insane goth… I don’t understand #1. Please PM and explain yes yes?^.^
I figured is like “You flip coin 9 times. All time, is land on heads. What is chances it land on tails on 10th time?” in which case answer is still 50/50.

Please to explain. Man have many many x and many many y. Why would be only 3 possible? Please do explain.

i do the small font again incase someone else doesn’t want the answer…

[size=42]With #2 we KNOW it is the 2nd child and either a girl or a boy, thus 50/50 chance.

However with the first we just know there are 2 children one being male, we do not know which child.

So 2 children here are all the combinations

Boy - Boy Boy-Girl Girl-Boy Girl-Girl

Four combinations 3 of which have girls 3 of which have boys - this would mean we have a 2/4 chance of being a girl (or 50/50) we can of course rule out G-G as one has to be a boy.

This leaves us with: Boy-Boy, Boy-Girl, Girl-boy

this means we have 3 combinations left and 2 of those are possiblities of having a girl this means the chance of the other child being female is 2/3

I know it doesn’t make much sense as you would think it is still 50/50 likely either b or g. But mathematically that would be how you work it out and thus makes the answer 2/3 [/size]

Congrats Insane_goth I couldn’t explain it better.

It’s definitely not 50/50. There are lots of statistical formules to solve these kind of questions. If you look at every coin flip independent of the other ones, then you may say you have 50/50 chance that it lands on its tail or head. But if you say “what’s the chance that it will land on its tail the 10th time AFTER it lands 9 times on its head”, then you get a whole new situation. If you want I’ll find the formulas for you but it’s surely not as simple as it looks at first sight

Talking about weird statistical facts: anyone know the birthday paradox? It goes like this: “how many people do you need in a room until there’s a 50/50 chance that two people have the same birthday?” [size=34]Answer: 23[/size]

I’m still puzzled and need a lie down! I can follow the reasoning of the BB, BG, GB logic…but I was always led to believe that in the law of probability every action should be looked at in isolation.

mystic. You wrong. NOt matter how many times you get heads peviously. If YOu flip coin in air, you still have 50/50 chance of landing on either. GO ASK MATH TEACHER IF YOU KNOW BELEIVE ME. English is least favorite class, Math is favorite ^.^

hehe. The question is a little mis-leading, the question has NOTHING to do with probability.

in a mathematical sence probability & chance cn be 2 different questions (i hate maths) probability (as moogle said) should be looked at isolated, thus the chance between 2 choices would be 50/50 all the time.

Chance however would require you to look at all variables and possibilities then base the answer on the combined mathematical outcome.

A mathematical equation of ALL possibilities and variations, not simple probability of that one child. I think that might be why some people find it a little difficult to grasp.

Sure I agree with the fact that you have always 50/50 chance of getting head/tail if you flip coins. But you mentioned a whole other scenario in which there are two related events: you said: suppose you get 9 heads AFTER EACH OTHER, then what’s the chance you get a tail the 10th time. If you want to solve this, you MUST also take into account the probability of first throwing 9 heads AFTER EACH OTHER. In this particular case, the 10th flip is NOT independent of the 9 previous flips (if you consider each case independent of each other you get indeed a 50/50 chance every time).
So let’s answer this question. What’s the chance that you’ll get 9 heads after each other? If we use the correct formula (binomial distributions) we get a chance of (1/2)^9 = 0.195%. To know the chance of getting a tail on the 10th flip, we must multiply this chance by 1/2, so we get 0.098%. Ask my professor of statistics if you don’t believe me

no, no must not. You is still flipping a coin. ANd You can show all formulas you want, is still 50/50 chance of getting either. We are not ask for chances of getting nine heads in row. That has been establish. ^.^

and insane goth, I still don’t get. WHY is that you eliminate one choice in first one!?

But we ARE talking about getting 9 heads after each other. You don’t need to change your original question now. I’ll quote it again:

You’re saying: FIRST you get 9 heads after each other. And after this, then what’s the chance you’ll get a tail the 10th time. In itself, this chance is indeed 50/50. But from the way you pose this question, it seems like you also need to consider the first 9 flips. And if you do this, the chance of getting a tail the 10th time after you get 9 heads is 0.098%. That’s irrefutable.

so so sorry. Allow me to say in perfect english grammar.

“You have flipped a quarter 9 times. Each time, the quarter has landed on heads. What are the chances mathmatically, that on your tenth flip, the same quarter will land on tails?”

Is better? NO?

that was all I was saying. ^.^You is overanalyse.

indeed, but that was not question. That is suppose to mislead person trying to solve math problem mystic. Is intentional. ^.^

well due to my tardiness lol i’ll post two. First person to get both correct will go next

they are easy i know but heck i can’t find my hard riddle book

Bouncing Bob was riding a particularly frisky horse when suddenly its bridle came off. As they raced down the road, a screaming Bob clung to the horse’s ears for dear life. Out of the corner of his eye, Bob saw a car coming, and realizing the horse was completely out of control, he panicked. Flailing his arms about, he accidentally caused the horse to come to an abrupt halt. What could Bouncing Bob have done to make the horse stop?

We are little creatures;
all of us have different features.
One of us in glass is set;
one of us you’ll find in jet.
Another you may see in tin,
and the fourth is boxed within .
If the fifth you should pursue,
it can never fly from you.
What are we?

They are easy when you know the answers! I’m glad you mislaid your hard riddle book. I think I may know the first one but I’m still thinking about the second.