In the popular casino game of craps, the pass line bet wins when a

Question

In the popular casino game of craps, the pass line bet wins when a pair of 6-sided dice is rolled and the sum is a seven or eleven. The pass line bet loses when the sum is two, three or twelve. Any other number from two to twelve that are not one of the aforementioned numbers becomes "the point" and no money is won or lost on that roll.

From the table below, select the expression that represents the probability of the pass line bet losing on the first roll and the probability of neither winning nor losing on the first roll.

Shankey94 · Accepted Answer

For a person to win : Sum on dice = 7,11
For a person to loose: Sum on dice = 2,3,12

Probability of loosing: Events where condition for sum is satisfied/ total possible events
Possible case for sum to be 2 = (1,1)
Possible case for sum to be 3 = (1,2),(2,1)
Possible case for sum to be 12 = (6,6)
total favorable cases = 4
total possible cases when two dices are rolled = 6^2 = 36

Hence Probability = 4/36 = 1/9

Probability of neither or winning a point: 1- (probability winning + loosing)
Probability of winning: 
Case for sum to be 7 = (1,6),(6,1),(2,5),(5,2),(3,4),(4,3)
Case for sum to be 11 (will be equal to the number of cases for 3 by symmetry) = (5,6),(6,5)
Total cases for win or loose: 6+2+4 = 12
Probability for neither = 1-(8+4)/36 = 2/3

Ghostrider3147 · Answer

Losing:1/9
Neither: 2/3( for 1st roll we have to consider "one" also)

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sanjayparihar16 · Answer

In the popular casino game of craps, the pass line bet wins when a pair of 6-sided dice is rolled and the sum is a seven or eleven. The pass line bet loses when the sum is two, three or twelve. Any other number from two to twelve that are not one of the aforementioned numbers becomes "the point" and no money is won or lost on that roll.

Total outcomes = 36
(1,1), (1,2) ...... (1,6)
(2,1), (2,2) ...... (2,6)
(3,1)...
...
...
(6,1), (6,2) ...... (6,6)

P(Losing) - Getting sum of 2 OR 3 OR 12 - Possible outcomes - (1,1),(1,2),(2,1),(6,6)
 P(Losing)  = Favorable Outcome / Total Outcome = 4/36 =  1/9

P (Nothing) = 1 - P(Losing) - P(Winning)

P(Winning)
Sum should be 7 OR 11
Possible outcome - (2,5),(5,2),(3,4),(4,3),(1,6),(6,1),(6,5),(5,6)
P (Winning) = 8/36 = 2/9

P (Nothing)  = 1 - 1/9 - 2/9 =  2/3

wadhwakaran · Answer

Total No. of possible cases are 36 (6^2)
Winning Points when sum is 7 or 11
Possible cases for winning points are (1,6) (6,1); (2,5) (5,2); (3,4) (4,3); (5,6) (6,5) =8
Losing points when sum is 2 or 3 or 12
Possible cases for losing points are (1,1); (6,6); (2,1) (1,2) =4

Remaining draw cases = 36-12=24
 P(Losing)=4/36=1/9
P(Neither)=24/36=2/3

SAHUSHIBASISH · Answer

Ans: 
A. Losing: 1/9
B. Neither: 2/3
Solution :
Sum NoOfEvent
2- 1
3-2
4-3
5-4
6-5
7-6
8-5
9-4
10-3
11-2
12-1
Total possible-36
Losing probability = 1/36 + 2/36 +1/36 = 1/9
Winning probability = 6/36+2/36= 2/9
Neither = 1- winning probability-losing probability 
=1- 2/9 - 1/9 = 2/3

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pintukr · Answer

The IR test focusses on PROBABILITY OF THE EVENT..

