機率題：飛機座位

用P(n)表示第1個人無隨機坐, 之後第2~n-1個人跟題目規則, 最後第n個坐到自己座位嘅機率
問題1就係求P(1000)

先考慮頭幾個簡單case
n=2時第1個人有1/2坐啱自己位, 1/2坐第2個人個位, P(2) = 1/2

n=3時第1個人有1/3坐啱自己位, 之後其他全部人坐啱
有1/3坐第2個人個位, 之後第2個人個情況同n=2時差唔多, 只係自己個位同第1個人個位調轉, 對第n個人無分別
有1/3坐第n個人個位
P(3) = 1/3 + 1/3*P(2) + 0 = 1/3 + 1/3*P(2) = 0.5

n=4時第1個人有1/4坐啱自己位, 之後其他全部人坐啱
有1/4坐第2個人個位, 之後第2個人個情況同n=3時差唔多, 只係自己個位同第1個人個位調轉, 對第n個人無分別
有1/4坐第3個人個位, 之後第3個人個情況同n=2時差唔多, 只係自己個位同第1個人個位調轉, 對第n個人無分別
有1/4坐第n個人個位
P(4) = 1/4 + 1/4*P(3) + 1/4*P(2) + 0 = 1/4 + 1/4*P(3) + 1/4*P(2) = 0.5

同樣方法諗
P(1000) = 1/1000 + 1/1000*P(999) + 1/1000*P(998) + ... + 1/1000*P(2)

用MI設 P(2)=P(3)=...=P(n)=0.5
P(n+1)
= 1/(n+1) + 1/(n+1)*P(n) + 1/(n+1)*P(n-1) + ... + 1/(n+1)*P(2)
= 1/(n+1) + (n-1)/(n+1) * 0.5
= 1/(n+1) * (1 + (n-1)*0.5)
= 1/(n+1) * (0.5n + 0.5)
= 0.5
所以P(1000)=0.5

我估佢地派膠姐

真係，唔知有咩意思
成班copy and paste 自己都無理解過
我自己唔識search咩

仲有話咩有飛一定坐到嗰啲，無飛上唔到機嗰啲唔講以為自己喺創意台/吹水台

第一題 998個人坐岩
淨番第一個唔見飛同埋你
咁唔係1/2？

題目睇唔明好小事 但你仲以為自己好醒 有啲可憐 你邊一日睇得明呢條題目再去諗下點樣解答

你鍾意啦
繼續活係你自己個世界

連登係咁
問個假設性數學問題都屌

自己個位俾人坐左點解唔係屌柒坐左你個位條柒頭叫佢死撚開？

全個post得你一個有問題

有趣
因為有編號最後其實當得2個位咁計就得
50％
一係有一係無

大佬我無諗到n=4個時坐錯位條友同比佢坐咗自己位嘅人，計算上，身份轉換係無問題

乘客遺失機票應該上唔到機, 完

第一題1/1000 不解釋
第二題係要計去到999上機既時候佢個位未俾人坐左既機率
有無人計到

我唔勁啦，#76個個巴打先勁

簡化咗做10問AI
This is an interesting probability problem! Let's think through it step by step:

First, let's consider what happens to the 10th person's seat:
If the first person doesn't take seat 10, it will be available for the 10th person.
If the first person takes seat 10, it won't be available.
The probability of the first person taking seat 10 is 1/10, as they choose randomly from all seats.
If the first person doesn't take seat 10 (which happens with probability 9/10), the 10th person will get their correct seat, regardless of what everyone else does.
If the first person does take seat 10 (probability 1/10), then the 10th person will only get their correct seat if they end up in the seat of the person who took their seat. This is essentially the same as if they were choosing randomly among the remaining seats.
In this case (first person took seat 10), the probability of the 10th person getting their correct seat is 1/9, as they're effectively choosing randomly from the 9 remaining seats.
We can now calculate the total probability: P(10th person gets correct seat) = P(first person doesn't take seat 10) + P(first person takes seat 10) * P(10th person gets correct seat in this case)
= 9/10 + 1/10 * 1/9
= 9/10 + 1/90
= 81/90 + 1/90
= 82/90
= 41/45
≈ 0.9111 or about 91.11%
Therefore, the probability of the 10th person having their own seat is 41/45 or approximately 91.11%.

