In craps, which of the following scenarios is least likely? (Remember, for each roll you're rolling two dice, so "rolling a 7" means any combination of dice that adds to 7.)

A. Roll an 8 on the first roll, and an 8 on the second roll to win the game.
B. Roll a 7 on the first roll to win the game.
C. Roll a 6 on the first roll, and a 6 on the second roll to win the game.
D. Roll a 4 on the first roll, and a 7 on the second roll to lose the game.
E. Roll a 2, 3, or 12 on the first roll to lose the game

Question

In craps, which of the following scenarios is least likely? (Remember, for each roll you're rolling two dice, so "rolling a 7" means any combination of dice that adds to 7.) 

A. Roll an 8 on the first roll, and an 8 on the second roll to win the game. 
B. Roll a 7 on the first roll to win the game. 
C. Roll a 6 on the first roll, and a 6 on the second roll to win the game. 
D. Roll a 4 on the first roll, and a 7 on the second roll to lose the game. 
E. Roll a 2, 3, or 12 on the first roll to lose the game

hockeychick23 · Answer

@jim_thompson5910 @ganeshie8

jim_thompson5910 · Answer

what probabilities have you computed so far?

hockeychick23 · Answer

well i got:
2: 1/36: 2.78%
3: 2/36: 5.56%

4: 3/36: 8.33%

5: 4/36: 11.10%

6: 5/36: 13.90%

7: 6/36: 16.70%

8: 5/36: 13.90%
9: 4/36: 11.10%
10: 3/36: 8.33%

11: 2/36: 5.56%

12: 1/36: 2.78%

jim_thompson5910 · Answer

good so far

jim_thompson5910 · Answer

let's look at "A. Roll an 8 on the first roll, and an 8 on the second roll to win the game. "

what is the probability here?

hockeychick23 · Answer

but I'm not sure what to do now

jim_thompson5910 · Answer

what's the probability of rolling a single 8?

hockeychick23 · Answer

5/36

jim_thompson5910 · Answer

yes

jim_thompson5910 · Answer

how about 2 eights?

hockeychick23 · Answer

25/1296

jim_thompson5910 · Answer

yes

jim_thompson5910 · Answer

25/1296 = 0.0192 = 1.92% (roughly)

jim_thompson5910 · Answer

"B. Roll a 7 on the first roll to win the game. "

that's roughly  6/36 = 16.70% and you already got that part

jim_thompson5910 · Answer

"C. Roll a 6 on the first roll, and a 6 on the second roll to win the game. "

this is found the same way done in part A

hockeychick23 · Answer

25/1296 again

jim_thompson5910 · Answer

yep which is roughly 1.92%

jim_thompson5910 · Answer

"D. Roll a 4 on the first roll, and a 7 on the second roll to lose the game."

same idea but with different numbers per roll

hockeychick23 · Answer

3/36*6/36= 18/1296

jim_thompson5910 · Answer

yep, 18/1296 = 0.01388 = 1.388% approx

jim_thompson5910 · Answer

"E. Roll a 2, 3, or 12 on the first roll to lose the game"

be careful about the keyword "or"

hockeychick23 · Answer

oh ok so what do i do different on this one?

jim_thompson5910 · Answer

you add the individual probabilities when it comes to "or"

this applies if the probabilities are mutually exclusive (they are in this case)

hockeychick23 · Answer

so 4/36 or 1/9

jim_thompson5910 · Answer

P(A or B) = P(A)+P(B) ... events A and B are mutually exclusive

P(A or B or C) = P(A)+P(B)+P(C) ... events A,B,C are mutually exclusive

jim_thompson5910 · Answer

yes, roughly 1.11%

jim_thompson5910 · Answer

now I'm going to list out the probabilities as percentages for A through E


A: 1.92%
B: 16.70%
C: 1.92%
D: 1.388%
E: 1.11%

hockeychick23 · Answer

so then the answer would be E right?

jim_thompson5910 · Answer

using percentages allows us to order the values

jim_thompson5910 · Answer

yes, 1.11% is the smallest of the 5

hockeychick23 · Answer

ok thanks!!

jim_thompson5910 · Answer

np