概率問題~Maths' question of Probability~

(a) In fact, it is easier to find the probability of the event "No match at all" so that we subtract this probability from 1 to obtain the required probability for this question.
Now, for this game, we can model it as shown in the diagram below:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Crazyprob1.jpg

In the game, there are N boxes each labelled a number from 1 to N. However, the box numbers are not shown to the player. On the other hand, the player has N balls each labelled a number from 1 to N which are known to the player.
Now the game is as follows:

The player drops the ball numbered 1 into a randomly selected box (number still unknown to him), and the box is selected by him.
Then the box with ball number 1 dropped is removed.
There are N-1 boxes and N-1 balls remained.
Repeat steps 1 to 3 until each ball has been separately dropped into the boxes.
So, what is the probability that all balls are dropped into a wrong box, i.e. the numbers on the balls do not match with the corresponding number on the box they are dropped?
Now, see when N = 2, the possible sequences that the player will select include: (1,2) and (2,1) in which only (2,1) will make all balls dropped to the wrong box.
So when N = 2, this prob = 1/2
When N = 3, possible sequences are: (1,2,3), (1,3,2), (2,3,1), (2,1,3), (3,1,2) and (3,2,1) in which (2,3,1) and (3,1,2) will make all balls dropped to the wrong box.
So when N = 3, this prob = 1/3
Extending further, the probability of all balls dropped to the wrong box are as follows:
N = 4 → P = 0.375
N = 4 → P = 0.3666
N = 4 → P = 0.3680
.
.
N = n → P = 1/e (when n is appreciably large and e is the natural number = 2.71...)
So when N = 52, which can also be regarded as large value in this case, the probability that no balls dropped into the correct box is 1/e.
Therefore, the probability that no cards matched in this question is also 1/e.
(b) First of all, the probability of getting a seven and an eight when throwing two dice are 1/6 and 5/36 respectively.
Now, let's see what events will make "a seven occurs before an eight":
Event 1: It appears 7 in the first throw.
Event 2: It does not appear 7 nor 8 in the first throw and appears 7 in the second throw.
Event 3: It does not appear 7 nor 8 in the first two throws and appears 7 in the third throw.
.
.
.
Event N: It does not appear 7 nor 8 in the first N-1 throws and appears 7 in the Nth throw.
So since the above events are mutually exclusive, the required probability can be summarised as follows:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Crazyprob2.jpg

with P being the total probability of all events as N tends to infinity which is calculated from the formula of geometric series sum to infinite number of terms.

a. assume each of the deck has 52 cards (without jokers), and assume "match" means both the number and "flower"( i.e. spade, heart, diamond and clubs) are equal.

Cards are then drawn one at a time from each deck: the first card can be any card, and the second card can have 1/52 probablitiy to equal the first cards in both number and flowers.

If "match" means only the number is matched, the first card can be any card, and there are 4 cards from second deck can be equal to the first card in terms of number, and so the probability is 4/52 = 1/13.


b. There are 7 occasions for number seven: 1,6; 2,5; 3,4; 4,3; 5,2; 6,1
And there are 5 occasion for number eight: 2,6; 3,5; 4,4 ; 5,3; 6,2
Given the dices are rolled until 7 or 8, there are total 12 occasions for 7 or 8.
And 7/12 is the possibility that seven occurs before an eight.