This result is quite interesting and perhaps counterintuitive. Despite the potential chaos introduced by the first person's random choice, the last person still has a very high chance of ending up in their assigned seat.

Would you like me to elaborate on any part of this explanation?

This is an interesting variation of the problem. Let's think through it step by step:

First, we need to consider what can happen to the 9th person's seat:
It could be taken by the 1st person (who chooses randomly)
It could be taken by someone else who was displaced
It could remain empty for the 9th person
The probability of the 1st person taking the 9th person's seat is 1/10.
If the 1st person doesn't take the 9th person's seat (probability 9/10), we need to consider two scenarios:
a) The 1st person took their own seat (probability 1/10)
b) The 1st person took someone else's seat (probability 8/10)
If scenario (a) occurs, everyone including the 9th person will get their correct seat.
If scenario (b) occurs, it creates a chain of displacements. The 9th person will get their seat unless this chain of displacements reaches their seat before they do.
The key insight is that in scenario (b), the 9th person has an equal chance of being in any position in this chain of displacements, including the last position (which would mean they get their own seat).
Given that there are 9 people involved in this chain (everyone except the 1st person), the probability of the 9th person being last in the chain (and thus getting their seat) is 1/9.
Now, let's calculate the total probability:

P(9th person gets correct seat) = P(1st person takes 9th seat) * 0 +
P(1st person doesn't take 9th seat) * [P(1st person takes own seat) * 1 +
P(1st person takes another seat) * P(9th person last in chain)]

= 1/10 * 0 + 9/10 * [1/10 * 1 + 8/10 * 1/9]
= 0 + 9/10 * [1/10 + 8/90]
= 9/10 * [9/90 + 8/90]
= 9/10 * 17/90
= 17/100
= 0.17 or 17%

Therefore, the probability of the 9th person having their own seat is 17%.

This result is quite different from the probability for the 10th person (which would be 90%), showing how the position in the seating order can significantly affect the chances of getting the correct seat.

4o mini 出既答案唔同

f(x): 有x個人，第一個人沒有對的位置，只能亂坐，剩下的人有對位則坐對位，無則亂坐，最終導致最後一個人坐對位的概率

f(2) = (1/2)x[1 + 0] = 1/2
f(3) = (1/3)x[1 + f(2) + 0] = (1/3)x(3/2) = 1/2
f(4) = (1/4)x[1 + f(3) + f(2) + 0] = (1/4)x(4/2) = 1/2
f(5) = (1/5)x[1 + f(4) + f(3) + f(2) + 0] = (1/5)x(5/2) = 1/2
…
f(1000) = (1/1000)x[1 + f(999) + f(998) + … + f(2) + 0]
= (1/1000)x(1000/2) = 1/2

g(1000): 有1000個人，第一個人亂坐，剩下的人有對位則坐對位，無則亂坐，最終導致最後一個人坐對位的概率

g(1000) = f(1000) = 1/2

對於 f(3) = (1/3)x[1 + f(2) + 0] 的解釋:
有1/3機會坐問題座位[不屬於這3人的座位]，有1/3機會坐尾二的人的座位，有1/3機會坐最尾的人的座位

總共有3人，尾二的人坐對位的概率:
h(3) = (1/3)x[1 + 0 + 1] = 2/3
同理:
h(4) = (1/4)x[1 + h(3) + 0 + 1] = (1/4)x(4x2/3) = 2/3
h(1000) = h(999) = h(998) = … = h(4) = h(3) = 2/3

樓上仲咁多人話50%啱

prob/combin用conditional probability+recursion簡化問題係頭一兩個prob course常用技巧嚟啦

頭豬智商過100嘅機率有幾高？

1. 1/2
2. 2/3

唔可以啊
因為只要其中一個人坐左 1 or 1000，其他既人都會跟位坐，唔會隨機架啦，所以1000上機個陣 1 or 1000一定會有其中一個位空左

冇錯

其實有個捷思：問題可以轉換成1號位定1000號位會被人坐先，兩者機率各0.5