# Weakest Link Probability Watch

Announcements

This discussion is closed.

Is it possible to work out the probability of winning on the Weakest Link?? Just a completely random thought

0

Report

#2

(Original post by

Is it possible to work out the probability of winning on the Weakest Link?? Just a completely random thought

**saiyamana**)Is it possible to work out the probability of winning on the Weakest Link?? Just a completely random thought

0

Report

#3

Well given that it's based on answering questions and stuff then I'd say no. Or at least not accurately at all.

0

Report

#5

the probabilties of any person winning are independent assuming they are random people....

0

Report

#6

(Original post by

the probabilties of any person winning are independent assuming they are random people....

**manps**)the probabilties of any person winning are independent assuming they are random people....

0

Report

#7

if I was on it the probablity would be 0.999999999..................... ..........................of being the first weakest link

0

Report

#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.

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

Newton.

0

Report

#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?!

0

Report

#10

(Original post by

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.

**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.

0

Report

#11

you have be smart enough to play dumb before the others notice. but if everyone adopts this tactic, then it wont work.

0

Report

#12

(Original post by

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.

**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.

Newton.

0

Report

#13

**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.

0

Report

#14

(Original post by

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.

**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.

Newton.

0

Report

#15

(Original post by

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?!

**[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?!

0

Report

#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.

0

Report

#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.

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

Report

#19

(Original post by

Finally! Thank You El Stevo

**[email protected]**)Finally! Thank You El Stevo

just to clarify, its 1/10 and can't be nothing other than 1/10 assuming independance and everyone is equal etc...

0

Report

#20

(Original post by

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.

**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.

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

X

new posts

Back

to top

to top