CrazyEngineers Archive

Old, but evergreen and popular discussions on CrazyEngineers, presented to you in read-only mode.

@Anoop Kumar • 07 Sep, 2012

One of the lifeline of KBC game aka (Who Wants to Be a Millionaire?) was

The Game: A question is asked and 4 option will be given out of which only option is correct and you have guess which is correct and there are some lifelines if case you have doubt to help you to guess correct answer. Other than some lifelines there is two life lines Double Dip and 50/50.

**Fifty-Fifty (50/50)**which is replaced by**in fifth season and still in current 6th season****Double Dip****.**The Game: A question is asked and 4 option will be given out of which only option is correct and you have guess which is correct and there are some lifelines if case you have doubt to help you to guess correct answer. Other than some lifelines there is two life lines Double Dip and 50/50.

**Question:**which lifeline is better**Double Dip**or**50/50**and why?**Double Dip:**This lifeline allows the contestant to make two guesses at a question.**50/50:**If the contestant uses this lifeline, the host will ask the computer to randomly remove and eliminate two of the "wrong" answers.
@Anil Jain • 07 Sep, 2012
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

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

@Sagar07 • 08 Sep, 2012
I agree with crazyboy

@Jeffrey Arulraj • 10 Sep, 2012
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

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

@zaveri • 08 Feb, 2013
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.

@Kaustubh Katdare • 08 Feb, 2013
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.

@Anoop Kumar • 09 Feb, 2013
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?

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?

@zaveri • 09 Feb, 2013
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.

but it is the rule of the game concerning, double dip, that really made it dangerous.

@Whats In Name • 12 Feb, 2013 • 1 like
It can't be based on the nature of contestant(his knowledge) but the nature of lifeline.

--

50-50:After using it,we are left with 2 choices.

Double-Dip:After using it,we are left with 3 choices.

--

So,Clearly 50-50 is better.

--

**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.

--

So,Clearly 50-50 is better.

@Garvit Garg • 24 Sep, 2014
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.....

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.....

@Shashank Moghe • 24 Sep, 2014
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.

@zaveri @durga

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.

@zaveri @durga

@Garvit Garg • 24 Sep, 2014
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....

@Shashank Moghe • 24 Sep, 2014

This is turning out to be an excellent mathematical exercise.

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?Garvit Gargwhenever 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....

This is turning out to be an excellent mathematical exercise.

@Garvit Garg • 24 Sep, 2014
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.??

@Garvit Garg • 24 Sep, 2014
@Shashank Moghe do look another thread on probability question..movie "21"...

@Shashank Moghe • 24 Sep, 2014

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?

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: https://www.stat.yale.edu/Courses/1997-98/101/condprob.htmGarvit Garg2nd 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.??

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?

@Shashank Moghe • 24 Sep, 2014
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.

@Garvit Garg 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.

@zaveri @durga 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?

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.

@Garvit Garg 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.

@zaveri @durga 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?

@durga ch • 24 Sep, 2014
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: @shashank and @Gravit have quite convincing explanations!

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: @shashank and @Gravit have quite convincing explanations!

@Shashank Moghe • 24 Sep, 2014

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.

durgathe 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.

@durga ch • 24 Sep, 2014
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

@Shashank Moghe • 24 Sep, 2014 • 1 like
You and @Garvit Garg 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.

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.

@Garvit Garg • 25 Sep, 2014
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...

@Anoop Kumar • 25 Sep, 2014
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 😘

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 😘

@Shashank Moghe • 25 Sep, 2014

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.

Garvit Gargtotal 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.

@Shashank Moghe • 25 Sep, 2014

Please explain this step. Not clear from the wordings. I assume the knowledge that you read the previous posts.Anoop Kumar(1/3*3/4)

@Garvit Garg • 25 Sep, 2014

who says that choosing right answer in first attempt is not allowed in double dip......we are finding overall winning probability in double dip....Shashank MogheB 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.

@Shashank Moghe • 25 Sep, 2014

We are looking at the worst case scenario. That's why you are not allowed to choose the right answer in the first case.Garvit Gargwho says that choosing right answer in first attempt is not allowed in double dip......we are finding overall winning probability in double dip....

@durga ch • 25 Sep, 2014
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

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

@Shashank Moghe • 25 Sep, 2014 • 1 like
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.

@Samba Sah • 03 Jul, 2015 • 1 like
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

For winning: either he will win in first attempt

probability of winning in first attempt = 1/4

probability of winning in second attempt = (first answer is wrong)

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 attemptprobability 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)
@ken1486 • 17 Feb, 2016
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.

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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