if anyone one can settle an argument....
20 cards in a deck no 1-20 whats the chance of pulling a 7 if you have 12 attempts
12/20 = 0.6 = 60%
I think, provided the cards are replaced each time.
Are you replacing cards after each go? ie if you pull a 2 on go one do you put it back in the deck
yes you replace cards so there is a 5% chance of pulling a 7 everytime.
afs if your right and one had 20 goes instead of 12 it would be 20/20 = 100%. that would not be right
Tis a good question, easy at first sight. I believe it envolves a long series of combinations & permutations ie on go one the ch ances you dont pull out a 7 are 19/20 * go 2 you dont pull out a 7 19/20 * go 3 lets say you do pull out a 7 1/20 ie 361/8000. Then add up all these very combinations all 12. I cant be arsed meself. Come to think of it there is probably an easier way, been out of the game a long time now!.
1 2 3 4 up to 12
1/20 +19/20*1/20+ 19/20*19/20*1/20 + 19/20*19/20?*19/20*1/20+.........
Quote from: AFS on March 26, 2009, 08:05:37 PM
12/20 = 0.6 = 60%
I think, provided the cards are replaced each time.
That's too simplistic
You have a 1/20 chance the first pick the first time, 1/19 the second pick and so on..
I think if you add them all up 1/20 + 1/19 + 1/18 etc you will end up with about 88/100 therefor a 88% chance...
Thats if my thinking is right..
Would be interested to know if thats correct or not.
Pat I reckon the method you have is correct if you are not replacing the drawn card.
19/20 *19/20 twelve times works out at 54% approx.
Thats 54% of a chance that you wont pull a 7 in 12 goes.
46% that you will.
20 picks would give you 35% it wont happen thus still only a 65% chance it will.
Would that be right??
1 - (19/20)^12 = 0.4596
ie the 46% that Rossie mentioned
Quote from: Death Valley on March 26, 2009, 08:32:54 PM
Tis a good question, easy at first sight. I believe it envolves a long series of combinations & permutations ie on go one the ch ances you dont pull out a 7 are 19/20 * go 2 you dont pull out a 7 19/20 * go 3 lets say you do pull out a 7 1/20 ie 361/8000. Then add up all these very combinations all 12. I cant be arsed meself. Come to think of it there is probably an easier way, been out of the game a long time now!.
1 2 3 4 up to 12
1/20 +19/20*1/20+ 19/20*19/20*1/20 + 19/20*19/20?*19/20*1/20+.........
This is correct. Works out at 46%.
Quote from: magickingdom on March 26, 2009, 07:59:57 PM
if anyone one can settle an argument....
20 cards in a deck no 1-20 whats the chance of pulling a 7 if you have 12 attempts
You won't pull a 7 in 12 trys - normally comes out ~ 15th iteration.
8, 4, and 6 are shoe ins.
Quote from: Rossie11 on March 26, 2009, 09:02:33 PM
19/20 *19/20 twelve times works out at 54% approx.
Thats 54% of a chance that you wont pull a 7 in 12 goes.
46% that you will.
20 picks would give you 35% it wont happen thus still only a 65% chance it will.
Would that be right??
46% is my answer too...
Quote from: magickingdom on March 26, 2009, 07:59:57 PM
if anyone one can settle an argument....
20 cards in a deck no 1-20 whats the chance of pulling a 7 if you have 12 attempts
You've a 1 in 20 chance every time or 5%
If you add the card back and you do it 12 times you've a 60% chance,
You do it 20 time you've a 100% chance.
Quote from: Gnevin on March 26, 2009, 10:48:50 PM
Quote from: magickingdom on March 26, 2009, 07:59:57 PM
if anyone one can settle an argument....
20 cards in a deck no 1-20 whats the chance of pulling a 7 if you have 12 attempts
You've a 1 in 20 chance every time or 5%
If you add the card back and you do it 12 times you've a 60% chance,
You do it 20 time you've a 100% chance.
correct, probability is 1, that does not mean it will happen as each draw is a new draw and the chances are still 1 in 20 for each draw.
1-(1 - 1/20)^12 which I believe is 46% chance of it occurring if you only draw 12 cards.
