# Probability Problem

Question asked by Prasad Ajinkya in #Coffee Room on Apr 4, 2011

Prasad Ajinkya · Apr 4, 2011

Rank B1 - LEADER

Sue and Bob take turns rolling a 6-sided die. Once either person rolls a 6 the

game is over. Sue rolls first, if she doesn't roll a 6, Bob rolls the die, if

he doesn't roll a 6, Sue rolls again. They continue taking turns until one of

them rolls a 6.

Bob rolls a 6 before Sue.

What is the probability Bob rolled the 6 on his second turn? Posted in: #Coffee Room

game is over. Sue rolls first, if she doesn't roll a 6, Bob rolls the die, if

he doesn't roll a 6, Sue rolls again. They continue taking turns until one of

them rolls a 6.

Bob rolls a 6 before Sue.

What is the probability Bob rolled the 6 on his second turn? Posted in: #Coffee Room

ISHAN TOPRE · Apr 4, 2011

Rank A2 - PRO

The answer is 1/18

vipandeep · Apr 4, 2011

Rank D2 - MASTER

is it 125/216 ?

Prasad Ajinkya · Apr 4, 2011

Rank B1 - LEADER

@ishutopre - no, could you explain the logic as to how you arrived at that no?

@vipandeep - no, although you seem to be thinking in the right direction!

@vipandeep - no, although you seem to be thinking in the right direction!

cooltwins · Apr 5, 2011

Rank B1 - LEADER

problablity that bob wins is 5^3/6^4 i.e. 125/1296

correct?

correct?

Prasad Ajinkya · Apr 5, 2011

Rank B1 - LEADER

Sorry cooltwins, perhaps its best that you discuss your approach, that way others can also help build the solution!!

cooltwins · Apr 5, 2011

Rank B1 - LEADER

No.of way bob wins the game= 5*5*5*1

and total no. of ways=6*6*6*6

so probability = 5^3/6^4

and total no. of ways=6*6*6*6

so probability = 5^3/6^4

cooltwins · Apr 5, 2011

Rank B1 - LEADER

or a simpler soln is:

if bob wins in 2nd round

probability=p(sue loses 1st round)*p(bob loses 1st round)*p(sue loses the 2nd round)*p(bob wins)

=5/6 * 5/6 * 5/6 * 1/6

if bob wins in 2nd round

probability=p(sue loses 1st round)*p(bob loses 1st round)*p(sue loses the 2nd round)*p(bob wins)

=5/6 * 5/6 * 5/6 * 1/6

ISHAN TOPRE · Apr 5, 2011

Rank A2 - PRO

I will modify the answer

There are two probabilities in the way so that Bob will have second chance.

Sue, Bob, Sue, Bob

Also the probability of Bob having 6 out of 6*4 attempts is 1=1/24

So 1/24

So is the answer 1/24?

Modified again as CT suggested. 😀.

There are two probabilities in the way so that Bob will have second chance.

Sue, Bob, Sue, Bob

Also the probability of Bob having 6 out of 6*4 attempts is 1=1/24

So 1/24

So is the answer 1/24?

Modified again as CT suggested. 😀.

cooltwins · Apr 5, 2011

Rank B1 - LEADER

but ishu in the question it clearly states that sue starts. 😐

Prasad Ajinkya · Apr 5, 2011

Rank B1 - LEADER

No, read the problem again. Ill IM you the hint 😀

Prasad Ajinkya · Apr 7, 2011

Rank B1 - LEADER

And here is the solution!

A: Sue wins

B: Bob wins

C: Bob wins in his second turn

D: Sue does NOT win in her first turn

We want to find the probability that Bob wins in his second turn, given that he wins in first place:

P(C|B) = P(C^B) / P(B)

Since B follows logically from C:

P(C|B) = P(C) / P(B)

The probability that the winner is determined in the fourth die toss (Bob’s second) can be easily calculated:

P(C) = (5/6) * (5/6) * (5/6) * (1/6) = 125/1296

Calculating P(B) is a little bit trickier, but not so much. First, we note that, with probability 1, either Sue or Bob must win the game at some point. (The probability that the game hasn’t before the nth die toss converges to zero as n grows. This doesn’t mean that it’s impossible that the game never ends.) And, since they can’t both win at the same time:

P(A) + P(B) = 1

Given that Sue doesn’t win in her first turn, Bob’s chances are exactly the same that they were for Sue before her first turn. So we have:

P(A) = P(B|D) = P(B^D)/P(D)

Since D follows logically from B:

P(A) = P(B|D) = P(B)/P(D) = P(B) * (6/5)

Replacing for P(A) in P(A) + P(B) = 1 gives:

P(B) * (6/5) + P(B) = 1

P(B) = 5/11

(Kudos to Nathanael Nerode for calculating the probability that Bob wins using geometric series!)

