Tough One - Probability

[shankar.ashwin]

A and B decide to play a game based on probability. They throw a fair dice alternately, whoever gets a 5 on his throw wins. What is the probability that A wins the game, if it is known that B starts the game.

A) 1/2
B) 1/6
C) 5/11
D) 2/3
E) 17/32

[Anurag@Gurome]

A and B decide to play a game based on probability. They throw a fair dice alternately, whoever gets a 5 on his throw wins. What is the probability that A wins the game, if it is known that B starts the game.

Note : In the following posts user123321 has provided the proper algebraic solution for this kind of problem. But for this p[articular problem we can use a walk-around method to guess the correct answer.

As the dice is a fair one, both of them will have almost equal chance of getting a 5 first. But as B starts the game, B will have slightly greater chance than A to get a 5 first.

Hence, probability that A wins given B has started the game will be slightly less than 1/2. Only option C works.

The correct answer is C.

[user123321]

since B started the game

the probability of A winning the game is...
1) if B looses first turn and A wins in second turn
5/6*1/6
2) if B looses,A looses,B looses and A wins in fourth turn
5/6*5/6*5/6*1/6
3).....

so adding probabilities of all individual possible events...
5/6*1/6 + 5/6*5/6*5/6*1/6 + ......
5/6*1/6(1+ (5/6)^2 + (5/6)^4 + ......) [infinite gp with common ratio less <1]
5/6*1/6*1/(1-25/36)
5/6*1/6*36/11
5/11

user123321

[qisma]

Confused....inclined to think that you should take the probability of B not winning (i.e. not throwing a 5, which is 5/6 chance when he throws the dice) and multiply by the probability of A winning (1/6 chance)....there could be a large sequence in which they continue to alternate plays until A finally throws a 5. But calculating all of these possible scenarios is not what we should do right?

Anurag, how did you arrive at the simple answer? Is the trick here not to consider the probability of throwing a 5, but the probability of winning?

[user123321]

we have infinite number of possibilities here...
1) B looses & A wins
2) B looses & A looses & B looses & A wins
.
.
infinity

The sum of these probabilities is what gives us the end result. And since the individual terms are nothing but terms of an infinite gp with mod(r)<1 we can use standard formula a/1-r.

HTH
user123321

[Whitney Garner]

Confused....inclined to think that you should take the probability of B not winning (i.e. not throwing a 5, which is 5/6 chance when he throws the dice) and multiply by the probability of A winning (1/6 chance)....there could be an large sequence in which they continue to alternate plays until A finally throws a 5. But calculating all of these possible scenarios is not what we should do right?

Anurag, how did you arrive at the simple answer? Is the trick here not to consider the probability of throwing a 5, but the probability of winning?

Hi shankar.ashwin & qisma!

A couple of notes on this problem:

(1) The GMAT would never ask you to compute a probability that relies on a Geometric Progression as this problem does.

BUT, if it did (which again, it WOULD NOT), one way we could guess this is to think about the chance of winning. We should expect that the person who goes first has a greater chance of winning (because if they roll a 5 then the game stops and the second person never gets a turn). But do we expect this advantage to be giant? Not really. So if we want the second player to win, we know that they have a disadvantage (eliminate A, D and E), but the disadvantage won't be as large as 1/6 chance of success (eliminate B).

(2) Another thing that tells me this problem isn't very GMAT-like is simply the presentation of the answer choices. When the GMAT gives answer choices that all look the same (all fractions, all whole numbers, etc) the convention is to list them in either increasing or decreasing order (as long as the task is not to find the largest or smallest). When answer choices are mixed in appearance (roots mixed with numbers mixed with fractions, etc), the order is a grab bag. We are not asked to find the largest or smallest here and the choices are all in fractions so they should be in order of size (in either direction) - but they are not.

(3) While it would be awesome to have a shortcut here, unfortunately user123321 hit the nail on the head with his assessment of the solution. And as he noted, you can use the standard form for the sum of a GP once you recognize the pattern, BUT this will NOT be simply a/(1-r) (where a is the initial value and r is the common ratio). Because we can simplify this serious to

(5/6)(1/6) * [ 1 + (5/6)^2 + (5/6)^4 + (5/6)^6......) [see user123321's post for this derivation]

we have a=(5/6)(1/6) and a common ratio r=5/6. BUT this geometric progression includes only the EVEN powers on r, so the Sum formula is actually:

a/(1-r^2)

[(5/6)(1/6)] / (1 - (5/6)^2)

(5/36) / (1-(25/36))

(5/36) / (11/36)

(5/36) x (36/11) = 5/11.

Hope this clears up a bit of the confusion!

Whit

Edit/Afterthought for my Point (3):
We could use the standard Geometric Progression Sum formula if we recognize that we can write the sequence as the following:

(5/6)(1/6) * [ 1 + (5/6)^2 + (5/6)^4 + (5/6)^6......)
(5/6)(1/6) * [ 1 + ((5/6)^2)^1 + ((5/6)^2)^2 + ((5/6)^2)^3......)

And therefore treat a=(5/6)(1/6) and the new r=(5/6)^2. But then, this is exactly where the formula

a/(1-r^2) comes from, but noticing the pattern allows us to forgo memorizing the sum formula with only even powers!!

[GmatMathPro]

Let B=the probability of B winning.

B=1/6+(5/6)(1-B)

solve for B=6/11

probability of A winning is 5/11

[Anurag@Gurome]

Sorry Guys!
My first analysis was wrong.
The solutions provided by user123321 and GmatMathPro are proper algebraic ones. I've edited the post to provide an walk-around strategy to solve the problem.

