Bloomberg LP Interview Question






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1
of 1 vote

Assume the probability of rolling the dice by me is 1/2. And similarly by you is 1/2.

That means each of us (you and me) have the same probability of winning.

- Anonymous December 21, 2009 | Flag Reply
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0
of 0 votes

yep

- rockersatish November 30, 2010 | Flag
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0
of 0 vote

i think answer should be equal to 1/6*1/2=1/12

- any December 22, 2009 | Flag Reply
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0
of 0 vote

I am guessing 5/36.

bcoz, for me to get a chance, you should not win, probability of which is 5/6.
Then, for me to win, I should roll a 6. For which the probability is 1/6.

Hence the total probability is 5/6 * 1/6 = 5/36.

Correct me if I am wrong.

- Ajay December 26, 2009 | Flag Reply
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0
of 0 votes

But he didn't say who would roll first...If the "I" roll first, then I agree with your answer. But if the "you" roll first, then it should be 1/6.

Please correct me if I am wrong.

- yzjinjin December 31, 2009 | Flag
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0
of 0 vote

This is a geometric series.

You can win on the first with probability 1/6.
You can win on the second with probability 4/6 * 1/6 (don't lose, then win).
Then 2/3 * 2/3 * 1/6 (don't lose twice, then win).

1/6 * Sum[n = 0 to inf, (2/3)^n] = 1/6 * 1/(1 - 2/3) = 1/6 * 3 = 1/2.

Intuitively that should make sense to. If the game is supposed to end and the roll is fair, you both have equal chances.

- Anonymous January 01, 2010 | Flag Reply
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0
of 0 votes

Never mind, you are alternating rolls.

Then you have to ask for the order.

I suppose if she goes first and you are limited to one round then 5/6 * 1/6 = 5/36 is the answer.
Nothing happens with 25/36 chance.

However, it's still a geometric series if you let the game go to infinity.

5/36 * Sum[n = 0 to inf, (25/36)^n] = 5/36 * 1/(1 - 25/36) = 5/36 * 36/25 = 1/5.

So think about going first. =P

- Anonymous January 01, 2010 | Flag
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of 0 vote

@above...your answer is supposed to be 5/36 * 36/11 = 5/11...
Other way is if I start the game then it will

1/6 (1+ (1/6)^2 + (1/6)^4+....) = 1/6 * 36/11 = 6/11.........

in either of case it will be almost 1/2....

- RajiniHassan January 02, 2010 | Flag Reply
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0
of 0 votes

Yes, I stand correct, 1 - 25/36 = 11/36 =D

After doing an infinite summation,

- Anonymous January 02, 2010 | Flag
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of 0 vote

If u start, 5/36 * Sum[n = 0 to inf, (25/36)^n] = 5/36 * 1/(1 - 25/36) = 5/36 * 36/11 = 5/11.

If i start, 1/6 * Sum[n = 0 to inf, (25/36)^n] = 1/6 * 1/(1 - 25/36) = 1/6 * 36/11 = 6/11.

- Rocky January 19, 2010 | Flag Reply
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0
of 0 vote

Its 5/11.

- Anales January 20, 2010 | Flag Reply
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Why we are making it so complex? we have 6 possible outcomes. All are mutually exclusive. Each has same chance (1/6). Favorable case is 1. So the probability is 1/6.

I am assuming the dice is rolled only once and that will decide I win or loose and not until I get 1 or 6. It is not mentioned in the question.

- Deepak Sharma February 02, 2010 | Flag Reply
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0
of 0 vote

probablity is 1/2..since is jsut winning and loosing on 1and 6 respectively and if rolled 2,3,4,5 den v roll again since no one won.

- Jai Mota February 15, 2010 | Flag Reply
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0
of 0 vote

Interesting question. If a turn is one roll for you and one roll for me, 1 in 6. However, if we say that the game continues until someone rolls a winning number, my chances go up to 1 chance in 2 since someone will win.

