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  • #61
  • Mar 26, 2007 8:07am Mar 26, 2007 8:07am
Quoting Samuri
The point is the two girls were born already.

Under the condition that they tell you one of the two children is a girl, the odds of they having two girls is then 1/3.
I guess I still don't understand the question.

If the two girls are born already, then the chances of having two girls is 100%.

You cannot calculate odds after the fact, only on future events.
  • #62
  • Mar 26, 2007 8:09am Mar 26, 2007 8:09am
Quoting Samuri
Mr. and Mrs. Smith have two children, and they tell you at least one of the children is a little girl. If the birth probability of a boy or a girl is the same @ 50%, what is the probability of Smiths having the other child is also a girl?
Problem:

OK, they have two children - done deal.

One is a girl.

Birth probability is 50/50.

What is probability of 2 girls?

Answer: Either 0% or 100%. Can't calculate probabilities on past events. The other child either is a girl or is not, it's done.
  • #63
  • Mar 26, 2007 8:10am Mar 26, 2007 8:10am
Quoting Samuri
The point is the two girls were born already.

Under the condition that they tell you one of the two children is a girl, the odds of they having two girls is then 1/3.
I agree with Darkstar's info. That is why I used the honest and fair coin reference. If I remember statistics class correctly from the 1970s it does not matter what the prior outcomes were there is always 50% chance on the coin. A baby would be skewed by birth stats. I did go to college 25 years ago so my mind is fading.
  • #64
  • Mar 26, 2007 8:13am Mar 26, 2007 8:13am
Quoting nvwine
I agree with Darkstar's info. That is why I used the honest and fair coin reference. If I remember statistics class correctly from the 1970s it does not matter what the prior outcomes were there is always 50% chance on the coin. A baby would be skewed by birth stats. I did go to college 25 years ago so my mind is fading.
You're correct, there are more boys born than girls, but the 50/50 was a given assumption in the problem.
  • #65
  • Mar 26, 2007 8:15am Mar 26, 2007 8:15am
OK, let's have a good one.

There are 40 lottery balls, numbered 1 - 40.

You pick 5 numbers. The balls come out randomly.

What are your odds of getting all 5 correct?
  • #66
  • Mar 26, 2007 8:15am Mar 26, 2007 8:15am
You were chosen to participate in a game show, the prize being a new car.
The car is hidden behind one of three doors, the other two doors having a monkey each behind.

You choose your door, then the host openes one door from the left two, one monkey jumps out. NOW, the host asks you if you stick to your door or change?

This simple (from first look) question confused a lot of people, and raised a lot of dispution many years ago.

What is your answer?
凹凸 only
  • #67
  • Mar 26, 2007 8:19am Mar 26, 2007 8:19am
Quoting Darkstar
EDIT:// Ohh and this might help http://mathforum.org/library/drmath/view/67124.html
I'm still trying to figure out the last answer for sure. Darkstar had to throw up a site that is right at my intellectual limit for quick understanding. I'm looking for my calculator.
  • #68
  • Mar 26, 2007 8:22am Mar 26, 2007 8:22am
Quoting blueruby
If the two girls are born already, then the chances of having two girls is 100%.

You cannot calculate odds after the fact, only on future events.
100% correct statement.

"What is your odds in guessing Smiths have two girls?"
This should be the question. I was wrong.
凹凸 only
  • #69
  • Mar 26, 2007 8:29am Mar 26, 2007 8:29am
Quoting Samuri
You were chosen to participate in a game show, the prize being a new car.
The car is hidden behind one of three doors, the other two doors having a monkey each behind.

You choose your door, then the host openes one door from the left two, one monkey jumps out. NOW, the host asks you if you stick to your door or change?

This simple (from first look) question confused a lot of people, and raised a lot of dispution many years ago.

What is your answer?
What is the question?

When you chose your door, your odds were 1 in 3.

If you pick again, your odds are now 1 in 2.

If, now you chose the same door (or the other door), your odds would be 1 in 2 that you chose correctly.

I think I remember this, and the answer was to change doors, but it makes no sense to me now. I think the belief was that 1 in 2 is better than 1 in 3, so you have to change otherwise you're keeping the 1 in 3 chance, but this is not correct.

