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I have created a "luck" game in the form of a multiple-choice quiz with 8 questions.

Each question has 4 possible nonsense answers. The points awarded for each "correct" answer vary between 1-3.

5/8 questions have more than 1 "correct" answer, although only one answer can be selected per question.

The possible points earned from each question are as follows:

A. 3/1/0/0 (where one answer is worth 3 points, one is worth 1 point, and two are worthless.)

B. 3/1/0/0

C. 2/1/0/0

D. 2/1/0/0

E. 2/1/0/0

F. 3/0/0/0

G. 3/0/0/0

H. 2/0/0/0

The highest achievable score is 20.

What is the probability of scoring 15 or higher, if answers are chosen at random?

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  • $\begingroup$ You are asking for $\Pr(X_A+\cdots+X_H\geq15)$ where the mentioned random variables are independent, and their distributions are obvious. It can be found by brute force. $\endgroup$ – drhab Apr 14 '17 at 7:15
  • $\begingroup$ "where one answer is worth 3 points, one is worth 1 point, and two are worthless" - this confuses me, could you clear this up a little bit? Also, you'll have to specify exactly what question has how many answers and how many of those are correct $\endgroup$ – vrugtehagel Apr 14 '17 at 7:15
  • $\begingroup$ The part about "nonsense" answers is somewhat confusing and probably should be eliminated from the question. The thought being, if they are "nonsense" answers, why are you awarding points for the "correct" ones. That seems like a contradiction of terms. $\endgroup$ – David Apr 14 '17 at 12:45
  • $\begingroup$ You should also make it so not all of the highest points for each question are the first answer. Reason being what if someone "randomly" chose all the first answers to the questions? They would get a "perfect" score! $\endgroup$ – David Apr 14 '17 at 13:44
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Since the score must be more than 14, we have to lose 5 points or less. As you can see for example in question A, we can either loose 2 marks, or 3 marks.(If we choose the option with 1 point, we have lost 2 points because the maximum mark possible is 3). So:

A: -2 -3

B: -2 -3

C: -1 -2

D: -1 -2

E: -1 -2

F: -3

G: -3

H: -3

We have to lose either 5 marks, 4 marks, 3 marks, 2 marks, 1 mark or loose no mark at all.

5 = 2 + 3 = 1 + 2 + 2 = 1 + 1 + 3 = 1 + 1 + 1 + 2

4 = 1 + 1 + 2 = 2 + 2 = 1 + 3

3 = 1 + 1 + 1 = 1 + 2 = 3

2 = 1 + 1 = 2

1 = 1

0 = 0

Now the question can be easily solved. For example, if we want to loose exactly 5 marks, we have to loose 1 question with 2 marks and 1 question with 3 marks, or two questions with 2 marks and one question with one mark, or 2 questions with 1 mark and 1 question with 3 marks. In the former state, We have 6 questions with the possibility of loosing two marks and 3 with the possibility of loosing 3 marks.(Of course we have to notice that there are questions with both of the possibilities.) So the overall number of states which we loose 5 marks in them is:

5: ($\binom{3}{1}\binom{2}{1} + \binom{2}{1}\binom{1}{1}) + (\binom{3}{1}\binom{5}{1}) + (\binom{3}{2}\binom{3}{1}) + (\binom{3}{3}\binom{3}{1})$

We will do the same for the other 4 numbers. Then add up these numbers.

Now we have to divide this number by the number of all states which is obviously $4^8$.

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  • $\begingroup$ Score must be 15 or higher, not strictly higher than 15. @Soroush khoubyarain $\endgroup$ – David Apr 14 '17 at 13:23
  • $\begingroup$ I have calculated scores above 14. $\endgroup$ – Soroush khoubyarian Apr 14 '17 at 17:29
  • $\begingroup$ Your comment states scores "more that 15". @Soroush khoubyarian $\endgroup$ – David Apr 15 '17 at 12:52
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Using computer simulation to simulate all possible combinations of answers to all 8 questions (there are 65,536 total possible outcomes), I am seeing 390 of them that total up to 15 or more points. So the probability is $390/65536 = 0.00595$ which is about $6/10th$ of $1$%. Not very good chances.

Just F.Y.I., I am including here the # of combinations for each possible score from 0 to 20:

$~~0 : ~~864$
$~~1 : 2160$
$~~2 : 3744$
$~~3 : 5832$
$~~4 : 7182$
$~~5 : 8247$
$~~6 : 8397$
$~~7 : 7668$
$~~8 : 6662$
$~~9 : 5132$
$10 : 3776$
$11 : 2525$
$12 : 1559$
$13 : ~~919$
$14 : ~~469$
$15 : ~~234$
$16 : ~~100$
$17 : ~~~~38$
$18 : ~~~~14$
$19 : ~~~~~~3$
$20 : ~~~~~~1$

The assumptions here are that all 8 questions must be answered but only 1 answer for each question can be chosen.

The simulation simply uses 8 nested loops which each go from 1 to 4 and those index 8 arrays which hold the points for each question-answer combo. The program took me like 5 minutes to write and runs in about 1/6th of a second.

Notice that the most likely score would be 6 (out of 20) and also notice that no 2 different scores are equally likely (although 2 and 10 are very close).

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