As for 20,
1-(1 - 1/20)^20 = 64% chance it occurring.
Quote from: thebigfella on March 27, 2009, 12:03:49 AM
1-(1 - 1/20)^12 which I believe is 46% chance of it occurring if you only draw 12 cards.
As for 20,
1-(1 - 1/20)^20 = 64% chance it occurring.
That makes no sense if you draw 20 times the odds are 100% of getting one.
Quote from: Gnevin on March 27, 2009, 12:12:44 AM
Quote from: thebigfella on March 27, 2009, 12:03:49 AM
1-(1 - 1/20)^12 which I believe is 46% chance of it occurring if you only draw 12 cards.
As for 20,
1-(1 - 1/20)^20 = 64% chance it occurring.
That makes no sense if you draw 20 times the odds are 100% of getting one.
Your not removing the cards after picking one and each event is independent, so it would make sense..... I could be wrong though :D
Quote from: Gnevin on March 27, 2009, 12:12:44 AM
Quote from: thebigfella on March 27, 2009, 12:03:49 AM
1-(1 - 1/20)^12 which I believe is 46% chance of it occurring if you only draw 12 cards.
As for 20,
1-(1 - 1/20)^20 = 64% chance it occurring.
That makes no sense if you draw 20 times the odds are 100% of getting one.
Clearly if you replace the card each time each draw is a new event. Therefore 20 goes would not be a 100% chance.
Think of rolling a dice.
It is not a 100% certainty that if you roll a di(c)e six times that you will hit the numbers 1,2,3,4,5 and 6 only once.
Ahh there's a reason why I dropped out of Maths after 1st year
Quote from: muppet on March 27, 2009, 12:55:23 AM
Quote from: Gnevin on March 27, 2009, 12:12:44 AM
Quote from: thebigfella on March 27, 2009, 12:03:49 AM
1-(1 - 1/20)^12 which I believe is 46% chance of it occurring if you only draw 12 cards.
As for 20,
1-(1 - 1/20)^20 = 64% chance it occurring.
That makes no sense if you draw 20 times the odds are 100% of getting one.
Clearly if you replace the card each time each draw is a new event. Therefore 20 goes would not be a 100% chance.
Think of rolling a dice.
It is not a 100% certainty that if you roll a di(c)e six times that you will hit the numbers 1,2,3,4,5 and 6 only once.
Their is a difference between change and certainty if their is a 1 in 20 chance of something happening if you do it 20 times chances says it should happen but doesn't always have to.
http://www.random.org/playing-cards/?cards=1&decks=1&spades=on&hearts=on&aces=on&twos=on&threes=on&fours=on&fives=on&sixes=on&sevens=on&eights=on&nines=on&tens=on&remaining=on
20 card , start drawing for the 7 of hearts.
Quote from: Gnevin on March 27, 2009, 01:07:02 AM
Quote from: muppet on March 27, 2009, 12:55:23 AM
Quote from: Gnevin on March 27, 2009, 12:12:44 AM
Quote from: thebigfella on March 27, 2009, 12:03:49 AM
1-(1 - 1/20)^12 which I believe is 46% chance of it occurring if you only draw 12 cards.
As for 20,
1-(1 - 1/20)^20 = 64% chance it occurring.
That makes no sense if you draw 20 times the odds are 100% of getting one.
Clearly if you replace the card each time each draw is a new event. Therefore 20 goes would not be a 100% chance.
Think of rolling a dice.
It is not a 100% certainty that if you roll a di(c)e six times that you will hit the numbers 1,2,3,4,5 and 6 only once.
Their is a difference between change and certainty if their is a 1 in 20 chance of something happening if you do it 20 times chances says it should happen but doesn't always have to.
http://www.random.org/playing-cards/?cards=1&decks=1&spades=on&hearts=on&aces=on&twos=on&threes=on&fours=on&fives=on&sixes=on&sevens=on&eights=on&nines=on&tens=on&remaining=on
20 card , start drawing for the 7 of hearts.
On that basis then 100% is not right as you can't be 100% certain a 7 will be drawn.
