 56 points profit,whats the chances I've finally got it cracked ? I'd say you might be onto something but need more results to say you've got it cracked. Could be a trend of losing shorteners which will even out. |
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Tame TT - What was the ROI on the series of bets ?( Just curious as to how "good" such a return is when someone "cracks it".)
For comparison purposes :- On laying 1st/2nd and 3rd favourites, where odds are "value" based on odds and RANK position in ratings, in order to gain 1 point.....the following stats are obtained using BF SP and my system ratings :- ROI 4.28100966 ( This is after using commission rate at 5%) NUMBER BETS 1174 TOTAL BF SP ODDS 5016.34 AVERAGE ODDS PER BET 4.27286201 NUMBER LOSING BETS 199 NUMBER WINNING BETS 975 =============================================================== The ROI on LAYING in PLACE market to a fixed liability of 1 point, using BF SP, and after commission at rate of 5%, is 14.52 %. The STATS for that are :_ ROI 14.52171435 NUMBER BETS 1174 TOTAL BF SP ODDS -1154.83 AVERAGE ODDS PER BET -0.98367121 NUMBER LOSING BETS -540 NUMBER WINNING BETS 634 ================================================================ Always useful to compare results....and would appreciate a note of your ROI on series of bets mentioned above. Thanks in anticipation  |
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Answer to the chance of the neighbours having 2 sons is 1/3, as most agreed upon. I couldn't explain it more simply than ror did early on. I agree it is just pedantic. Doesn't help to pick a winner.
Dunlaying, the handshakes problem was not posted by me. I believe the answer was 6+5+4+3+2+1=21. I couldn't understand any of your maths, but I did see in the bottom/right of your post 125/3. I have reason to suspect that answer is incorrect. Methane Magnet: the number of persons required before you would expect coinciding birthdays is 25 or 26. I have never calculated the answer, but I remember seeing it before. I was surprised that it was so few. The answer to your question as to how many children must be born before a Chinese child arrives is 1 according to my friend, Wing Chan. |
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I'm not being pedantic here. I have referred the following simple problem to the Maths Departments of Uni of Sydney and NSW Uni. I got opposite responses:
For the accountants, Mathematicians and self-appointed sages, I require a ruling; $1/50c? In words, what is one dollar divided by fifty cents? Obviously most people answer 2. |
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The answer is of course 2.
Now let all the prophets come out of the woodwork with nonsense about it depends which pocket it's in or something else that doesn't make sense in the real world. |
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Answer to the chance of the neighbours having 2 sons is 1/3, as most agreed upon.
Consensus rules I guess because intelligence is more important than wisdom to some. I'll leave this thought with you, so you can see just how ridiculous the accepted answer is - When told there is one son by the neighbour, knowing they have two children most people will respond with "so you have two sons or a girl as well?" There can only be two possible reponses. There aren't other responses unless you want to include the dog or the neighbour's boss. Professor's sometimes outsmart themselves, so they actually become dumber than the average joe. It's called "intelligence regression theory". Give you an example, I highly intelligent person I know with a Bachelor's and Master's in Litrature, went to a movie and saw things that were not intended by the director. In fact she missed the entire point of the story. She was looking for things that weren't there, because she had been trained that way. She's highly intelligent, but wisdom has escaped her. |
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BTO
Nobody disputes there are only 2 responses. 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) 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. 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. |