Total possibilities for determining the Sum with roll of a pair of dice = 36 (6x6)

Given, Win = Sum (7 or 11)
 Loss = Sum (2 or 3 or 12)
 'Point' ie, neither Win nor Loss = Sum (remaining numbers in the Range  )

p(Win) = p(Sum (7 or 11)).............(1)
Possible combinations = (1,6), (6,1), (2,5), (5,2), (3,4), (4,3), (5,6), (6,5)
so, p(Win) = 8/36 = 2/9

p(Loss) = p(Sum (2 or 3 or 12)).............(2)
Possible combinations = (1,1), (1,2), (2,1), (6,6)
so, p(Loss) = 4/36 = 1/9

p(Point, ie, Neither Win nor Loss) = p(Sum (remaining numbers in the Range  )).............(3)
so, p(Point, ie, Neither Win nor Loss) = 1 - ( 2/9 + 1/9)
 = 1- 3/9
 = 1- 1/3
 = 2/3

Hence, the answers

p(Losing) = 1/9
p(Point, ie, Neither Win nor Loss) = 2/3

Archit3110 · Answer

total pairs of 2 die ; 36
winning pairs 
sum is 7 : ( 1,6) ( 2,5) ( 3,4) (4,3) ( 5,2) ( 6,1) ; 6
sum is 11 : ( 5,6) ( 6,5) 2

lossing pairs 
sum is 2 : (1,1) ;1
sum is 3: 1,2 ' 2,1 ;2
sum is 12 " 6,6 ' 1
lossing pairs 4
P ( 4/36 ) '  1/9

winning + lossing ; 8+4 ; 12 
36-12 ; 24 pairs of no loss or win
24/36 ;  2/3

Kushchokhani · Answer

When 2 dice are rolled, possibilities-
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

Sum of 7 possibilities:
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = total 6 possibilities

Sum of 11 possibilities:
(5,6), (6,5) = total 2 possibilities

Therefore, Total winning chances=8/36

Sum of 2 possibilities:
(1,1) = 1 possibility

Sum of 3 possibilities:
(1,2), (2,1) = total 2 possibilities

Sum of 12 possibilities:
(6,6) = 1 possibility

Therefore, Total losing chances=4/36=1/9 (1st column)

So, neither chances=36-12=24
=>24/36=2/3 (2nd column)

Mark 1st column in front of 1/9, and 2nd column in front of 2/3

UtkarshAnand · Answer

Given that Pass Line Loses (L) when sum of the numbers on two dices is 2, 3 or 12.
Outcomes for Losing (L) -  1+1  (for 2);  1+2, 2+1  (for 3);  6+6  (for 12).
Total outcomes for Losing (L) - 4
Total possible outcomes - 36 (6*6)
Therefore, Probability of Losing(L) - Total outcomes for Losing/Total possible outcomes = 4/36 or  1/9 .  Answer .

P(N) i.e. Probability of Neither = 1-   where P(L) = Probability of Losing and P(W) = Probability of Winning.
We have P(L) = 1/9
So calculating Probability of Winning(W):
Given that Wins (W) occur when sum of the numbers on two dices is 7 or 11.
Outcomes for winning (W) -  6+1, 1+6, 5+2, 2+5, 4+3, 3+4  (for 7);  6+5, 5+6  (for 11).
Total outcomes for winning (W) - 8
Total possible outcomes - 36 (6*6)
So, P (W) = Total outcomes for winning / Total possible outcomes = 8/36 or  2/9 .

Now Probability of Neither, P(N) = 1-   = 1 -   = 1 - 1/3 =  2/3 .  Answer .

av1901 · Answer

A pair of 6 sided dice are rolled
Total number of outcomes = 6*6 = 36

Winning outcomes: Sum = 7/11
Losing outcomes: Sum = 2/3/12
Neither outcomes: Sum = 4/5/6/8/9/10

Let us examine the first case:  Probability (Losing on the first roll)

Losing outcomes: Sum = 2/3/12

Sum = 2: Possible outcomes = (1,1)
Sum = 3: Possible outcomes = (1,2) (2,1)
Sum = 12: Possible outcomes = (6,6)

Probability = Number of possible outcomes / Total number of outcomes = 4/36 = 1/9

