Looking at the numbers in Table 6–1, we seethat most of the results are “near” 15 , in that they are between 12 and 18 . We can get a better feeling for the details of these results if weplot a graph of the distribution of the results. We count the number ofgames in which a score of k was obtained, and plot this number for each k . Such a graph is shown in Fig. 6–2. Ascore of 15 heads was obtained in 13 games. A score of 14 heads was also obtained 13 times. Scores of 16 and 17 were each obtained more than 13 times. Are we to conclude that there is some bias toward heads?Was our “best estimate” not good enough? Should we conclude now that the “mostlikely” score for a run of 30 tosses is really 16 heads? But wait! In all the games taken together, there were3000 tosses. And the total number of heads obtained was 1493 . The fraction of tosses that gave heads is 0.498 , very nearly, but slightly less than half. We should certainlynot assume that the probability of throwing heads is greaterthan 0.5 ! The fact that one particular set of observations gave 16 heads most often, is a fluctuation. We still expect thatthe most likely number of heads is 15 .
从表6-1，我们可以看到，大部分的结果都在15附近，也就是说，在12到18之间。对这些结果，如果我们画一张其分布图的话，那么，我们对其就会有更好的感觉。对于某一得分k，我们统计它出现的局数，然后，为每一个得分，画出它出现的局数，这样一幅图就是图6-2。正面为15的得分，在13个局中都出现过。正面为14的得分，出现了13次。16和17都超过了13次。我们是否要得出结论说：对正面有偏见呢？我们“最好的估计”，是不是还不够好呢？我们现在是否可以得出结论说，在每30次抛掷中，是否正面为16，才是最可能的得分？但是请等一下！把这所有的局加起来，总共有3000次抛掷，其中正面为1493，那么，正面的分数就是0.498，非常接近一半，但还是要少一点。我们当然不能假定，抛掷为正面的概率，要大于0.5！某一组观察给出正面的数字为16，这是最多的，但这一事实，只是一个波动。我们仍然期待，正面的最可能的数字，还是15。

Fig. 6–2.Summary of the results of 100 gamesof 30 tosses each. The vertical bars show the number of games in which a scoreof k heads was obtained. The dashed curve shows the expected numbersof games with the score k obtained by a probability computation. 图6-2 100局结果的总结，每局30次抛掷。垂直坐标,是正面得分k所出现的局数。虚线是通过概率计算，所得到的得分k，所画成的线。

We may ask the question: “What is theprobability that a game of 30 tosses will yield 15 heads—or 16 , or any other number?” We have said that in a game of one toss, theprobability of obtaining one head is 0.5 , and the probability of obtaining no head is 0.5 . In a game of two tosses there are four possible outcomes: HH, HT , TH , TT . Since each of these sequences is equally likely, we conclude that(a) the probability of a score of two heads is 1/4 , (b) the probability of a score of one head is 2/4 , (c) the probability of a zero score is 1/4 . There are two ways of obtaining one head, but only one ofobtaining either zero or two heads.
我们可能会问：“在一局30次抛掷中，会产生15次正面、16次正面、或任何其他数的概率，是多少？”我们已经说过，只抛一次，那么正反面的概率各是0.5。抛掷两次，有四种可能的结果，正正、正负、负正、负负。由于这些后果是平等的，所以我们可以得出结论：（a）两个正面的概率是1/4，（b）一个正面的概率是2/4，（c）零正面的概率是1/4。得到一个正面的方式，有两种，但是，得到零个或两个正面的方式，只有一种。

Consider now a game of 3 tosses. The third toss is equally likely to be heads or tails.There is only one way to obtain 3 heads: we must have obtained 2 heads on the first two tosses, and then heads on the last. Thereare, however, three ways of obtaining 2 heads. We could throw tails after having thrown two heads (oneway) or we could throw heads after throwing only one head in the first twotosses (two ways). So for scores of 3 -H , 2 -H , 1 -H , 0 -H we have that the number of equally likely ways is 1 , 3 , 3 , 1 , with a total of 8 different possible sequences. The probabilities are 18 , 38 , 38 , 18 .
现在我们考虑一局三次抛掷。第三次抛掷，正反面的概率同等。只有一种方式，可以得到三次正面，即前面得到了两次正面，最后又得到一次正面。然而，得到两次正面，有三种方式。我们可先得到两次正面（只一种方式），然后再得到一次反面，或者，我们可以在前面两次抛掷中，先得到一正一反（共两种方式），然后再得到一次正面。于是，在8次不同顺序的可能结果中，对于三次正面、两次正面、一次正面、和零次正面，我们就有得到它们的方式的数目，即1，3，3，1，其中的有些方式，是等价的。相应的概率就是1/8，3/8，3/8，1/8。

Fig. 6–3.A diagram for showing the numberof ways a score of 0, 1, 2, or 3 heads can be obtained in a game of3 tosses. 图6-3 在一局3次抛掷中，能得到不同的正面次数的图解。

