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Prasad Ajinkya
Prasad Ajinkya • Apr 4, 2011

Probability Problem

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?
ISHAN TOPRE
ISHAN TOPRE • Apr 4, 2011
The answer is 1/18
vipandeep
vipandeep • Apr 4, 2011
is it 125/216 ?
Prasad Ajinkya
Prasad Ajinkya • Apr 4, 2011
@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!
cooltwins
cooltwins • Apr 5, 2011
problablity that bob wins is 5^3/6^4 i.e. 125/1296

correct?
Prasad Ajinkya
Prasad Ajinkya • Apr 5, 2011
Sorry cooltwins, perhaps its best that you discuss your approach, that way others can also help build the solution!!
cooltwins
cooltwins • Apr 5, 2011
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
cooltwins
cooltwins • Apr 5, 2011
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
ISHAN TOPRE
ISHAN TOPRE • Apr 5, 2011
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. 😀.
cooltwins
cooltwins • Apr 5, 2011
but ishu in the question it clearly states that sue starts. 😐
Prasad Ajinkya
Prasad Ajinkya • Apr 5, 2011
No, read the problem again. Ill IM you the hint 😀
Prasad Ajinkya
Prasad Ajinkya • Apr 7, 2011
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
Sachin Jain
Sachin Jain • Apr 8, 2011
Is answer 275/1296 ?
Reya
Reya • Apr 8, 2011
@blunderboy: see the above post!!Answer is already given 😀
Sachin Jain
Sachin Jain • Apr 8, 2011
@ 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 😔
vibhor_one
vibhor_one • Jan 14, 2013
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
rahul69 • Jan 14, 2013
vibhor_one
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?
My guess is 200....😕
vibhor_one
vibhor_one • Jan 18, 2013
how you solved it? please explain I am not getting the same answer
rahul69
rahul69 • Jan 18, 2013
vibhor_one
how you solved it? please explain I am not getting the same answer
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_one
vibhor_one • Jan 20, 2013
yes answer is correct, but how you worked it out? I am not getting the same answer
rahul69
rahul69 • Jan 20, 2013
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: {(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
vibhor_one • Jan 20, 2013
thanks this solution really helped me to understand the concept
Saidu Konneh
Saidu Konneh • Feb 4, 2013
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

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