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berty2001 | 16:06 Wed 11th Aug 2004 | Quizzes & Puzzles
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After seeing 'What are the odds?' question, is it true that actually the chances of rolling ten sixes in a row,say, on a dice is juts one in six because each role has its own probabilty as is not affected in anyway by what has gone before, or what is going to happen after it?
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The probability of 1 dice being a 6 is 1 in 6. The probability that the 10 dices are all 6 is (6 multiplied by itself 10 times) which is approx 60,466,176. I am not sure how correct this is !!!
Chances of throwing 10 consecutive sixes You make the first throw There is 1 chance in 6 that it will be a six There is 1 chance in 6 that the next throw will also produce a six, so there is 1 chance in 36 that you will throw two sixes. Again, there is 1 chance in 6 that the next throw will produce a six, so there is 1 chance in 216 that you will throw three sixes. And so on rolling 5 sixes 1 chance in (6*6*6*6*6) = 77761296 rolling 10 sixes 1 chance in (6*6*6*6*6*6*6*6*6*6) = 60466176 Chances of throwing 10 the same - without defining beforehand what that number is to be The first throw defines the number, and so does not need to be taken into the calculation 1 chance in 10077696
Before you start rolling, the chances of getting 10 successive sixes is a large number against, as Pinus says. But having rolled 9 sixes, the probability of the 10th roll producing a six is simply one in six - as the die itself is unaware of what has gone before! (Incidentally, "die" is not a typo; it's the proper word - "dice" is the plural.)
We�re into conditional probability now. If we know that there is already a run of any length, the probability of adding one more to that run is always 1 in 6 The probability of adding two more is always 1 in 36, and so on ========= Do you remember the old question about the children in a household? We are told that a household has two children, but not what sex they are. What is the probability of there being a boy and a girl? The possible ways of having two children are Boy first, younger girl Boy first, younger boy Girl first, younger boy Girl first, younger girl Two of the four possibilities give one boy and one girl, so the probability is � that the children are boy and girl BUT What if we knock at the door and it is answered by a girl? There are then only three possibilities for the children Boy first, younger girl Girl first, younger girl Girl first, younger boy So the probability is 2/3 that the children are boy and girl. The change in the probability has come about because we have more information about the situation ==== Maddening how mathematics works, isn�t it!
To clarify one (perhaps rather subtle) point. If you roll a die and get 9 6's in a row, the probability that you get a 6 the next time is NOT 1/6. It IS 1/6 if you are told that the die is fair, or unbiased. But, unless you are so told, it is extremely unlikely that the die is fair, so the probability is somewhat less than 1/6. Evaluating the correct probability requires multiple integrals, and is perhaps a bit recondite for a family forum.
pinus - nice story about the kids, but disagree with the logic of the first bit. The ages are not a relevant factor. To prove the point, they could be twins; 2 boys, 2 girls, or one of each. Does the probability then become 3/7? I think the logic is more along the lines of the probability that the second child is not the same sex as the first is 1/2. Happy to be proved wrong if indeed I am. :o)
Yes but don't worry too much I've tried explaining this to my mum several times and she refuses to believe that the chances of consecutive numbered balls coming out on the lottery are the same as non consecutive balls...although I have to say the probability that she is right and it is a fix is probably quite high! To quote Terry Pratchett "1 in 100 chances crop up 9 times out of 10" or something like that!
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For those who haven';t read the curious incident of the dog in the night there is an exmaple similar to the boys and girl story. Basically if you are given a chocie of three box and in one has money. You choose a box, but don;'tget to see what is inside. Oneof the other boxes is opened and it does not coantin money. you have the opportunity to change you mind or stick with yourbox. What do you do?
The example mentioned by berty2001 is a variation of a famous one in probability called the Monty Hall problem. You're on a game show and you're shown three doors. Behind one of them is a car and behind the others there's nothing. You have to choose one of the doors. When you make your choice the game show host doesn't tell you whether it's the one with the car or not. Instead he opens one of the other two doors and shows you that there's nothing behind it. He then asks you whether you want to stick with your original choice or whether you want to choose the remaining closed door instead. The question then, is what should you do? And the answer is change your original choice- you'll have a greater probability of winning the car.
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Correct, but can't rember the equation to work out why it is better probablity. will have to go home and read it again.
If you'd rolled 9 sixes in a row i'd do a normal hypothesis test of the binomial approximation (np, npq) at the 99% significance level to prove your die was rigged. Although the probably of a Type I error (assuming the die is rigged when it isn't) would be so horrifically high i might as well just not bother.
BenDToy The problem is stated with ages to show the difference between boy+girl and girl+boy Once you know the sex of one child, you know that the option of two children of the other sex is ruled out. If there are two children in the household, and you know that one of them is a girl, the two children cannot both be boys Similarly, if you know that one of them is a boy, there cannot be two girls Even with twins, one is older than the other, so the problem as stated is still valid. As a mere bachelor I don't know if having non-identical twins boy/girl is different from having girl/boy :} The crucial point is that once you have more information you're looking at a different question. But of course, if you're working the maths, you assume "other things being equal"
pinus - I think we should agree to differ here. It is very cumbersome to explain the logic in writing. Suffice to say that I agree with the 1/2 bit, but not the logic applied to get it.
I can't understand BenDToy's logic, which seems to have been influenced by some sort of poisonous treacle

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