Fig. 6–4.A diagram like that of Fig. 6–3, for agame of 6 tosses. 图6-4与图6-3类似，不过是一局6次抛掷。
The argument we have been making can besummarized by a diagram like that in Fig. 6–3. It isclear how the diagram should be continued for games with a larger number oftosses. Figure 6–4 showssuch a diagram for a game of 6 tosses. The number of “ways” to any point on the diagram is justthe number of different “paths” (sequences of heads and tails) which can betaken from the starting point. The vertical position gives us the total numberof heads thrown. The set of numbers which appears in such a diagram is known asPascal’s triangle. The numbers are also known as the binomialcoefficients, because they also appear in the expansion of (a+b)n. If we call n the number of tosses and k the number of heads thrown, then the numbers in the diagram areusually designated by the symbol (nk) . We may remark in passing that the binomial coefficients can also becomputed from
C(n,k)=n!/(k!(n−k)!),(6.4)
where n! , called “n -factorial,” represents the product (n)(n−1)(n−2)⋯(3)(2)(1).
我们所做的讨论，可以通过图6-3来总结。对于更大的抛掷数目，这个图如何继续，是很清楚的。图6-4就是这样一个图，一局六次抛掷。在图上，到任何点的“方式”的数目，正是不同的“路径”（正面和反面的顺序）的数目，这些路径，可以从开始点选择。垂直的位置，给了我们正面的总数。在这样一幅图中所出现的数字集合，被称为帕斯卡三角形。这些数字，也被称为二项式的系数，因为它们也出现在 (a+b)n 的展开式中。如果我们把n称为抛掷的次数，k称为正面的次数，那么，图解中的数字，通常就可通过符号C（n,k）来指定。我顺便说一下，这个二项式的系数，可以通过公式6.4来计算：
C(n,k)=n!/(k!(n−k)!),(6.4)
这里，n!被称为“n的阶乘”，表示着 (n)(n−1)(n−2)⋯(3)(2)(1)的积。

说明，由于网页排版的原因，所以，公式可能会变形，所以，建议读者以理解为主，想弄清楚公式的，最好去下载pdf。

We are now ready to compute the probability P(k,n) of throwing k heads in n tosses, using our definition Eq. (6.1). The total number of possible sequences is 2n (since there are 2 outcomes for each toss), and the number of ways of obtaining k heads is (nk) , all equally likely, so we have
P(k,n)=C(n,k)/2n.(6.5)
现在，我们就可以使用方程（6.1），来计算n次抛掷中，出现k次正面的概率P(k,n)了。总的可能的排列数是2的n次方，（由于每次抛掷，都有两种可能的结果），而获得ｋ次正面的方式，有C（n,k）个，它们都是同等可能的，所以我们就有：
P(k,n)=C(n,k)/2的n次方. (6.5)

Since P(k,n) is the fraction of games which we expect to yield k heads, then in 100 games we should expect to find k heads 100⋅P(k,n) times. The dashed curve in Fig. 6–2 passesthrough the points computed from 100⋅P(k,30) . We see that we expect to obtain a score of 15 heads in 14 or 15 games, whereas this score was observed in 13 games. We expect a score of 16 in 13 or 14 games, but we obtained that score in 15 games. Such fluctuations are “part of the game.”
由于P(k,n)是这个游戏的分数，在1局游戏中，我们期待产生k次正面，那么，在100局中，我们就期待找到k次正面100⋅P(k,n)次{？}。在图6-2中，虚线通过的点，就是从100⋅P(k,30)计算出来的。我们看到，对于一局15次正面这个得分，我们期待，在14或15个局中能得到，而实际上，是在13个局中看到了它。对于一局16次正面，我们期待在13或14个局中观察到，但是，我们在15个局中看到了它。这种波动就是这个“游戏的一部分”。

The method we have just used can be applied to the most general situation in which there are only two possible outcomes of a single observation. Let us designate the two outcomes by W (for “win”) and L (for “lose”). In the general case, the probability of W or L in a single event need not be equal. Let p be the probability of obtaining the result W . Then q , the probability of L , is necessarily (1−p) . In a set of n trials, the probability P(k,n) that W will be obtained k times is
P(k,n)=C(n,k)pkqn−k.(6.6)
This probability function is called the Bernoulli or, also, the binomial probability.
我们刚刚所使用的方法，可以应用于一种更普遍的情况，就是一次观察，只有两种可能结果的情况。让我们用W（代表“胜”）和L（代表“败”）来指代这两种结果。一般来说，在一次事件中，W的概率与L的概率，不需要是同等的。让p代表结果为W的概率，q为L的概率，那么q就必然是（1-p）。在一系列的n次试验中，W将会被得到k次的概率P(k,n)就是：
P(k,n)=C(n,k)（p的k次方）(q的n−k次方）.(6.6)
这个概率公式，被称为伯努利概率，也叫二项式概率。