[shankar.ashwin]

I am not quite able to comprehend what GmatMathPro has done,could someone clarify?

1/6 is the Prob. B wins in the first chance,
I don't get what (5/6)(1-B) is. (1-B) would be Prob losing or no result.

PS. user123321 great soln.
And as Whit mentioned, though unlikely to see such problems in the GMAT, the topics tested are pretty much in the Scope of the GMAT IMO. This was a challenge problem in one of a coaching institute my friend attends. THought of sharing it here.

[Anurag@Gurome]

And as Whit mentioned, though unlikely to see such problems in the GMAT, the topics tested are pretty much in the Scope of the GMAT IMO.

Here is an approach so that no one needs to remember the "formula" for the sum of infinite geometric progression.

From user123321's solution,

• --> P = 5/36 + (5/36)*(5/6)^2 + (5/36)*(5/6)^4 + (5/36)*(5/6)^6 + ...
  --> P = 5/36 + [(5/6)^2][5/36 + (5/36)*(5/6)^2 + (5/36)*(5/6)^4 + ...]
  --> P = 5/36 + [(5/6)^2]*P
  --> P = 5/36 + (25/36)*P
  --> P - (25/36)*P = 5/36
  --> (11/36)*P = 5/36
  --> P = (5/36)*(36/11) = 5/11

In fact this is how you derive the formula for the sum of infinite geometric progression.

[GmatMathPro]

I am not quite able to comprehend what GmatMathPro has done,could someone clarify?

1/6 is the Prob. B wins in the first chance,
I don't get what (5/6)(1-B) is. (1-B) would be Prob losing or no result.

Notice that once B fails to get a 5 on his first roll, A is in the exact same position B was at the beginning of the game: taking his turn and staring down a potentially infinite sequence of alternating rolls. Hence, at this point, the probability of A winning is IDENTICAL to what B's probability of winning was when they started the game. In other words, the conditional probability of A losing given that B fails on his first rule is exactly the same as 1-B.

So, you are correct that 1-B is the probability of B losing at the beginning, but his probability of losing changes if he doesn't get it done on the first roll. If he doesn't, now he's in the same crappy position A was in in the first place: having to watch the other guy take his turn and hoping to hell that he doesn't get a 5. So on the (5/6)(1-B) part, think of the (1-B) part as representing "the probability of the guy who's rolling right now eventually losing".

Does that help?

[shankar.ashwin]

Hmm..still unsure here..
In other words, the conditional probability of A losing given that B fails on his first rule is exactly the same as 1-B.
But doesn't 1-B include the probability A wins as well?
As, B= B wins, and 1-B can be no result and probability of A winning?

Notice that once B fails to get a 5 on his first roll, A is in the exact same position B was at the beginning of the game: taking his turn and staring down a potentially infinite sequence of alternating rolls. Hence, at this point, the probability of A winning is IDENTICAL to what B's probability of winning was when they started the game. In other words, the conditional probability of A losing given that B fails on his first rule is exactly the same as 1-B.

So, you are correct that 1-B is the probability of B losing at the beginning, but his probability of losing changes if he doesn't get it done on the first roll. If he doesn't, now he's in the same crappy position A was in in the first place: having to watch the other guy take his turn and hoping to hell that he doesn't get a 5. So on the (5/6)(1-B) part, think of the (1-B) part as representing "the probability of the guy who's rolling right now eventually losing".

Does that help?

[GmatMathPro]

First, note that the probability of no result is zero, because we are assuming they are willing to play this game infinitely many times, so we don't have to consider that. But I realize this is not your main point of contention.

It may help to use a different variable, instead of A and B, because it is counterintuitive that 1-B should ever be A's chance of losing.

Let P=the probability of whoever's turn it is to roll ultimately winning. If we define P this way, 1-P must be the probability of the guy who has to wait to roll ultimately winning. Don't think of it as A or B, think of it as the guy going now, and the guy not going now.

The guy going now could either win on the roll coming up with probability of 1/6 OR that guy could not win on that roll with probability of 5/6 AND then win eventually somewhere down the line. But once we posit that he did not win on the first roll, he now becomes the guy waiting to roll. And the guy waiting to roll always has a win probability of 1-P. So, P=1/6+(5/6)(1-P).

Also, note that A and B's probabilities of winning are just alternating back and forth as 5/11 and 6/11 as they take turns rolling.

Roll 1:
A=5/11 B=6/11

Roll 2(if necessary):
A=6/11 B=5/11 because they've essentially just switched places

Roll 3(if necessary):
A=5/11 B=6/11

Roll 4(if necessary):
A=6/11 B=5/11

Obviously the past rolls have no effect on future rolls, so every new roll it's like they're starting all over again, with the guy going first having the slight advantage.

So, B will win either on the first roll, with probability of 1/6 or they get to the second roll with probability 5/6, but if they get to the second roll, B's probability of winning has switched to 5/11, because now he has to wait for A to go. So B wins with probability of 1/6+(5/6)(5/11), which equals 6/11. But 5/11 is just one minus his probability of winning on the first roll, or 1-B. Of course we wouldn't have these values available if we were solving this problem from scratch, but it does clarify (i hope) why it makes sense to use 1-B for B's probability of winning eventually if he doesn't win immediately.

[shankar.ashwin]

AH! Get it know. Thanks a lot for the elaborate explanation. Appreciate it.
Thanks again