如何解决3个骰子问题？

概率也称为可能性。这意味着机会的数学，即可能发生的事件的交易。该值被指定为从零到一。在数学中，概率已被证明可以猜测事件发生的可能性。基本上，概率是预期某事发生的范围。

可能性

为了更正确地理解概率，以掷骰子为例，可能的结果是 1、2、3、4、5 和 6。出现任何有利结果的可能性是 1/6。由于发生任何事件的概率是相同的，因此得到任何可能数字的概率相似，在这种情况下，它是 1/6 或 50/3。

概率公式

Probability of an event = {Number of favourable events } ⁄ {number of total events}

P(A) = {Number of ways A occurs} ⁄ {Total number of events}

骰子

骰子是一个小方块，边缘有 1 到 6 个标记或点，在游戏中用于给出随机数字。骰子是小的、可投掷的块，带有标记的边，可以在几个数字中暂停。它们被用来产生随机数字，通常作为备牌游戏以及骰子游戏、棋盘游戏、角色扮演游戏和机会游戏的一部分。

一个惯用的骰子是一个立方体，它的六个面中的每一个面都带有从一到六个不同数量的图形。当可投掷或滚动时，骰子会暂停，在其较高的一侧出现一个从 1 到 6 的随机数，每个事件的发生概率相同。骰子也可能有凹形或不均匀的形状，并且可能有用数字或字符而不是点子标记的面。加载的骰子被抽出来以服务于某些结果而不是其他结果，以用于逃避或娱乐。

如何解决3个骰子问题？

解决方案：

Possibility for throwing six sided three dice will be 1, 2, 3, 4, 5 and 6 dots in each (three) dies.

Three dice are roll at the same time, number of attainable outcome can be 6³ = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its dots.

Now, let’s consider the possible sums from rolling three dice. The smallest attainable sum occur when all of the dice are the smallest, or one each. This gives a sum of three when we are throwing three dice i.e., (1, 1, 1). The greatest number on a die is six, which means that the greatest possible sum happens when all three dice are sixes is 18 i.e., (6, 6, 6). When n dice are rolled, the minimal attainable sum is n and the greatest attainable sum is 6n. There is only one way when three dice can total 3,

• 3 ways for 4
• 6 for 5
• 10 for 6
• 15 for 7
• 21 for 8
• 25 for 9
• 27 for 10
• 27 for 11
• 25 for 12
• 21 for 13
• 15 for 14
• 10 for 15
• 6 for 16
• 3 for 17
• 1 for 18

Forming Sums

As shown above, for three dice the possible sums include every number from three to 18. Calculate probability by using add up plan and acknowledging that we are considering ways to separate a integer into absolutely three whole numbers. For example, to obtain a sum of three there is only one way i.e., 3 = 1 + 1 + 1. Since each die is individualistic from the others, a sum such as four can be acquired in three different ways:

• 1 + 1 + 2
• 1 + 2 + 1
• 2 + 1 + 1

Forward add up arguments can be used to find the number of ways of creating the other sums. The partitions for each sum follow:

• 3 = 1 + 1 + 1
• 4 = 1 + 1 + 2
• 5 = 1 + 1 + 3 = 2 + 2 + 1
• 6 = 1 + 1 + 4 = 1 + 2 + 3 = 2 + 2 + 2
• 7 = 1 + 1 + 5 = 2 + 2 + 3 = 3 + 3 + 1 = 1 + 2 + 4
• 8 = 1 + 1 + 6 = 2 + 3 + 3 = 4 + 3 + 1 = 1 + 2 + 5 = 2 + 2 + 4
• 9 = 6 + 2 + 1 = 4 + 3 + 2 = 3 + 3 + 3 = 2 + 2 + 5 = 1 + 3 + 5 = 1 + 4 + 4
• 10 = 6 + 3 + 1 = 6 + 2 + 2 = 5 + 3 + 2 = 4 + 4 + 2 = 4 + 3 + 3 = 1 + 4 + 5
• 11 = 6 + 4 + 1 = 1 + 5 + 5 = 5 + 4 + 2 = 3 + 3 + 5 = 4 + 3 + 4 = 6 + 3 + 2
• 12 = 6 + 5 + 1 = 4 + 3 + 5 = 4 + 4 + 4 = 5 + 2 + 5 = 6 + 4 + 2 = 6 + 3 + 3
• 13 = 6 + 6 + 1 = 5 + 4 + 4 = 3 + 4 + 6 = 6 + 5 + 2 = 5 + 5 + 3
• 14 = 6 + 6 + 2 = 5 + 5 + 4 = 4 + 4 + 6 = 6 + 5 + 3
• 15 = 6 + 6 + 3 = 6 + 5 + 4 = 5 + 5 + 5
• 16 = 6 + 6 + 4 = 5 + 5 + 6
• 17 = 6 + 6 + 5
• 18 = 6 + 6 + 6