Quote from: thebigfella on March 27, 2009, 01:27:50 AM
Quote from: Gnevin on March 27, 2009, 01:07:02 AM
Quote from: muppet on March 27, 2009, 12:55:23 AM
Quote from: Gnevin on March 27, 2009, 12:12:44 AM
Quote from: thebigfella on March 27, 2009, 12:03:49 AM
1-(1 - 1/20)^12 which I believe is 46% chance of it occurring if you only draw 12 cards.
As for 20,
1-(1 - 1/20)^20 = 64% chance it occurring.
That makes no sense if you draw 20 times the odds are 100% of getting one.
Clearly if you replace the card each time each draw is a new event. Therefore 20 goes would not be a 100% chance.
Think of rolling a dice.
It is not a 100% certainty that if you roll a di(c)e six times that you will hit the numbers 1,2,3,4,5 and 6 only once.
Their is a difference between change and certainty if their is a 1 in 20 chance of something happening if you do it 20 times chances says it should happen but doesn't always have to.
http://www.random.org/playing-cards/?cards=1&decks=1&spades=on&hearts=on&aces=on&twos=on&threes=on&fours=on&fives=on&sixes=on&sevens=on&eights=on&nines=on&tens=on&remaining=on
20 card , start drawing for the 7 of hearts.
On that basis then 100% is not right as you can't be 100% certain a 7 will be drawn.
Their is a difference between chance and certainty . This is about the chance of something happening in a series of independent events. The chance of a 7 on any draw particular will always remain 1 in 20.
Quote from: Gnevin on March 27, 2009, 12:12:44 AM
Quote from: thebigfella on March 27, 2009, 12:03:49 AM
1-(1 - 1/20)^12 which I believe is 46% chance of it occurring if you only draw 12 cards.
As for 20,
1-(1 - 1/20)^20 = 64% chance it occurring.
That makes no sense if you draw 20 times the odds are 100% of getting one.
The answer for 20 is 64%.
If you did this exercisee 100 times, you would pick a 7 within 20 goes in close enough to 64 times - I bet.
100% chance (statistically), means it will definitely occur (it doesnt mean it should or might, it means it will). Which clearly is wrong.
Quote from: Gnevin on March 27, 2009, 01:31:23 AM
Their is a difference between chance and certainty . This is about the chance of something happening in a series of independent events. The chance of a 7 on any draw particular will always remain 1 in 20.
There is but i dont think you understand the difference:)
100% chance means that it will always happen, no matter how often you run the test every time you draw 20 cards you will get a 7.
If you draw a card & replace it then it is the same probability each time
You arent guaranteed to get a 7, no matter how may times you do it
If I have remembered correctly, calculating the probabilty a least one occurance of k independent events is calucaulated by
1 - (1 - α)k
where each of the events has a probablity α of occuring. Thus you can't get 100%.
I first thought that it had to be 20/1 as each event was independent but now I'm not so sure. THe probablity that you will draw a seven is 20/1 for each event, so the probability that you would draw a seven on the first go is exactly the same for the 12th go. A bit like if you toss a coin 25 times and get 24 tails in a row the probalbiltiy that you will get a tails on the 25th go? 2/1. But here there area a fixed number of events. So if we said there are two draws what are the chances of drawing a 7 in either draw? The ans is not 20/1. If you toss a coin 20 times what are the chances of tossing a tails at some point during the 20 events? It's not 2/1...the chances are much greater than that...and that is the question that's being asked here...I think ???
going to throw a spanner in the works here, are these 20 cards drawn randomly from a standard pack of 52 cards?
if so, you need to factor in the probability that there may not be a 7 in the deck of 20 drawn, or am i just being silly?
Quote from: stpauls on March 27, 2009, 09:49:36 AM
going to throw a spanner in the works here, are these 20 cards drawn randomly from a standard pack of 52 cards?
if so, you need to factor in the probability that there may not be a 7 in the deck of 20 drawn, or am i just being silly?
That's the leaving cert question. We are still at Junior Cert so our cards are numbered 1 - 20.
We should have started with a coin.
Quote from: muppet on March 27, 2009, 10:17:33 AM
Quote from: stpauls on March 27, 2009, 09:49:36 AM
going to throw a spanner in the works here, are these 20 cards drawn randomly from a standard pack of 52 cards?
if so, you need to factor in the probability that there may not be a 7 in the deck of 20 drawn, or am i just being silly?
