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14 years ago
#21
(Original post by Newton)
The probability that you win any one round is (1/(11-n)) where n is the certain round number i. e. the first or the second or the third. It is obvious by the equation that the probability increases as the game progresses, and this is also apparent in the actual game as the number of contestants to compete against decreases respectively. Hence, the probability of winning the first round is (1/11-1)=(1/10), the propability of winning the second is (1/11-2)=(1/9). Since you want to find out what is the probability of winning the first AND the second AND the third AND... AND the last round, if we let W be the event that we win the whole game is P(W)=(1/10)*(1/9)*(1/8)*... *1

Which in fact means that P(W)=(1/10!)=0.00000027557

Newton.
Are you saying that the probability of winning is 0.00000027557?!
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14 years ago
#22
He made a mistake. We've been over this.
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14 years ago
#23
I'm crap at Maths but aren't there nine contestants?
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14 years ago
#24
(Original post by LH)
I'm crap at Maths but aren't there nine contestants?
hahaha if that's the case, it wouldn't surprise me....maths people don't generally watch the Weakest Link
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14 years ago
#25
(Original post by LH)
I'm crap at Maths but aren't there nine contestants?
yup there are...

so in theory its 1/9, but does it still count as a win if they don't bank any money?
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14 years ago
#26
(Original post by LH)
I'm crap at Maths but aren't there nine contestants?
If there are N contestants then the probability of winning is 1/N. Happy now ?
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14 years ago
#27
(Original post by AntiMagicMan)
If there are N contestants then the probability of winning is 1/N. Happy now ?
that is... beautiful!
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14 years ago
#28
Yes. It should be called the Fundamental Theorem of The Weakest Link.
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14 years ago
#29
It's on now!
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