Specific Probabilities

Probability of an event = {Number of favourable events } ⁄ {number of total events},or 216. The results are:

• Possibility of getting a sum of 3: 1/216 = 0.0046 × 100= 0.5%
• Possibility of getting a sum of 4: 3/216 = 0.0138 × 100 = 1.4%
• Possibility of getting a sum of 5: 6/216 = 0.0277 × 100 = 2.8%
• Possibility of getting a sum of 6: 10/216 = 0.0462 × 100 = 4.6%
• Possibility of getting a sum of 7: 15/216 = 0.069 × 100 = 7.0%
• Possibility of getting a sum of 8: 21/216 = 0.097 × 100 = 9.7%
• Possibility of getting a sum of 9: 25/216 = 0.115 × 100 = 11.6%
• Possibility of getting a sum of 10: 27/216 = 0.125 × 100 = 12.5%
• Possibility of getting a sum of 11: 27/216 = 0.125 × 100 = 12.5%
• Possibility of getting a sum of 12: 25/216 = 0.115 × 100 = 11.6%
• Possibility of getting a sum of 13: 21/216 = 0.097 × 100 = 9.7%
• Possibility of getting a sum of 14: 15/216 =0.069 × 100 = 7.0%
• Possibility of getting a sum of 15: 10/216 = 0.0462 × 100 = 4.6%
• Possibility of getting a sum of 16: 6/216 = 0.0277 × 100 = 2.8%
• Possibility of getting a sum of 17: 3/216 = 0.013 × 100 = 1.4%
• Possibility of getting a sum of 18: 1/216 = 0.0046 × 100 = 0.5%

As can be seen, the extreme values of 3 and 18 are least probable. The sums which are in the middle are most probable.

示例问题

问题1：三个骰子一起扔。求总数为 5 的概率。

解决方案：

Three dice are roll at the same time, number of attainable outcome can be 6³ = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its dots.

Number of favourable events of getting a total of 5 = 6

i.e., (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (2, 1, 2) and (1, 2, 2)

So, probability of occurring a total of 5

P(A) = {Number of favourable events } ⁄ {number of total events}

= 6/216

= 1/36

问题2：三个骰子一起扔。求总数接近 5 的概率。

解决方案：

Three dice are roll at the same time, number of attainable outcome can be 6³ = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its dots.

Number of affair of occurring a total of atmost 5 = 10

i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1) and (1, 2, 2).

Therefore, possibility of happening a total of atmost 5

P(E) = {Number of favourable events } ⁄ {number of total events}  

= 10/216

= 5/108

问题3：三个骰子放在一起。找出总数至少为 5 的概率。

解决方案：

Three dice are roll at the same time, number of attainable outcome can be 6³ = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its dots.

Number of affair of happening a total of less than 5 = 4

i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1) and (2, 1, 1).

Therefore, possibility of happening a total of less than 5

P(E) = {Number of favourable events } ⁄ {number of total events}  

= 4/216

= 1/54

Therefore, possibility of occurring a total of at least 5 = 1 – P (occurring a total of less than 5)

= 1 – 1/54

= (54 – 1)/54

= 53/54

问题4：三个骰子放在一起。求总数为 6 的概率。

解决方案：

Three dice are roll at the same time, number of attainable outcome can be 6³ = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its dots.

Number of affair of happening a total of 6 = 10

i.e. (1, 1, 4), (1, 4, 1), (4, 1, 1), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1) and (2, 2, 2).

Therefore, possibility of happening a total of 6

P(E) = {Number of favourable events } ⁄ {number of total events}

= 10/216

= 5/108

问题5：三个骰子放在一起。求总数为 8 的概率。

解决方案：

Three dice are roll at the same time, number of attainable outcome can be 6³ = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its dots.

Number of affair of happening a total of 8 = 21

i.e. (2, 2, 4), (4, 2, 2), (2, 4, 2), (1, 5, 2), (2, 5, 1), (5, 1, 2), (2, 1, 5), (1, 2, 5), (3, 3, 2), (3, 2, 3), (2, 3, 3) and so on…

Therefore, possibility of happening a total of 8

P(E) = {Number of favourable events } ⁄ {number of total events}

= 21/216