GMAT Problem on Dice

Perhaps the hardest of the bunch, this question requires that we find a solution that doesn’t take much longer than 2 minutes. A quick way – or at least relatively quick way – is to determine the number of instances in which our roll of one die will yield more than two die. Intuitively, this is a good place to start because the number of instances in which rolling a great number with one die than with two is much smaller than vice versa.

The only way to roll higher on one die is if the magicians rolls between 2 and 5, inclusive, with two dice. Were he to roll a six with two dice than there is no way he could eclipse that number by rolling one die.

Below is the probability of rolling a certain number with two dice.

‘2’ – 1/36

‘3’ – 2/36

‘4’ – 3/36

‘5’- 4/36

Now the only numbers a magician can roll with the one die and win is between 3 and 6, inclusive. The chances of rolling any are always 1/6.

Next, we have to combine the probability distribution relating to two dice with that relating to the one.

The chances of a magician rolling any given number with one die are 1/6. So let’s start with the lowest number he can roll: a ‘3.’ To win with this roll, he will have to roll a ‘2’ with two dice, the odds of which are 1/36. So {1/6} * {1/36} = 1/216.

So the chances of him rolling a ‘3’ with one die and winning are 1/216.

Let’s repeat this logic for the next roll, ‘4.’ Chances of rolling are 1/6. Only way a ‘4’ wins is if he rolls a ‘2’ or a ‘3’ with two dice.

Odds of rolling a ‘2’ with two dice + odds of rolling a ‘3’ with two dice = {1/36} + {2/36} = {3/36}. Combine this with the odds of rolling a ‘4’ ( which is 1/6): {3/216}.

If he rolls a ‘5’ with one die, he can win if he rolls a ‘4’ with two dice, the probability is 3/36. He can also win if he rolls a ‘2’ or a ‘3’ with two dice, the number outcomes we just found: 3/36. So we add {3/36} + {3/36} = {6/36} * {1/6} (odds of rolling a 5 with one die) = {6/216}

Next, if he rolls a ‘6’ on one die, he can beat ‘2’ through ‘5’ with the two dice. Number of ways to roll a ‘5’ = 4/36. Combining this with the odds of rolling a ‘6’ on one die with the odds or rolling ‘2’, ‘3’, ‘4’, or ‘5’ with the two dice we get: 10/216.

You may be wondering why I left the denominator as 216. Well, this allows us to add up all the instances he can possibly win:

1/216 + 3/216 + 6/216 + 10/216 = 20/216 = 5/54 (Answer B)

The least sum of the numbers of the two dice in the right hand is 2. The maximum sum of the

numbers on these two dice can be 5, for the left hand dice to be able to have a greater number.

Therefore the number on the dice in the left hand must be 3 or more.

Number of ways to get a sum of 2 with the two dice in the right hand= 1

Number of ways to get a sum of 3 with the two dice in the right hand= 2

Number of ways to get a sum of 4 with the two dice in the right hand= 3

Number of ways to get a sum of 5 with the two dice in the right hand= 4

Therefore required probability = P(two right dice have a sum of 2)*P(left dice has 3) + P(right dice

have sum of 2 or 3) P(left dice has 4) + P(two right dice have a sum of 2,3, or 4) P(left dice has 5) +

P(two right dice have a sum of 2,3,4,or 5)*P(left dice has 6)

= (1/36) (1/6) + (3/36) (1/6) + (6/36) (1/6) + (10/36) (1/6)

= (1/216) * (1+3+6+10)

= 20/216

= 5/54

Option B

For this question I drew a 6x6 grid showing all the possible sums that you can get with two dice. If the number on one die must be greater than the sum the number on the other two, it can not be 1 or 2. (Henceforth I will be writing the probability of each event in brackets with the event) If we get 3 (1/6) it is greater than a sum of 2 which occurs only once in the table: 1+1 (1/36). Similarly if we get 4 (1/6) it is greater than 2 and 3 which occur a total of three times 1+2, 2+1, 1+1 (3/36). For 5 (1/6) 2, 3, and 4 are present in six places in the table (6/36) and for 6 (1/6) 2, 3, 4 and 5 appear 10 times (10/36).

Therefore the total probability will be (1/6 * 1/36) + (1/6 * 3/36) + (1/6 * 6/36) + (1/6 * 10/36) = 5/54

Total no. of observations=6^3 as there are 3 dice and in each dice no. of possible outcomes=6 Now suppose the no. in his left hand comes 1, then 1 cannot be greater than the sum of no.'s in his right hand. Same is with 2 because in the right hand the minimum no.'s can come is 1 and 1 and their addition=2 which is not greater than 2. If 3 comes in left hand, then 1,1's addition is smaller than 3 so when 3 comes, 1 favourable outcome is there. If 4 comes in left hand, (1,1),(1,2)(2,1) are the favourable outcomes i.e. 3 favourable outcomes. If 5 comes in left hand then (1,1),(1,2),(2,1),(3,1),(1,3),(2,2) i.e. 6 favourable outcomes. If 6 comes then (1,1),(1,2)(2,1),(3,1),(1,3),(4,1),(1,4),(3,2),(2,2),(2,3) i.e. 10 favourable outcomes. So total no. of favourable outcomes=1+3+6+10=20 Total no. of possible outcomes=6^3=216 Probability=20/216=5/54

favourable no of cases are 20(for left die value 3-1 case,for left die value 4-3 cases,for left die value 5-6 cases,for left die value 6-10 cases)...where as total no of cases are 256...so the solution will be 20/256=5/54

(1/6)(10/36 +6/36 +3/36 +1/36)=5/54

Total number of combinations = 216. the left hand must be at least 3 to make it greater than the least sum at the right. LEFT HAND: 3 there is only 1 sum less than it (1+1), 4 is 3 sums, 5 is 6 sums, 6 is 10 sums. (1+3+6+10)/216 = 5/54

P = 1/36x4/6+2/36x3/6+3/36x2/6+4/36x1/6

Ok, i get it right using probability, but when i tried different idea i get something wrong i tried to use something like truth table as another method and i calculated it as follow we have a truth table from all one to all six 1 1 1 1 1 2 . . 6 6 6 with total space equal 6^3=108 so how many times the left one will be greater will be for 3 11 for 4 11 12 21 for 5 11 12 21 13 31 22 for 6 11 12 21 13 31 22 41 14 23 32 so the total number of case will be 1+3+6+10=20 so the solution will be 20/108 something is missing here any clue :)

This requires just some presence of mind. when 3 die are thrown, number of possibilities-108 now on one hand, its either 1 2 3 4 5 6 1 cant be larger than any number making probability 10 only with the 108 outcomes(5 with each dice) answer= 10/108 = 5/54