While playing a certain dice game, Chris wins if the sum of

Question

While playing a certain dice game, Chris wins if the sum of the two dice is 7, at which point the game is over. If the game allows Chris three rolls in an attempt to win, what is the probability that Chris will win?

A. 1/2
B. 17/36
C. 91/216
D. 11/36
E. 25/216

I could not understand how to approach this problem.
Can someone throw some light on how to solve this.

regan1sp · Accepted Answer

Total outcomes possible: 36
Total outcomes possible with sum 7: 6
Probability to win when rolled once = 6/36
Probability not to win when rolled once = 30/36 = 5/6
Probability to win in three attempts= 1- Probability will not to win in all three attempts
 = 1- (5/6* 5/6*5/6) 
 = 91/216

wunsun · Answer

There are 36 ways that a pair of dice can land. To have a sum of 7, there are 6 ways: 1-6, 2-5, 3-4, 4-3, 5-2 and 6-1

Therefore, the chance to have a sum of 7 is  \frac{1}{6} .

Making a sum of 7 on the first time is  \frac{1}{6}=\frac{36}{216} 
Making a sum of 7 on the second time is  \frac{5}{6}*\frac{1}{6}=\frac{5}{36}=\frac{30}{216} 
Making a sum of 7 on the third time is  \frac{5}{6}*\frac{5}{6}*\frac{1}{6}=\frac{25}{216}

Add them all together to get  \frac{91}{216}

Answer is C.

plaverbach · Answer

I understand the solution. But I have a meaning problem:
 How many dices could be rolled (max) ? Three pairs (6) or three dices (3)

I understand that he rolls the 
First dice (from 1 to 6). No thing I can do.
Second dice (1 to 6) = 1/6 of chance of getting a 7.
If the sum is below 7:
Third dice (1 to 6). I dont know how to calculate! =)

My final question is: is this problem ambiguous?

marco2 · Answer

I think question sould be read as follows: Charlie has three rolls with two dices each of them.
Hence every roll minimum score would be 2 (1+1) and maximum 12 (6+6).
Probability of having 7 in the first roll is 1/6 (6 out of 36).
You can follow one of the two ways explained here above. Both works.
Hope it helps.

ramalo · Answer

Why dont be consider a 4th option where on none of the 3 rolls he gets a 7? shouldnt we be adding   ^3 to the 91/216?

rajendra12 · Answer

I am not able to understand above approach. (please elaborate it)

however I have got answer using inclusion exclusion principle.....
Considering 3 rolls of 2 dice
A= event that I get sum equal to 7 in first roll
B= event that I get sum equal to 7 in 2nd roll
C= event that I get sum equal to 7 in 3rd roll

in every roll we have six ways to get sum equal to 7 (i.e. 1,6 or 6,1 and so on)

|A∪B∪C| = |A| + |B| + |C| - |A∩B| - |B∩C| - |A∩C| + |A∩B∩C|
 =  1/6+1/6+1/6-1/6*1/6-1/6*1/6-1/6*1/6+1/6*1/6*1/6 
 =  91/216

IS this way correct?

GMATinsight · Answer

Probability of winning game = 1- Probability of losing game

Probability of losing game = (Probability of not getting sum 7 in any of the three attempts)

Ways of getting sum 7 = (1,6)(2,5)(3,4)(4,3)(5,2)(6,1)= 6 ways
Total ways of getting outcome on two dice =6*6=36
Probability of getting sum as 7 in any attempt =6/36=1/6
Probability of NOT getting sum as 7 in any attempt = 1-(1/6)= 5/6

Probability of losing game =(5/6)*(5/6)*(5/6)=125/216
I.e. Probability of wining game = 1-(125/216) = 91/216

Answer : Option C

mvictor · Answer

oh man..I understood that he has to throw 3 dice..not 3 times....
so i took..
probability that it won't work..
1*5/6 x 5/6 = 25/36
and thus, 11/36 is probability that out of 3 dice, he will get 7..

Mathivanan Palraj · Answer

There are 6 possibilities for the sum to be 7. (1,6) (2,5) (3.4) and the three reverse order pairs.
The probability of a win is 6 over 36 or 1/6 and the probability of a loss is 5/6
Now, there are three possible cases: W or LW or LLW
so, 1/6 + 5/6 * 1/6 + 5/6 * 5/6 * 1/6 = 91/216

mrinali · Answer

The sum of two dice can be 7 in six ways - (1,6) , (2,5), (3,4), (4,3) , (5,2) and (6,1)
Total combinations of the numbers on throwing two dice = 36
Thus probability of getting a sum of 3 = 6/36 = 1/6
Probability of not getting a sum of 7 = 1-1/6 = 5/6

Now Chris has 3 attempts to get the desired sum. If he gets sum 7 in 1st attempt than the result is achieved and the other two attempts are not required -> P1= 1/6

If Chris does not gets sum 7 in 1st attempt, then he needs another attempt and if he gets six then prob -> P2= 5/6*1/6 (P(not getting sum 7 in 1st attempt)* P(get sum 7 in 2nd attempt))

Similarly, a 3rd attempt would be needed if Chris fails to gets sum 7 in previous attempts -> P3 = 5/6*5/6*1/6

Total prob = P1 +P2+ P3 = 1/6 + 5/6*1/6 * 5/6*5/6*1/6 = (36+30+25)*6^3 = 91/126

Bambi2021 · Answer

1 - P(He doesnt win in 3 attempts) = P(at least one = he wins)

1 - (30/36)^3

(30/36)^3 = 5^3*6^3 / 6^3*6^3 = 5^3 / 6^3 = 125/216

1 - 125/216 = 91/216

Kinshook · Answer

While playing a certain dice game, Chris wins if the sum of the two dice is 7, at which point the game is over.

If the game allows Chris three rolls in an attempt to win, what is the probability that Chris will win?

Cases when the sum of 2 dices is 7= {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} : 6 cases
Total cases when 2 dices are rolled = 6*6 = 36 cases

The probability that Chris wins when 2 dices are rolled = 6/36 = 1/6 
The probability that Chris does not win when 2 dices are rolled = 1 - 1/6 = 5/6

The probability that Chris does not win when three rolls are allowed = (5/6)^3 = 125/216
The probability that Chris wins when three rolls are allowed = 1 - (5/6)^3 = 1 - 125/216 = 91/216

IMO C

Adit_ · Answer

It wasn't clear to me whether three rolls meant rolling 3 individual die and leading upto a value of 7 and hence was pondering for quite a while thinking what went wrong. I think the question can be better framed that each roll indeed comprised of 3 rolls each of two dice. Or I just plain misinterpreted. But this case is much simpler:
It is 1-Prob(not winning in any roll).
Getting a sum of 7 in a pair of dice = 1/6. Thus losing = 5/6
Losing all 3 turns = 5/6^3 = 125/216
Now 1-125/216 = 91/216.

Answer:  Option C

____________________________

Bunuel  is there a question where individually 3 dice are thrown similar to this question and where a sum of 7 in the first occurence would be considered as a win?
Will the answer be 1/6+2/9+1/36 = 5/12 for such a question

While playing a certain dice game, Chris wins if the sum of

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