That's the leaving cert question. We are still at Junior Cert so our cards are numbered 1 - 20.
We should have started with a coin.
probably would have been an easier question alright!! ;D
Quote from: stpauls on March 27, 2009, 09:49:36 AM
going to throw a spanner in the works here, are these 20 cards drawn randomly from a standard pack of 52 cards?
if so, you need to factor in the probability that there may not be a 7 in the deck of 20 drawn, or am i just being silly?
Think of it this way, you have a bag of 20 marbles, each number 1 to 20. Each marble is then replaced after each draw and thus each draw is independent of each other.
In your example, you have 2 mutually exclusive events, "if 20 cards are drawn from a deck and none are a 7, then you cannot draw a 7 from the 20". Might have to go get the old GCSE text books out ;)
Quote from: thebigfella on March 27, 2009, 10:28:34 AM
Quote from: stpauls on March 27, 2009, 09:49:36 AM
going to throw a spanner in the works here, are these 20 cards drawn randomly from a standard pack of 52 cards?
if so, you need to factor in the probability that there may not be a 7 in the deck of 20 drawn, or am i just being silly?
Think of it this way, you have a bag of 20 marbles, each number 1 to 20. Each marble is then replaced after each draw and thus each draw is independent of each other.
In your example, you have 2 mutually exclusive events, "if 20 cards are drawn from a desk and none are a 7, then you cannot draw a 7 from the 20". Might have to go get the old GCSE text books out ;)
this is more like an a level question so will have to see if those books are still lying about somewhere!! :D
Quote from: stpauls on March 27, 2009, 10:32:28 AM
Quote from: thebigfella on March 27, 2009, 10:28:34 AM
Quote from: stpauls on March 27, 2009, 09:49:36 AM
going to throw a spanner in the works here, are these 20 cards drawn randomly from a standard pack of 52 cards?
if so, you need to factor in the probability that there may not be a 7 in the deck of 20 drawn, or am i just being silly?
Think of it this way, you have a bag of 20 marbles, each number 1 to 20. Each marble is then replaced after each draw and thus each draw is independent of each other.
In your example, you have 2 mutually exclusive events, "if 20 cards are drawn from a desk and none are a 7, then you cannot draw a 7 from the 20". Might have to go get the old GCSE text books out ;)
this is more like an a level question so will have to see if those books are still lying about somewhere!! :D
I done Advanced Maths at GCSE ;)
After a bit of searching through maths books admiittedly, I'm getting the same result as bigfella, ie 46%
1st pick, 95% chance of not picking the card (think we're all agreed there!)
2nd pick - on its own is again a 95% but multiply this by the first 95% to calculate chance over the 2 cards = 90.25%
3rd pick, multiply the 90.25% by a further 95% and so on
So....
Pick No Chance it Won't Happen Chance It Will
1 pick 0.95 5%
2 0.9025 10%
3 0.857375 14%
4 0.81450625 19%
5 0.773780938 23%
6 0.735091891 26%
7 0.698337296 30%
8 0.663420431 34%
9 0.63024941 37%
10 0.598736939 40%
11 0.568800092 43%
12 0.540360088 46%
Quote from: Treasurer on March 27, 2009, 10:59:50 AM
After a bit of searching through maths books admiittedly, I'm getting the same result as bigfella, ie 46%
1st pick, 95% chance of not picking the card (think we're all agreed there!)
2nd pick - on its own is again a 95% but multiply this by the first 95% to calculate chance over the 2 cards = 90.25%
3rd pick, multiply the 90.25% by a further 95% and so on
So....
Pick No Chance it Won't Happen Chance It Will
1 pick 0.95 5%
2 0.9025 10%
3 0.857375 14%
4 0.81450625 19%
5 0.773780938 23%
6 0.735091891 26%
7 0.698337296 30%
8 0.663420431 34%
9 0.63024941 37%
10 0.598736939 40%
11 0.568800092 43%
12 0.540360088 46%
So based on the same rule used to work out the probablity for 12 independent events, apply it to 20 and the answer is 64% (to the nearest % ;)).