6–3The random walk 6-3 随机行走
There is another interesting problem inwhich the idea of probability is required. It is the problem of the “randomwalk.” In its simplest version, we imagine a “game” in which a “player” startsat the point x=0 and at each “move” is required to take a step either forward(toward +x ) or backward (toward −x ). The choice is to be made randomly, determined, for example,by the toss of a coin. How shall we describe the resulting motion? In itsgeneral form the problem is related to the motion of atoms (or other particles)in a gas—called Brownian motion—and also to the combination of errors inmeasurements. You will see that the random-walk problem is closely related tothe coin-tossing problem we have already discussed.
还有另外一个有趣的问题，在其中也要用到概率的想法。这个问题就是随机行走。我们来看它的一个最简单的版本，我们想象，在一局游戏中，一个玩家从点x=0开始，而他的每一次“移动”，都要走一步，要么向前（向+x），要么向后（向-x）。选择是随机做出的，例如，通过抛掷一枚硬币来决定。我们如何才能描述所导致的运动呢？这个问题，就其基本形式而言，与气体中原子的运动—即布朗运动有关，也与测量中错误的组合有关。你将会看到，随机行走这一问题，与我们已经讨论过的抛硬币问题，紧密相关。

First, let us look at a few examples of arandom walk. We may characterize the walker’s progress by the net distance DNtraveled in N steps. We show in the graph of Fig. 6–5 threeexamples of the path of a random walker. (We have used for the random sequenceof choices the results of the coin tosses shown in Fig. 6–1.)
首先，让我们看几个随机行走的例子。对于一个行走者的进程，我们通过在N部中所走的净步数DN，来刻画。我们用图6-5中的曲线，显示了一个随机行走者的三个路径的例子。对于随机选择的这个次序，我们使用了图6-1中的抛硬币的结果。

Fig. 6–5.The progress made in a randomwalk. The horizontal coordinate N is the total number of steps taken; the vertical coordinate DNis the net distance moved from the starting position. 图6-5 一个随机行走的进程。水平坐标N是所走的总步数；垂直坐标DN是距开始点的净距离。

What can we say about such a motion? Wemight first ask: “How far does he get on the average?” We must expectthat his average progress will be zero, since he is equally likely to go eitherforward or backward. But we have the feeling that as N increases, he is more likely to have strayed farther from the startingpoint. We might, therefore, ask what is his average distance travelled in absolutevalue, that is, what is the average of |D| . It is, however, more convenient to deal with another measure of“progress,” the square of the distance: D2 is positive for either positive or negative motion, and is therefore areasonable measure of such random wandering.
对于这样一个运动，我们能说什么呢？我们可以首先问：“他平均能走多远？”我们应该期待，他的平均进展是零，因为他往前走和往后走的可能性是同等的。但是我们感觉到，随着N的增加，他更可能偏离起点更远。因此，我们可以问，他旅行的平均距离的绝对值是什么，也就是说，|D|的平均值是什么？然而，对于进展，还有另外一种测量方法，即距离的平方，处理这种方法会更方便些。D2对于正的运动和负的运动，都是正的，从而，是这种随机徘徊运动的一种合理的测量方式。

We can show that the expected value of D2Nis just N , the number of steps taken. By “expected value” we mean the probablevalue (our best guess), which we can think of as the expected averagebehavior in many repeated sequences. We represent such an expected valueby ⟨ D2N⟩ , and may refer to it also as the “mean square distance.” After onestep, D2 is always +1 , so we have certainly ⟨D21⟩=1 . (All distances will be measured in terms of a unit of one step. Weshall not continue to write the units of distance.)
我们可以指出，D2N的期待的值，正是N，即已经走了的步数。通过“期待的值”，我们的意思是说：可能的值（我们最好的猜测），它是在很多重复的次序中，我们所能期待的平均表现。我们用 ⟨ D2N⟩，来代表这样一个期待的值，也把它认为是“均方距离”。在一步之后，D2总是+1，所以，我们当然就有 ⟨D21⟩=1。（所有的距离，都将依据一步为单位，来测量。我们将不继续写距离的单位。）

The expected value of D2Nfor N>1 can be obtained from DN−1 . If, after (N−1) steps, we have DN−1 , then after N steps we have DN=DN−1+1or DN=DN−1−1. For the squares,
当N>1时的D2N的期待的值，可以从DN−1获得。如果在（N-1）步后，我们有DN−1，那么，在N步后，我们就有DN=DN−1+1或 DN=DN−1−1。对于平方，则有：

In a number of independent sequences, we expect to obtain each valueone-half of the time, so our average expectation is just the average of the twopossible values. The expected value of D2Nis then D2N−1+1 . In general, we should expect for D2N−1its “expected value” ⟨D2N−1⟩ (by definition!). So
在一系列独立的次序中，我们期待一半时间得到这个值，一半时间得到另一个值，于是，我们的平均期待值，就正是这两个可能值得平均。的期待值因此就是D2N−1+1。一般来说，对于D2N−1，我们应该期待它的“被期待的值” ⟨D2N−1⟩ (通过定义!). 于是：

We have already shown that ⟨D21⟩=1 ; it follows then that
我们已经指出了⟨D21⟩=1；于是就有：

a particularly simple result!
一个特别简单的结果！

If we wish a number like a distance, ratherthan a distance squared, to represent the “progress made away from the origin”in a random walk, we can use the “root-mean-square distance” Drms：
在一个随机行走中，如果我们希望用距离这个数、而不是用距离的平方，来代表离开原点的进展，我们可以用“均方根距离”Drms：