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Double Dip - That gives you probability of .58 that answer would be correct; however 50/50 gives probability of .5
Also, in 50/50 they remove the two obvious (easily identified) choice, and in double dip if you can identify one obvious wrong, then you get 2 attempts out of 3, which is bit more probable,
-CB
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I agree with #-Link-Snipped-#
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both depends if we have a slight doubt then they are of some value
if we can't just differentiate between two answer it is better to have double dip
if our head draws blank it is better to go for 50/50
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Double dip is based on the 50-50 lifeline, and has replaced the same, but according to the rule, that once the contestant has opted for that lifeline then he/she cannot quit the game, makes the lifeline dangerous and risky.
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Wow, I totally missed this discussion! I've had the same question too; but the choice becomes easier when you've 'doubts'. It's still a gamble; but I'd prefer 50/50 over double-dip.
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Lets talk on fact of probability...
in 50/50: no doubt chances of winning is 50%.
in double dip: you have chance to select two options out of four. so technically it is also 50% probability.
But,when you are confused between two answers which could be correct. double dip is better it is sure shot rather than giving only one answer in case of 50/50.
is there any case in which 50/50 could be better rather than double dip?
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from mathematical point of view both lifelines provide 50 % chances of winning.
but it is the rule of the game concerning, double dip, that really made it dangerous.
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It can't be based on the nature of contestant(his knowledge) but the nature of lifeline.
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Worst Case Scenario:
50-50:After using it,we are left with 2 choices.
Double-Dip:After using it,we are left with 3 choices.
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So,Clearly 50-50 is better.
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Lets talk on fact of probability...
50/50 --- 0.5 probability....
but in double dip.....when you are choosing first option, at that time, probability of getting a right answer is 0.25....if its wrong than you are remaining with 3 options...so now probability of getting right answer is 0.33....so total probability of getting a right answer is 0.25+0.33=0.58 > 0.5...
so always double dip over 50/50.....
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Let A = probability that the second answer in the double dip is the right one.
B = probability that the first answer was wrong
P(A|B) = P (A intersection B)/P(B) ------ probability that the second answer is right after the first one was wrong
Now, P (A) = 1/3 (since only 3 options remain once you have got the first one wrong)
P(B) = 3/4 (Since there is only one right answer, the probability of getting the first answer wrong is 75%)
P(A intersection B) = Probability that the second answer is right after the first one is wrong = 3/9 = 1/3 (This is tricky to imagine, so just count the possible answers one might give from options A,B,C,D: assuming B is the right answer. These options can be: AB, AC, AD, CA, CB, CD, DA, DB, DC. Of these, 3 options get the right answer in the second attempt. Hence 1/3 probability. [Note: If you select B in the first chance, this analysis does not hold good. We are considering the worst case scenario]).
Thus, by the formula,
P(A|B) = (1/3)/(3/4) = 4/9 < 50%
Hence, clearly, 50:50 is a better lifeline to use, mathematically.
This analysis assumes you are totally unaware of the answer and are purely playing on chance.
#-Link-Snipped-# #-Link-Snipped-#
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whenever i use to have confusion in probability i use to go for finding odds to lose it....probability of losing the question is probability of losing in 1st attempt AND losing it in 2nd attempt....(3/4)*(2/3)= 0.5 actually....now m more convinced with this approach....so equal probabilities....
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Garvit Garg
whenever i use to have confusion in probability i use to go for finding odds to lose it....probability of losing the question is probability of losing in 1st attempt AND losing it in 2nd attempt....(3/4)*(2/3)= 0.5 actually....now m more convinced with this approach....so equal probabilities....
I guess you can have multiplication of that sort only when the events are mutually exclusive. Give the dependence on the first answer being wrong, can the second selection be mutually independent of the first event?
This is turning out to be an excellent mathematical exercise.
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2nd event depends on first event....thats why probability of losing=probability of wrong answer in first attempt * probability of wrong answer in second attempt when first attempt has been done....P = P(A)*P(B/A)= (3/4)*(2/3)....right.??
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#-Link-Snipped-# do look another thread on probability question..movie "21"...
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Garvit Garg
2nd event depends on first event....thats why probability of losing=probability of wrong answer in first attempt * probability of wrong answer in second attempt when first attempt has been done....P = P(A)*P(B/A)= (3/4)*(2/3)....right.??
Well, I am justified with my answer. But it is not "who", but "what" is right that I am looking for here. I find your answer simple, yet my answer "by the rules". Please check this link: <a href="https://www.stat.yale.edu/Courses/1997-98/101/condprob.htm" target="_blank" rel="nofollow noopener noreferrer">www.stat.yale.edu</a>
I am having trouble differentiating if in this case, P(B/A) and P(A intersection B) is the same thing.
Because,
P(A|B) = "being right on the second attempt after I have already been wrong on the first attempt"
P (A intersection B) = "Being wrong in the first attempt AND being right in the second".
What is the difference?
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I guess I have found my answer.
P(A|B) = "being right on the second attempt after I have already been wrong on the first attempt" needs to be evaluated as an independent event. Hence this probability comes out to be 1/3.
We need to find the answer to P (A intersection B), and that will be our solution.
Again,
Let A = probability that the second answer in the double dip is the right one.