This is slightly off topic but would help me settle an old discussion ;)
Say a couple have seven children, all girls. Base this on the premise that the chances of boy/girl are 50/50 each time. Am I right in saying that the overall chances of having 7 girls is 0.004%?
But each event, excluding any biological/physiological factors that may skew the probability, is still 50/50?
I would have it at 0.0078%
(1/2) to power of 7
Quote from: scud on March 27, 2009, 11:24:47 AM
This is slightly off topic but would help me settle an old discussion ;)
Say a couple have seven children, all girls. Base this on the premise that the chances of boy/girl are 50/50 each time. Am I right in saying that the overall chances of having 7 girls is 0.004%?
But each event, excluding any biological/physiological factors that may skew the probability, is still 50/50?
I know this isn't what you're asking, but afaik the probablility of having a girl is actually less than 50, it's something like 49% - don't ask me why though.
Quote from: Bogball XV on March 27, 2009, 11:44:07 AM
Quote from: scud on March 27, 2009, 11:24:47 AM
This is slightly off topic but would help me settle an old discussion ;)
Say a couple have seven children, all girls. Base this on the premise that the chances of boy/girl are 50/50 each time. Am I right in saying that the overall chances of having 7 girls is 0.004%?
But each event, excluding any biological/physiological factors that may skew the probability, is still 50/50?
I know this isn't what you're asking, but afaik the probablility of having a girl is actually less than 50, it's something like 49% - don't ask me why though.
Men (historically) more likely to die before reproduction?
Quote from: Rossie11 on March 27, 2009, 11:38:17 AM
I would have it at 0.0078%
(1/2) to power of 7
Agreed - 0.00078 is the correct answer.
Quote from: muppet on March 27, 2009, 01:11:36 PM
Quote from: Bogball XV on March 27, 2009, 11:44:07 AM
Quote from: scud on March 27, 2009, 11:24:47 AM
This is slightly off topic but would help me settle an old discussion ;)
Say a couple have seven children, all girls. Base this on the premise that the chances of boy/girl are 50/50 each time. Am I right in saying that the overall chances of having 7 girls is 0.004%?
But each event, excluding any biological/physiological factors that may skew the probability, is still 50/50?
I know this isn't what you're asking, but afaik the probablility of having a girl is actually less than 50, it's something like 49% - don't ask me why though.
Men (historically) more likely to die before reproduction?
Yeah, would imagine it's evolution working its wonders again.
Look at these figures from Russia:
Quote0-14 years: 14.8% (male 10,644,833/female 10,095,011)
15-64 years: 71.5% (male 48,004,040/female 52,142,313)
65 years and over: 13.7% (male 5,880,877/female 13,274,173) (2009 est.)
Quote from: Bogball XV on March 27, 2009, 02:19:14 PM
Quote from: muppet on March 27, 2009, 01:11:36 PM
Quote from: Bogball XV on March 27, 2009, 11:44:07 AM
Quote from: scud on March 27, 2009, 11:24:47 AM
This is slightly off topic but would help me settle an old discussion ;)
Say a couple have seven children, all girls. Base this on the premise that the chances of boy/girl are 50/50 each time. Am I right in saying that the overall chances of having 7 girls is 0.004%?
But each event, excluding any biological/physiological factors that may skew the probability, is still 50/50?
I know this isn't what you're asking, but afaik the probablility of having a girl is actually less than 50, it's something like 49% - don't ask me why though.
Men (historically) more likely to die before reproduction?
Yeah, would imagine it's evolution working its wonders again.
Look at these figures from Russia:
Quote0-14 years: 14.8% (male 10,644,833/female 10,095,011)
15-64 years: 71.5% (male 48,004,040/female 52,142,313)
65 years and over: 13.7% (male 5,880,877/female 13,274,173) (2009 est.)
Not sure about Russia but in countries like India and China the male to female live birth ratio (6:5) is much more skewed towards male children. More to do with aborting female children than anything evolutionary.
Loved these sort of questions at school. A bit rusty now though.
Reminds me of the one I once read...if there are thirty people in a room what are the chances that two of them share the same birthday (i.e. the same date but not the same year)? Can't remember the exact anwer but it's higher than you might at first think.
