24 point game Experience Series 6:
Look at the 24 o'clock in the eyes of probability
Author Ococo 2003/1/30 http://ococo.vip.myrice.com
24 o'clock game is used to use four cards to use the addition and subtraction and divide the first to get 24 to win (where A represents 1; J represents 11; Q represents 12; K represents 13; Dadi Wang represents 5).
When you do 24 games, how many times you need to know if you have a solution? For the sake of simplicity, only four different situations are discussed here.
First, look at the two counts of addition and subtraction and division, seeing there are several situations:
A BA * BA - BA / BB - AB / A
Remove (B A), and (B * a) equivalent (B * a) equivalent (B A), there are 6 situations.
From the four numbers, 4 * 3/2 = 6 may be known according to probability statistics.
Therefore, the first step is calculated with 4 * 3/2 * 6 = 36 possible.
Now, there is a calculated number, and the other two in 4 numbers, a total of 3 numbers. Two and select an operational way from it is removed. This operation has 3 * 2/2 * 6 = 18 possible (principle the same). The last two counts have 2 * 1/2 * 6 = 6 possible.
Therefore, in general, there are 36 * 18 * 6 = 3888 possible situations in each group. Each solution is one of these 3888 situations. Computer solve is also attempted one by one in this 3888 cases.
So how many of the groups is there. This is simple. When the four numbers are not the same, there are 13 * 12 * 11 * 10 = 17160 groups.
Ask all the affiliation of all the situation, that is, 17160 * 3888 = 66718080 cases, doing this for the current household computers, a piece of dish, instantly completed.
The following is a distribution of four numbers without repeating 1-13:
The number of 108 groups did not understand.
Other circumstances:
SUM (1-10) = 356 // Indicates a number of solutions in 356 groups in (1-10) between SUM (11-20) = 121 SUM (21-30) = 80 SUM (31-40) = 29 SUM (41-50) = 8 SUM (51-60) = 5 SUM (61-70) = 2 SUM (71-80) = 2 SUM (81-90) = 2 Sum (91-100) = 1 SUM (101-110) = 0 SUM (111-120) = 1