So we can calculate the original conditional probability:

P(C|B) = P(C) / P(B) = (125/1295) / (5/11) = 275/1296

A: Sue wins

B: Bob wins

C: Bob wins in his second turn

D: Sue does NOT win in her first turn

We want to find the probability that Bob wins in his second turn, given that he wins in first place:

P(C|B) = P(C^B) / P(B)

Since B follows logically from C:

P(C|B) = P(C) / P(B)

The probability that the winner is determined in the fourth die toss (Bob’s second) can be easily calculated:

P(C) = (5/6) * (5/6) * (5/6) * (1/6) = 125/1296

Calculating P(B) is a little bit trickier, but not so much. First, we note that, with probability 1, either Sue or Bob must win the game at some point. (The probability that the game hasn’t before the nth die toss converges to zero as n grows. This doesn’t mean that it’s impossible that the game never ends.) And, since they can’t both win at the same time:

P(A) + P(B) = 1

Given that Sue doesn’t win in her first turn, Bob’s chances are exactly the same that they were for Sue before her first turn. So we have:

P(A) = P(B|D) = P(B^D)/P(D)

Since D follows logically from B:

P(A) = P(B|D) = P(B)/P(D) = P(B) * (6/5)

Replacing for P(A) in P(A) + P(B) = 1 gives:

P(B) * (6/5) + P(B) = 1

P(B) = 5/11

(Kudos to Nathanael Nerode for calculating the probability that Bob wins using geometric series!)

So we can calculate the original conditional probability:

P(C|B) = P(C) / P(B) = (125/1295) / (5/11) = 275/1296

Sachin Jain · Apr 8, 2011

Rank C2 - EXPERT

Is answer 275/1296 ?

Reya · Apr 8, 2011

Rank A3 - PRO

@blunderboy: see the above post!!Answer is already given 😀

Sachin Jain · Apr 8, 2011

Rank C2 - EXPERT

@ Praveena

Hey thanks for pointing out but i just saw the problem and did not see the posts on second page.

I just saw the posts on first page and was not aware of post containing solution.

My bad 😔

Hey thanks for pointing out but i just saw the problem and did not see the posts on second page.

I just saw the posts on first page and was not aware of post containing solution.

My bad 😔

vibhor_one · Jan 14, 2013

Rank C3 - EXPERT

A candidate is required to answer 6 out of 10 questions, which are divided into two groups each containing 5 questions, and he is nott permited to attempt more than 4 from the group. In how many ways can he makw up his choice?

rahul69 · Jan 14, 2013

Rank B2 - LEADER

My guess isvibhor_oneA candidate is required to answer 6 out of 10 questions, which are divided into two groups each containing 5 questions, and he is nott permited to attempt more than 4 from the group. In how many ways can he makw up his choice?

**200**....😕

vibhor_one · Jan 18, 2013

Rank C3 - EXPERT

how you solved it? please explain I am not getting the same answer

rahul69 · Jan 18, 2013

Rank B2 - LEADER

I suppose u must be knowing the right answer, so is my answer correct? If it is correct then there is sense in explaining, otherwise not, and what is the answer u r getting?vibhor_onehow you solved it? please explain I am not getting the same answer

vibhor_one · Jan 20, 2013

Rank C3 - EXPERT

yes answer is correct, but how you worked it out? I am not getting the same answer

rahul69 · Jan 20, 2013

Rank B2 - LEADER

Well here is the solution: We have to select 6 questions from 10 questions divided in 2 groups of 5 and not more than 4 for each group,

So using combinations: {

=

=

=

Here, we got the answer.

Hope it helps 😀

So using combinations: {

**(5C4)*(5C2) + (5C3)*(5C3) + (5C2)*(5C4)**}=

**{5*10+10*10+10*5}**=

**{50+100+50}**=

**200**Here, we got the answer.

Hope it helps 😀

vibhor_one · Jan 20, 2013

Rank C3 - EXPERT

thanks this solution really helped me to understand the concept

Saidu Konneh · Feb 4, 2013

Rank E2 - BEGINNER

The fact that Sue starts the game and that anyone playing a six ends the game, it implies that the total number of rolls is four(4), provided Bob plays six during his second roll.

So now: The probability of Bob playing a six in his second roll is equalled to the number of six's divided by the total number of rolls in the game: 1/4

So now: The probability of Bob playing a six in his second roll is equalled to the number of six's divided by the total number of rolls in the game: 1/4