I always enjoyed things like this too. Derren Brown talks alot about this sort of thing in one of his books and its quite interesting. The one I used to like to use with people was the three door prize. Say you are on a game show and at the end you get to choose from three doors to win a prize. Behind one door is the prize. Behind the other doors are booby prizes. You pick a door, the game show host then opens one of the other doors to reveal a booby prize. The host then asks you would you like to change your door. Should you change? This appeared in 21 there last year
Quote from: David McKeown on March 27, 2009, 05:50:34 PM
I always enjoyed things like this too. Derren Brown talks alot about this sort of thing in one of his books and its quite interesting. The one I used to like to use with people was the three door prize. Say you are on a game show and at the end you get to choose from three doors to win a prize. Behind one door is the prize. Behind the other doors are booby prizes. You pick a door, the game show host then opens one of the other doors to reveal a booby prize. The host then asks you would you like to change your door. Should you change? This appeared in 21 there last year
I hate that one. One of my flatmates last year was doing a Maths PhD and I remember him trying to explain how this works to me. Still none the wiser :-\
Quote from: David McKeown on March 27, 2009, 05:50:34 PM
I always enjoyed things like this too. Derren Brown talks alot about this sort of thing in one of his books and its quite interesting. The one I used to like to use with people was the three door prize. Say you are on a game show and at the end you get to choose from three doors to win a prize. Behind one door is the prize. Behind the other doors are booby prizes. You pick a door, the game show host then opens one of the other doors to reveal a booby prize. The host then asks you would you like to change your door. Should you change? This appeared in 21 there last year
it doesnt matter whether you change your door its 50:50 at that stage
Quote from: magickingdom on March 27, 2009, 08:37:21 PM
Quote from: David McKeown on March 27, 2009, 05:50:34 PM
I always enjoyed things like this too. Derren Brown talks alot about this sort of thing in one of his books and its quite interesting. The one I used to like to use with people was the three door prize. Say you are on a game show and at the end you get to choose from three doors to win a prize. Behind one door is the prize. Behind the other doors are booby prizes. You pick a door, the game show host then opens one of the other doors to reveal a booby prize. The host then asks you would you like to change your door. Should you change? This appeared in 21 there last year
it doesnt matter whether you change your door its 50:50 at that stage
Not right, the odds have changed from 1/3 to 2/3 in your favour.
I cheated.
Quote from: magickingdom on March 27, 2009, 08:37:21 PM
Quote from: David McKeown on March 27, 2009, 05:50:34 PM
I always enjoyed things like this too. Derren Brown talks alot about this sort of thing in one of his books and its quite interesting. The one I used to like to use with people was the three door prize. Say you are on a game show and at the end you get to choose from three doors to win a prize. Behind one door is the prize. Behind the other doors are booby prizes. You pick a door, the game show host then opens one of the other doors to reveal a booby prize. The host then asks you would you like to change your door. Should you change? This appeared in 21 there last year
it doesnt matter whether you change your door its 50:50 at that stage
Apparently not, it took me ages to get my head around it but you have a 2/3 chance if you switch.
This helped:
http://en.wikipedia.org/wiki/Monty_Hall_problem
Quote from: Maiden1 on March 27, 2009, 04:19:30 PMNot sure about Russia but in countries like India and China the male to female live birth ratio (6:5) is much more skewed towards male children. More to do with aborting female children than anything evolutionary.
I don't know, but I'd be shocked if abortions were the reason for this - despite the recent economic advances in both countries in recent years I can't see that a remotely significant proportion of expectant parents would be in a position to know the sex prior to birth.
Brilliant puzzle that.
Best explanation I read was this one:
It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three (vos Savant 1990). In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. The game host then opens 999,998 of the other doors revealing 999,998 goats—imagine the host starting with the first door and going down a line of 1,000,000 doors, opening each one, skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat. A rational player should switch. Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one
Quote from: Bogball XV on March 27, 2009, 11:57:18 PM
Quote from: Maiden1 on March 27, 2009, 04:19:30 PMNot sure about Russia but in countries like India and China the male to female live birth ratio (6:5) is much more skewed towards male children. More to do with aborting female children than anything evolutionary.