Keeping your door or switching, your odds are 1 in 2.
  • #70
  • Mar 26, 2007 8:35am Mar 26, 2007 8:35am
Quoting blueruby
Problem:

OK, they have two children - done deal.

One is a girl.

Birth probability is 50/50.

What is probability of 2 girls?

Answer: Either 0% or 100%. Can't calculate probabilities on past events. The other child either is a girl or is not, it's done.
I'll go for a gender challenged child. A sex change candidate. So I say we can get a 50/50. Just kidding.
  • #71
  • Mar 26, 2007 8:45am Mar 26, 2007 8:45am
probability is confusing sometimes..is that why most traders lose?
  • #72
  • Mar 26, 2007 8:45am Mar 26, 2007 8:45am
Quoting nvwine
I'll go for a gender challenged child. A sex change candidate. So I say we can get a 50/50. Just kidding.
No kidding! These days, 100% is not equal to 100%! Even that can change! LOL!

Two girls today..... a girl and a boy tomorrow....
  • #73
  • Mar 26, 2007 8:54am Mar 26, 2007 8:54am
Quoting hilmy83
probability is confusing sometimes..is that why most traders lose?
I would say most traders lose the same way I do.

When I have my head on straight, I make 1 - 2% per day average, that's my goal (consistency, not money).

When I am impusive and jump in before the signals are there, break the rules, don't apply proper money management, let greed, or fear of missing out, or fear of leaving money on the table take over, I lose. Then if I try harder, I lose more. I've blown my account because I really liked one trade and loaded up (gambling), trying to be a hero.

So, you don't need to be an expert at math, statistics, or anything, but you do have to follow money management or you're toast. I have to make myself consider that I could be wrong, no matter how sure I am about a trade, it can go wrong, and I have to force myself to consider the outcome.

I think a person has to find their strengths and apply them with discipline, face their weaknesses and make rules to prevent allowing them to take over.
  • #74
  • Mar 26, 2007 8:54am Mar 26, 2007 8:54am
Quoting blueruby

Keeping your door or switching, your odds are 1 in 2.
Don't forget the host, he knows where the monkey is.

If you agree that your first choice has a 1/3 odd to be correct, you definitely should change, cause the host will help you to win.

Take three cards, you easily find it out what you should do.
凹凸 only
  • #75
  • Mar 26, 2007 8:56am Mar 26, 2007 8:56am
Quoting blueruby
OK, let's have a good one.

There are 40 lottery balls, numbered 1 - 40.

You pick 5 numbers. The balls come out randomly.

What are your odds of getting all 5 correct?

Anyone?
  • #76
  • Mar 26, 2007 8:57am Mar 26, 2007 8:57am
Quoting blueruby

I think a person has to find their strengths and apply them with discipline, face their weaknesses and make rules to prevent allowing them to rule.
This is more important than math's skill. Lessons from tears and sweat.
凹凸 only
  • #77
  • Mar 26, 2007 9:04am Mar 26, 2007 9:04am
Quoting blueruby
Anyone?
1/40 * 1/39 * 1/38 * 1/37 * 1/36 = 1 / 78960960

correct?
凹凸 only
  • #78
  • Mar 26, 2007 9:14am Mar 26, 2007 9:14am
Quoting blueruby
Anyone?
1/40*1/39*1/38*1/37*1/36*5*4*3*2*1 or 40!/(35!*5!)
(with n!=n*(n-1)*(n-2)*..2*1)

Assume the 5 numbers in strict order->1/40*1/39*1/38*1/37*1/36
However the number are not in strict order, but can be rearranged in 5*4*3*2*1 ways.
  • #79
  • Edited 9:50am Mar 26, 2007 9:15am | Edited 9:50am
Quoting Samuri
1/40 * 1/39 * 1/38 * 1/37 * 1/36 = 1 / 78960960

correct?
My mind is on holiday so I can't figure it out but I think you are wrong. In the California Super Lotto the odds of picking 5 correctly out of 47 numbers is 1,592,937 to 1. My logic is based on this web site.
  • #80
  • Mar 26, 2007 9:39am Mar 26, 2007 9:39am
Quoting Samuri
Don't forget the host, he knows where the monkey is.

If you agree that your first choice has a 1/3 odd to be correct, you definitely should change, cause the host will help you to win.

Take three cards, you easily find it out what you should do.
So how does this apply to trading?
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