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A certain board game is played by rolling a pair of fair six
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05 Feb 2014, 10:44
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Difficulty:
65% (hard)
Question Stats:
55% (01:17) correct 45% (01:30) wrong based on 189 sessions
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A certain board game is played by rolling a pair of fair six-sided dice and then moving one's piece forward the number of spaces indicated by the sum showing on the dice. A player is "frozen" if her opponent's piece comes to rest in the space already occupied by her piece. If player A is about to roll and is currently six spaces behind player B, what is the probability that player B will be frozen after player A rolls?
In this game a player is "frozen" whenever another player lands on the spot where the first player already has her piece. Then the question asks for the probability that player B will be frozen by player A, who's 6 spaces behind. What does it depend on? It depends on whether player A gets a roll of exactly 6 on the dice. So what's the probability of getting 6 on the dice? First, find total possible outcomes for the roll. Since 6 numbers on each die, total number of possible rolls = 6 x 6 = 36. There are 36 possible rolls. How many of them would allow player A to move exactly 6 spots? Well, any roll that adds up to 6. So we could have 1/5, 2/4 and 3/3. But remember that since there are 2 dice, there are 2 ways to get 1/5 and 2/4 because either die could get the 1 or the 5, etc. So there's 1/5, 5/1, 2/4, 4/2 and 3/3. That's 5 rolls that add up to exactly 6. So 5 out of the 36 possible rolls would allow player A to "freeze" player B, which gives a probability of 5/36
Re: A certain board game is played by rolling a pair of fair six
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23 Jun 2014, 02:01
maggie27 wrote:
gandhigauravz wrote:
A certain board game is played by rolling a pair of fair six-sided dice and then moving one's piece forward the number of spaces indicated by the sum showing on the dice. A player is "frozen" if her opponent's piece comes to rest in the space already occupied by her piece. If player A is about to roll and is currently six spaces behind player B, what is the probability that player B will be frozen after player A rolls?
A. 1/12 B. 5/36 C. 1/6 D. 1/3 E. 17/36
no. of possible outcomes = 6*6 = 36
no. of outcomes that result a total of 6 (as A is 6 spcaes behind B) = 5 ( (1,5), (2,4), (3,3), (4,2), (5,1) )
So, the probability = 5/36 (Option B)
Hi , Can u please explain me that from where did the rolling of die B comes into picture.
There are no die B or A. There are players A and B. Player A is six spaces behind player B and the question asks about the probability that player A be on the same spot as B, after rolling a pair of fair six-sided dice. So, the question basically asks about the probability of getting 6 when rolling a pair of fair six-sided dice.
Re: A certain board game is played by rolling a pair of fair six
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24 Jun 2014, 10:33
gandhigauravz wrote:
no. of possible outcomes = 6*6 = 36
no. of outcomes that result a total of 6 (as A is 6 spcaes behind B) = 5 ( (1,5), (2,4), (3,3), (4,2), (5,1) )
So, the probability = 5/36 (Option B)
As the question says that we roll the pair of dice together and not one by one.... thus shouldn't be (1,5) and (5,1) be considered not ordered and as one outcome rather than two different outcomes.... (same for (2,4) and (4,2)
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Re: A certain board game is played by rolling a pair of fair six
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24 Jun 2014, 23:40
goodyear2013 wrote:
A certain board game is played by rolling a pair of fair six-sided dice and then moving one's piece forward the number of spaces indicated by the sum showing on the dice. A player is "frozen" if her opponent's piece comes to rest in the space already occupied by her piece. If player A is about to roll and is currently six spaces behind player B, what is the probability that player B will be frozen after player A rolls?
In this game a player is "frozen" whenever another player lands on the spot where the first player already has her piece. Then the question asks for the probability that player B will be frozen by player A, who's 6 spaces behind. What does it depend on? It depends on whether player A gets a roll of exactly 6 on the dice. So what's the probability of getting 6 on the dice? First, find total possible outcomes for the roll. Since 6 numbers on each die, total number of possible rolls = 6 x 6 = 36. There are 36 possible rolls. How many of them would allow player A to move exactly 6 spots? Well, any roll that adds up to 6. So we could have 1/5, 2/4 and 3/3. But remember that since there are 2 dice, there are 2 ways to get 1/5 and 2/4 because either die could get the 1 or the 5, etc. So there's 1/5, 5/1, 2/4, 4/2 and 3/3. That's 5 rolls that add up to exactly 6. So 5 out of the 36 possible rolls would allow player A to "freeze" player B, which gives a probability of 5/36
Here are a couple of posts you might find interesting. They discuss throwing three dice:
Re: A certain board game is played by rolling a pair of fair six
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26 Jun 2014, 08:32
@blackiscorpio : there isn't much difference between rolling 2 dice one after the other and rolling them simultaneously.. the outcome of one die is independent of the outcome of other... just like tossing two coins or tossing the same coin twice makes no difference, ( for example, probability of getting two heads is 1/2 * 1/2 , either way)
When we roll 2 dice the total number of outcomes is 36 and it includes both 2,4 and 4,2 ( and of course 5,1 and 1,5) .. these are separate independent outcomes.
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55% (01:17) correct 