Let us examine the second case:  Probability (Neither winning nor losing on the first roll)

Neither outcomes: Sum = 4/5/6/8/9/10

Sum = 4: Possible outcomes = (1,3) (3,1) (2,2)
Sum = 5: Possible outcomes = (1,4) (4,1) (2,3) (3,2)
Sum = 6: Possible outcomes = (1,5) (5,1) (2,4) (4,2) (3,3)
Sum = 8: Possible outcomes = (2,6) (6,2) (3,5) (5,3) (4,4)
Sum = 9: Possible outcomes = (3,6) (6,3) (4,5) (5,4)
Sum = 10: Possible outcomes = (4,6) (6,4) (5,5)

Probability = Number of possible outcomes / Total number of outcomes = 24/36 = 2/3

Losing: 1/9 ; Neither: 2/3

gokul5544 · Answer

the probability of the pass line bet losing on the first roll is 2/9
and the probability of neither winning nor losing on the first roll is 2/3

Sajjad1994 · Answer

Hello Everyone!

Welcome to all of you.

OA to this question is:

Losing:  1/9
 Neither:  2/3

The same is posted in the question. Everyone did a good job.

Arm007 · Answer

Two dice are rolled here, so the sample space contains 36 e. 6*6 possible outcomes.

The outcomes which favor winning pass line bet are the ones which yield 7 or 11 when summed up. These are {{1,6},{2,5},{3,4},{4,3},{5,2},{6,1},{5,6},{6,5}}. So 8 outcomes out of 36 are in favor.

The outcomes which are favor losing pass line bet are the ones which yield 2, 3 ,or 12 when summed up. These are {{1,1},{1,2},{2,1},{6,6}}. 4 outcomes out of 36 are not in favor.

So probability of line bet losing on the first roll is: 4/36 -> 1/9

probability of neither winning nor losing on the first roll: 1-P(winning in first roll) - P(losing in first roll)
= 1-1/9-8/36
= 2/3

Sajjad1994 · Answer

Official Explanation

There are four sequences that would lose on the first roll: Both ones (to sum to two); a one on the first and a two on the second (to add to three); a two on the first and a one on the second (again, summing to three); and two sixes (to sum to 12). Therefore, because 4 out of the 36 total possibilities reduces to 1/9, that answer is B.

Answer: Losing: 1/9

To neither win nor lose, the four losing situations must be added to the winning situations. Those are

This totals 8 winning pairings, meaning that 12 pairings (8 win plus 4 lose) end the game on the first roll. The other 24, then, form the Neither category, and 24/36 reduces to 2/3 making D the correct answer.

Answer: Neither: 2/3

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JDLaw · Answer

To make the calculation easier,
Total possible outcomes= 6*6=36 (6 ways die1*6 ways die2)
winning possibilities = (1,6)(2,5)....=8 possibilities
Loosing possibility= (1,1)(1,2)(2,1)(6,6)= 4 possibilities

Therefore total winning Probability= Winning possibility/ total possibility=8/36=2/9
Total loosing probability= Loosing poss/total poss=4/36=1/9 ...(Ans 1)
Total probablity of not winning or loosing= 1- prob of winning-prob of loosing=1-2/9-1/9=6/9=2/3 ...(Ans 2)

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Losing	Neither
		\(\frac{1}{12}\)
		\(\frac{1}{9}\)
		\(\frac{3}{10}\)
		\(\frac{11}{18}\)
		\(\frac{2}{3}\)
		\(\frac{13}{18}\)

In the popular casino game of craps, the pass line bet wins when a

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Losing	Neither
		1/12
		1/9
		3/10
		11/18
		2/3
		13/18

Losing	Neither
		1/12
		1/9
		3/10
		11/18
		2/3
		13/18

Losing	Neither
		1/12
		1/9
		3/10
		11/18
		2/3
		13/18

GMAT Two-part Analysis (TPA) Questions

In the popular casino game of craps, the pass line bet wins when a

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My Rewards