Simone
BAN USER@skumaringuva (for some reason the Reply button doesn't work...)
No, the "sums" are not equally distributed from 0 to 220 since the probability of having very low or very high numbers is extremely low.
Is like for instance when you roll 2 dices with 6 faces: the probability of having 12 is 1/36, while the probability of having 7 is 1/6, because 7 can be obtained in 6 different ways (1+6, 2+5, ...), while 12 only with 6+6.
I'm not counting the only the cards that have a value multiple of 7.
Since there is clearly not enough information in the problem (it never says what type of cards they are, neither what numbers are present on their faces), I assumed the problem required to count the number of cards face-up, and that's what my solution is doing.
Of course my interpretation could be wrong.
PLEASE NOTE:
>>>>
I wrote the following answer when the question was different since the guy who created this post
has recently edited the question.
In the original question, there was no information on the type of cards present in the deck, and it was very unclear whether the candidate was supposed to add the value of each card facing up or to count the total number of cards facing up.
Since the problem was very unclear, in this solution I'm counting the probability of the total number of cards with the face up is a multiple of 7.
Since now the poster has changed the question, the assumptions made for this answer are no longer valid. I'm still leaving this answer because it's correct if you want to count the number of cards with the face up.
>>>>
Assuming the probability of face-up is p=0.5.
Let āNā be the total number of cards, in this case 50.
I will use the "^" as the power operator, not as the bitwise XOR.
The fact that the cards are thrown at the same time is not relevant, this problem is like flipping a coin 50 times and counting how many times a certain side was up.
The probability of having K cards face-up is the probability of having the first K cards face-up (p^K) and the remaining N-K cards face down ((1-p)^(N-K) = p^(N-K) becase p=0.5) times the number of unique permutations (N! / (K! * (N-K)!)).
Another way to think of all the possible unique permutations is the following. Imagine you have 50 coins and you want to find the number of subsets having 7 coins (without ordering): the way to compute that number is by using the binomial coefficient (wikipedia / Binomial_coefficient), which is exactly N! / (K! * (N-K)!), or, with 50 and 7 ==> 50! / (7! * 43!) = 99884400.
Now we just need to sum the probability of every different K which is a multiple of 7, that is 0, 7, 14, 21, 28, 35, 42, 49.
P = (1 + 99884400 + 937845656300 + 67327446062800 + 88749815264600 + 2250829575120 + 536878650 + 50) / 2^50
= 159266573321921 / 2^50
= 0.141457
Lol in python it's 4 lines :D
- Simone June 04, 2016