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Old 09-11-2005, 07:44 AM   #101
Sever
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Quote:
Originally posted by Aelia Jusa:
Isn't it that the chance of having 12 babies all being female is very small, but if you have 11 girls then the chance that your next one will be female is still 50% because probability has no memory? Like, if you toss a coin whether heads or tails comes up has no bearing on whether you tossed heads the last time you tossed it. Or something? [img]graemlins/blush.gif[/img]
That's about the truth of it, yes. But how often does a 50/50 probability continue unbroken for 12 straight instances?
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Old 09-11-2005, 07:47 AM   #102
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Quote:
Originally posted by Azred:
Combinations. Suppose you have k objects and you want to arrange n of them, the formula is kCn = k!/(n!*(k-n)!) Example: 10C5 = 10!/(5!*5!) = 252.

The probability of answering correctly with a guess is .25; hence, the probability of guessing incorrectly is .75. If you guess 3 correctly that means 7 incorrectly, or (.25^3) * (.75^7). Since each question is independent, you wind up multiplying the probabilities together.

Getting three questions correct means arranging 3 out of 10. From the above forumla, 10C3 = 10!/(3! * 7!) = 120 ways of answering 3 correctly.

Putting the forumulas together gives...
...the likelihood of answering n questions correctly = 10Cn * (.25^n) * (.75^(10-n)). For 3 questions 120 * .25^3 * .75^7 = .250282288.

For the cumulative probabilities, obviously add.

Simple, yes? [img]graemlins/petard.gif[/img] Now, if only I could answer some of those ridiculous questions. [img]graemlins/laugh3.gif[/img]
[img]graemlins/saywhat.gif[/img] I was with you all the way up to suppose!

[ 09-11-2005, 07:48 AM: Message edited by: Sever ]
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Old 09-11-2005, 08:03 AM   #103
ocellis
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Quote:
That's about the truth of it, yes. But how often does a 50/50 probability continue unbroken for 12 straight instances?
Well i'm no math wiz but this is how i'd explain it:

Every time you get a kid it has a probability of 1/2 for being female. This is fixed no matter how often you repeat the.. er.. experiment

Now the probability that you get two girls in a row is:
Probability for first girl multiplied with the Probability for the second girl. So:
1/2*1/2 = 1/4 = 25%

So if you want to calculate the probability for 12 girls in a row you'll have to calculate:
1/2*1/2* ... * 1/2 = 1/2^12 = 0.00024%

[ 09-11-2005, 08:04 AM: Message edited by: ocellis ]
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Old 09-12-2005, 04:04 PM   #104
Melcheor
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Azred, trying to teach the binomial distribution on a computer gaming forum probably isn't going to be very succesful. I got 2 right. By your table that seemed on the unlucky side of probable, but only just!
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Old 09-12-2005, 04:25 PM   #105
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Wow, I actually won one!
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Old 09-12-2005, 07:57 PM   #106
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Quote:
Originally posted by Melcheor:
Azred, trying to teach the binomial distribution on a computer gaming forum probably isn't going to be very succesful. I got 2 right. By your table that seemed on the unlucky side of probable, but only just!
You're probably correct, there. Anyone who wants further information on the topic should pick up a probability/statistics textbook...for a little light reading. [img]graemlins/petard.gif[/img]
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Old 09-12-2005, 10:05 PM   #107
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for some reason, it always takes me 61 seconds to complete the test...
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Old 09-13-2005, 12:23 AM   #108
Sever
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Quote:
Originally posted by Azred:
quote:
Originally posted by Melcheor:
Azred, trying to teach the binomial distribution on a computer gaming forum probably isn't going to be very succesful. I got 2 right. By your table that seemed on the unlucky side of probable, but only just!
You're probably correct, there. Anyone who wants further information on the topic should pick up a probability/statistics textbook...for a little light reading. [img]graemlins/petard.gif[/img] [/QUOTE]Ok, ok. I get the hint...
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Old 09-13-2005, 05:17 AM   #109
Melcheor
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Right, what seems to happen to me is that I know 2 answers and then guess all of the other 8 wrong. Thats happened twice now. Probability of guessing all 8 wrong (no combinations needed, thank god) is 0.75^8 = 0.1001129150... Essentially 1 in 10.

Bear in mind that this has happened twice out of three goes and you get the probability of this happening as (3C2)x(0.1^2)x(0.9) = 0.027

This is damn unlucky. I should win some sort of wooden spoon or booby prize. Hmmmm, booby prize...

(edit - can do maths but can't spell)

[ 09-13-2005, 05:18 AM: Message edited by: Melcheor ]
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Old 09-13-2005, 09:36 AM   #110
Sever
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No booby prize for you. I came last out of 27 people (including at least 1 mock person) for the first two days that this quiz was running...

And nobody gave me a horse's ass! [img]tongue.gif[/img]
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