- Wandering programmer March 01, 2010 | Flag Reply
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0
of 0 vote

To sum up.

Answer is 1/6 providing question doesnot mention they keep on rolling the dice until someone wins.

However If question said so..then the above answer was correct.

This is a geometric series.

You can win on the first with probability 1/6.
You can win on the second with probability 4/6 * 1/6 (don't lose, then win).
Then 2/3 * 2/3 * 1/6 (don't lose twice, then win).

1/6 * Sum[n = 0 to inf, (2/3)^n] = 1/6 * 1/(1 - 2/3) = 1/6 * 3 = 1/2.

Intuitively that should make sense to. If the game is supposed to end and the roll is fair, you both have equal chances.

- kapoor.utd March 06, 2010 | Flag Reply
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0
of 0 vote

read the above answer was correct as "Below Answer"
Sorry guys for Typo.

- kapoor.utd March 06, 2010 | Flag Reply
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0
of 0 vote

P(youWin)=p(Iwin)=p=1/6
p(youLose)=p(youLose)=q=5/6
if you start,
prob(me winning)=qp+qqqp+qqqqqp+...=qp(1+q^2+q^4+...)=qp/(1-q^2)=q/(1+q)=5/11
if I start,
prob(me winning)=1-5/11=6/11

- Neerav March 14, 2010 | Flag Reply
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0
of 0 vote

Apparently the one starting at first will have the high chance to win....

- Anonymous April 09, 2010 | Flag Reply
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if I rolls first
P(you winning game ) = (1/6)*((5/6)/ (1-(5/6)^2) = 5/11

if YOU rolls first
then P( You winning game) = 6/11

- Anonymous June 20, 2010 | Flag Reply
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of 0 vote

{{IF I STARTS FIRST :1/6+5/6*5/6*1/6+...=6/11;
IF HE STARTS FIRST:5/6*1/6+5/6*5/6*5/6*1/6+...=5/11;}}

- jaggu September 15, 2010 | Flag Reply
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0
of 0 vote

I think it is 1/2

Ans:(1/2)*(1/6) + (1/2)*(5/6) = 6/12 = 1/2
If I rolls If other rolls

- rockersatish November 30, 2010 | Flag Reply
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0
of 0 vote

>>If I roll a 1, I win.

Does this mean, you lost? {It is important}
Otherwise, I only have to consider, outcome of my turn, right?

- Anonymous December 19, 2010 | Flag Reply
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if there are two persons A and B,..
probability of A rolling first is 1/2..
probability of A winning is 1/2*1/6=1/12..
then probability of B wining is 1/2*5/6*1/6=5/72

- dinesh February 22, 2011 | Flag Reply
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0
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if there are two persons A and B,..
probability of A rolling first is 1/2..
probability of A winning is 1/2*1/6=1/12..
then probability of B wining is 1/2*5/6*1/6=5/72

- dinesh February 22, 2011 | Flag Reply
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0
of 0 vote

5/36 is the correct answer, Think conditional probability...

- Anon March 24, 2011 | Flag Reply
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0
of 0 vote

5/36 is the correct answer, Think conditional probability...

- Anon March 24, 2011 | Flag Reply
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of 0 vote

Correct me if I am wrong..

Chance of A starting first is 1/2
Chance of A rolling 1 is 1/6
Total Probability that A wins is 1/2 * 1/6 = 1/12

Probability that A not winning is 1-1/12 = 11/12

Now for B, A should not have won and B must roll a 6
So its 11/12 * 1/6= 11/72

- nagarajmailing May 21, 2011 | Flag Reply
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of 0 vote

P(first wins)=p. When the first players misses (5 out of 6 times) the second finds himself in the same position the other was just before. As a consequence, holds: 1 - p = (5/6)*p. Therefore, P(first wins)=6/11, P(second wins)=5/11.

- GodOfCode April 01, 2013 | Flag Reply


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