I don't know, but I'd be shocked if abortions were the reason for this - despite the recent economic advances in both countries in recent years I can't see that a remotely significant proportion of expectant parents would be in a position to know the sex prior to birth.
Post-natal 'abortions' perhaps :-\
Quote from: AFS on March 28, 2009, 02:09:09 AM
Quote from: Bogball XV on March 27, 2009, 11:57:18 PM
Quote from: Maiden1 on March 27, 2009, 04:19:30 PMNot sure about Russia but in countries like India and China the male to female live birth ratio (6:5) is much more skewed towards male children. More to do with aborting female children than anything evolutionary.
I don't know, but I'd be shocked if abortions were the reason for this - despite the recent economic advances in both countries in recent years I can't see that a remotely significant proportion of expectant parents would be in a position to know the sex prior to birth.
Post-natal 'abortions' perhaps :-\
China had a one child policy for ages...Think they still might have...Lots of female children were aborted as couples wanted males to carry on the family name ...so to speak...
http://en.wikipedia.org/wiki/One-child_policy
Quote from: AFS on March 27, 2009, 08:50:03 PM
Quote from: magickingdom on March 27, 2009, 08:37:21 PM
Quote from: David McKeown on March 27, 2009, 05:50:34 PM
I always enjoyed things like this too. Derren Brown talks alot about this sort of thing in one of his books and its quite interesting. The one I used to like to use with people was the three door prize. Say you are on a game show and at the end you get to choose from three doors to win a prize. Behind one door is the prize. Behind the other doors are booby prizes. You pick a door, the game show host then opens one of the other doors to reveal a booby prize. The host then asks you would you like to change your door. Should you change? This appeared in 21 there last year
it doesnt matter whether you change your door its 50:50 at that stage
Apparently not, it took me ages to get my head around it but you have a 2/3 chance if you switch.
This helped:
http://en.wikipedia.org/wiki/Monty_Hall_problem
i'm with the people who cant get that. if you choose your door from 3 and AFTER you choose you door the host opens a dud one i cannot possibly see how your chances improve if you change doors. the fact is your left with 2 doors one a dud one not 50:50
Quote from: magickingdom on March 28, 2009, 02:05:22 PM
i'm with the people who cant get that. if you choose your door from 3 and AFTER you choose you door the host opens a dud one i cannot possibly see how your chances improve if you change doors. the fact is your left with 2 doors one a dud one not 50:50
Its a good one alright.
Maybe it could be explained like this.
You've three doors to pick from to win the big prize. You choose A.
Then before any doors are opened, you are given the choice to change to BOTH B and C. So by sticking with A, you'd have a 1/3 chance of winning but by choosing B and C, you have 2/3 chance of winning. So given the choice any logical person would choose B and C to give themselves the 67% chance, rather than stick with A and 33%.
Now you know that one of B and C contains the booby prize. At least one of them must have a booby prize.
So the host shows you that B has a booby prize. Does that mean your chances of winning have reduced from 67% to 50%?? Even though you knew before B was opened that one of your doors had the booby prize?
Quote from: Hound on March 28, 2009, 03:56:02 PM
Quote from: magickingdom on March 28, 2009, 02:05:22 PM
i'm with the people who cant get that. if you choose your door from 3 and AFTER you choose you door the host opens a dud one i cannot possibly see how your chances improve if you change doors. the fact is your left with 2 doors one a dud one not 50:50
Its a good one alright.
Maybe it could be explained like this.
You've three doors to pick from to win the big prize. You choose A.
Then before any doors are opened, you are given the choice to change to BOTH B and C. So by sticking with A, you'd have a 1/3 chance of winning but by choosing B and C, you have 2/3 chance of winning. So given the choice any logical person would choose B and C to give themselves the 67% chance, rather than stick with A and 33%.
Now you know that one of B and C contains the booby prize. At least one of them must have a booby prize.
So the host shows you that B has a booby prize. Does that mean your chances of winning have reduced from 67% to 50%?? Even though you knew before B was opened that one of your doors had the booby prize?
I liked the explanation I posted above but that is a very good one as well Hound.
Being explained on Horizon on BBC2 right now :)