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Thread starter 14 years ago
#1
Is it possible to work out the probability of winning on the Weakest Link?? Just a completely random thought 0
14 years ago
#2
(Original post by saiyamana)
Is it possible to work out the probability of winning on the Weakest Link?? Just a completely random thought 1/10 presuming there are 10 people.
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14 years ago
#3
Well given that it's based on answering questions and stuff then I'd say no. Or at least not accurately at all.
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14 years ago
#4
it increases the further you get 0
14 years ago
#5
the probabilties of any person winning are independent assuming they are random people....
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14 years ago
#6
(Original post by manps)
the probabilties of any person winning are independent assuming they are random people....
crank out a probability tree then.... 0
14 years ago
#7
if I was on it the probablity would be 0.999999999..................... ..........................of being the first weakest link 0
14 years ago
#8
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.
0
14 years ago
#9
what?! of course it's 1/10. The chance that a particular person wins has got to be 1/10. Why would a randomly chosen person have a better chance of winning than another randomly chosen person?!
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14 years ago
#10
(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.
have you taken into account in the last few rounds voting is tactical ? if you are the strongest link the weaker players conspire to vote you off.
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14 years ago
#11
you have be smart enough to play dumb before the others notice. but if everyone adopts this tactic, then it wont work.
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14 years ago
#12
(Original post by evariste)
have you taken into account in the last few rounds voting is tactical ? if you are the strongest link the weaker players conspire to vote you off.
Your point is valid; there are many other points like this aswell, which we need to take into consideration when working out the latter. Although we do not take them into consideration as we are just talking about an average and a random person of winning the whole game irrespectible of his level of knowledge.

Newton.
0
14 years ago
#13
(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.
But you don't need to win every round to win the weakest link, all you need to do is not get voted off each round and then win the final.
0
14 years ago
#14
(Original post by Nima)
But you don't need to win every round to win the weakest link, all you need to do is not get voted off each round and then win the final. In theory, you could come 2nd to last in every round up until the final round, come last in that, then win the final, and you've won the weakest link.
Let P(W) be the probability that you stay for the next round.

Newton.
0
14 years ago
#15
(Original post by [email protected])
what?! of course it's 1/10. The chance that a particular person wins has got to be 1/10. Why would a randomly chosen person have a better chance of winning than another randomly chosen person?!
That's what I was thinking! 0
14 years ago
#16
if you pick a person prior to the game knowing 'nothing about no-one' then i'm on [email protected]'s side... 1/10.
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14 years ago
#17
Finally! Thank You El Stevo
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14 years ago
#18
Er Nima it would be

P(Win) = 9/10*8/9*7/8*6/7*5/6*4/5*3/4*2/3*1/2 = 1/10

As is clearly obvious from the fact that 10 people enter the studio each with an equal chance of winning, therefore 1/10.
0
14 years ago
#19
(Original post by [email protected])
Finally! Thank You El Stevo
no problem... now wheres that tenner just to clarify, its 1/10 and can't be nothing other than 1/10 assuming independance and everyone is equal etc...
0
14 years ago
#20
(Original post by AntiMagicMan)
Er Nima it would be

P(Win) = 9/10*8/9*7/8*6/7*5/6*4/5*3/4*2/3*1/2 = 1/10

As is clearly obvious from the fact that 10 people enter the studio each with an equal chance of winning, therefore 1/10.
No, because in the round of 3 you can't be voted off - But, my calculation was only going on Newton's idea of doing it. I still think it's 1/10.

EDIT: Ooopsy, for some reason I've just noticed that you do get voted off in the round of 3, ahem, what was I thinking? How else does it go down to 2, lol.

So yes, by both methods, it is 1/10. 0
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