B = probability that the first answer was wrong
P(A|B) = 1/3
P(B) = 3/4 (Since there is only one right answer, the probability of getting the first answer wrong is 75%)
P(A intersection B) = Probability that the second answer is right AND the first one is wrong = P(B)*P(A|B) = (1/3)*(3/4) = 1/4
Even more, 50:50 is a better lifeline to use, mathematically.
#-Link-Snipped-# in your case, you have a probability of 50% on getting the second answer wrong after you have gone wrong with the first answer. That does not mean that you have 50% probability of getting the answer right. Remember, you have 2 wrong answers and only one right answer in the remaining 3 options. The ratio of right:wrong is 1:2, hence you are getting twice the probability as I am getting.
Hope this helps clear the fog of probability.
#-Link-Snipped-# #-Link-Snipped-# tag all who have an interest in mathematics.
This brings me to another interesting question: Do the probability(right) and (wrong) need to add upto 1, in case of conditional probability situations?
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the contestant would get a second chance in double dip only when the contestant failed to answer right in my first attempt. right?
so considering that, after 1st chance probability of contestant 'guessing' a right answer is reduced to 1/3 ie 33%
where as 50-50% : the contestant gets to choose from 2 options hence 1/2 ie 50%
EDIT: #-Link-Snipped-# and @Gravit have quite convincing explanations!
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durga
the contestant would get a second chance in double dip only when the contestant failed to answer right in my first attempt. Those are not independent events right?
so considering that, after 1st chance probability of contestant 'guessing' a right answer is reduced to 1/3 ie 33%
where as 50-50% : the contestant gets to choose from 2 options hence 1/2 ie 50%
But in this case, we have to evaluate the conditional probability. You have just evaluated the probability of the second event, as if it were an independent event.
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coming to think about it, a contestant can loose on a question if he fails in both attempts of double dip. that is 3/4 * 2/3 = 50% :-s
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You and #-Link-Snipped-# are both right at that. But did you read through the explanation I tagged you in? The odds of going wrong on both attempts is 0.5. But as per my calculations, the chance of going right on the second attempt AFTER you have gone wrong on the first attempt is 0.25.
I want someone to open my eyes, or be an accomplice in wondering at the awe of this result. Either ways. I am amazed how such a seemingly trivial problem is keeping me from doing office work 😁 Which obviously means either I undervalued its complexity or I am looking too much into minute (and incorrect) details.
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total possibilities to answer in double dip-AB AC AD BA BC BD CA CB CD DA DB DC....in which right answer ll be in options AB BA BC BD Cb DB....probability= 0.5...might this solution ll help...
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Statistically both answer give you 50% probability.
50-50: 1/2 , straight forward.
Double-Dip: 1/4 + (1/3*3/4) = 1/4+1/4 = 1/2.
Because here you are choosing 1/3 or remaining 3/4 😘
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Garvit Garg
total possibilities to answer in double dip-AB AC AD BA BC BD CA CB CD DA DB DC....in which right answer ll be in options AB BA BC BD Cb DB....probability= 0.5...might this solution ll help...
B is the right answer. Any options with B as the first answer will not proceed to the second chance. You are right in the first chance, this is not the scenario we are considering.
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Anoop Kumar
(1/3*3/4)
Please explain this step. Not clear from the wordings. I assume the knowledge that you read the previous posts.
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Shashank Moghe
B is the right answer. Any options with B as the first answer will not proceed to the second chance. You are right in the first chance, this is not the scenario we are considering.
who says that choosing right answer in first attempt is not allowed in double dip......we are finding overall winning probability in double dip....
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Garvit Garg
who says that choosing right answer in first attempt is not allowed in double dip......we are finding overall winning probability in double dip....
We are looking at the worst case scenario. That's why you are not allowed to choose the right answer in the first case.
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ok i have not watched the show , and dont know what happens to incorrect answer after first attempt,
is it just an assumption we are making that for second trail we have only 3 options to choose from?
just a random thought
first attempt i choose an incorrect answer ie 3/4
second attmept (if i am allowed to choose from all 4 , keeping aside the fact that the contestant now knows one of them is incorrect) 3/4?
3/4*3/4???
i am quite sure this is not the case, but just some random thought crossed my mine while I was jogging
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Well, I have not watched the show in a very long time either. In double dip, after the first answer is wrong, you obviously are left with three options to choose the right one from, on the second attempt.
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50/50 : probability is 0.5
double dip : (probability of winning in first attempt) + (probability of winning in 2nd attempt)
😔0.25)+(0.75*0.33)
😔0.25)+(0.2475)
:0.4975
OR means + in probability , AND means * in probability
For winning: either he will win in first attempt or in second attempt
probability of winning in first attempt = 1/4
probability of winning in second attempt = (first answer is wrong) and (second answer is right) = (3/4 * 1/3)
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1. 50-50 option: -
Sample space – 2
Probability of success P(s) = ½ = 0.5
Probability of failure P(f) = ½ = 0.5
2. Double dip option: -
· In, first attempt
Sample space – 4
Probability of success P(s) = 1/4 = 0.25
Probability of failure P(f) = 3/4 = 0.75
Given that failed in first attempt,
· In, second attempt
Sample space – 3
Probability of success P(s) = 1/3 = 0.34
Probability of failure P(f) = 2/3 = 0.67
Either success in first attempt P(s)=0.25 or success in second attempt P(s)=0.34. so, in both cases the probability of success in double dip option is lesser than that of in 50-50 option.
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