(博奕論)北海道大學入學試題,1至100選最大數字

岩岩加左好多 randamoness去個舊有stat度,無論我sim左幾多次,去到600-700人左右都係揀1係最好,唔知係咪因為同個stat的pool size有關

我唔係讀stat出身所以出黎個結果我唔識理解

目測有嘢搞錯咗
明明1同其他嘅機率差唔多
應該冇乜理由會特別優勝過人

樓上用咩software同咩procedure整呢個simulation？

岩岩加左好多 randamoness去個舊有stat度,無論我sim左幾多次,去到600-700人左右都係揀1係最好,唔知係咪因為同個stat的pool size有關

我唔係讀stat出身所以出黎個結果我唔識理解

目測有嘢搞錯咗
明明1同其他嘅機率差唔多
應該冇乜理由會特別優勝過人

樓上用咩software同咩procedure整呢個simulation？

我用stat個結果當係一個binomal distribution 的pdf function,,再係最初既stat加d
random既結果上去

搵淨係出現一次,再乘返對應個outcome分數, 睇下唔同人數參與之下邊個得到分數最大

用python plot的,d code放左上github,雖然應該打得幾柒下

https://github.com/jackyko1991/Hokkaido-Test

岩岩加左好多 randamoness去個舊有stat度,無論我sim左幾多次,去到600-700人左右都係揀1係最好,唔知係咪因為同個stat的pool size有關

我唔係讀stat出身所以出黎個結果我唔識理解

目測有嘢搞錯咗
明明1同其他嘅機率差唔多
應該冇乜理由會特別優勝過人

樓上用咩software同咩procedure整呢個simulation？

我用stat個結果當係一個binomal distribution 的pdf function,,再係最初既stat加d
random既結果上去

搵淨係出現一次,再乘返對應個outcome分數, 睇下唔同人數參與之下邊個得到分數最大

用python plot的,d code放左上github,雖然應該打得幾柒下

https://github.com/jackyko1991/Hokkaido-Test

冇睇code, 不過best choice唔係應該count 每次sim既winner, i.e. largest unique number咩? 即係計下邊個數贏得最多次

岩岩加左好多 randamoness去個舊有stat度,無論我sim左幾多次,去到600-700人左右都係揀1係最好,唔知係咪因為同個stat的pool size有關

我唔係讀stat出身所以出黎個結果我唔識理解

目測有嘢搞錯咗
明明1同其他嘅機率差唔多
應該冇乜理由會特別優勝過人

樓上用咩software同咩procedure整呢個simulation？

我用stat個結果當係一個binomal distribution 的pdf function,,再係最初既stat加d
random既結果上去

搵淨係出現一次,再乘返對應個outcome分數, 睇下唔同人數參與之下邊個得到分數最大

用python plot的,d code放左上github,雖然應該打得幾柒下

https://github.com/jackyko1991/Hokkaido-Test

冇睇code, 不過best choice唔係應該count 每次sim既winner, i.e. largest unique number咩? 即係計下邊個數贏得最多次

你岩,唔使乘個分數

即係binomial process, 個pdf用返stat的結果,搵出淨係中一次邊個數字的probability係最高

想知而家有幾多人參加第二輪w
呢個遊戲好好玩

[size=999] 2 [/size=999]
喺份卷到寫到最大

想知而家有幾多人參加第二輪w
呢個遊戲好好玩

184 only

玩既人愈多,應該揀既數字就愈細,但係又唔係揀0-10果d數字,點解會咁

好多on9...

利申:大學篇論文係關於博弈論

玩既人愈多,應該揀既數字就愈細,但係又唔係揀0-10果d數字,點解會咁

因為唔係全部玩家都經過思考去玩

玩既人愈多,應該揀既數字就愈細,但係又唔係揀0-10果d數字,點解會咁

可唔可以講下點解開頭set咗 ran_player 等於 int(my_data.size*randomness_ratio)
跟住下面又set咗 num_of_player 等於 800?

玩既人愈多,應該揀既數字就愈細,但係又唔係揀0-10果d數字,點解會咁

可唔可以講下點解開頭set咗 ran_player 等於 int(my_data.size*randomness_ratio)
跟住下面又set咗 num_of_player 等於 800?

哦，我明啦
你唔係加randomness咁簡單，而係整咗個mixture出來

玩既人愈多,應該揀既數字就愈細,但係又唔係揀0-10果d數字,點解會咁

可唔可以講下點解開頭set咗 ran_player 等於 int(my_data.size*randomness_ratio)
跟住下面又set咗 num_of_player 等於 800?

哦，我明啦
你唔係加randomness咁簡單，而係整咗個mixture出來

我既數學只限於UG minor math程度

純粹想玩下個stat,愈多人玩應該愈準

個pdf應該係hidden function來,有d人類既心理因素係入面,但我無學過game theory只好用d基本的probability玩下

大陸上唔到google. 我有個質素答案，三位數既

愈多人 安全範圍愈細係岩
假設大家會思考完去投
遊遊戲分為part1 撞/唔撞 part2 大/細

假設2個玩家
就會揀100因爲一係贏一係輸

3個玩家
就會有幾個情況
A. 100 99(100以外嘅數字) 98(100,99以外嘅數字)
B. 100 100 99(100以外嘅數字)
C. 100 100 100
D. 99 99 98(100,99以外嘅數字)
投100就係抱住唔撞號碼+最大嘅心態
投99就為搏另外2個投100
投98就為避開危險區100-99

有人問點解97-1會冇人揀
因為97-1雖然離開危險區，但數字愈黎愈細愈不利
part 2

愈參加人數愈黎愈多 危險區都會愈黎愈大

太長唔想用中文打

The strategy of choosing 100 is a weakly dominant strategy.

Consider a player, Alice. We only need to consider two cases:

Case 1: The largest number chosen by the rest of the group is strictly less than 100. Then, Alice can always win by choosing 100, but if she may lose if she chooses a number less than 100.

Case 2: The largest number chosen by the rest of the group is 100. Then, Alice will lose no matter which number she chooses, so she is indifferent between all possible strategies.

The conclusion means that all (rational) players should choose 100, leading to a Nash equilibrium outcome in which all players gain zero point.

Logic又唔得
英文又差
柒少陣啦......

假如3個人玩
然後你偷睇到A填100，B填99
咁你應該填乜

緊係填99然後比填99嘅人見到你填99，

咁佢得返100呢個選擇轉

然後揀100嗰兩條友攬炒

等睇round 2結果

假如3個人玩
然後你偷睇到A填100，B填99
咁你應該填乜

緊係填99然後比填99嘅人見到你填99，

咁佢得返100呢個選擇轉

然後揀100嗰兩條友攬炒

簡99再話比b知我寫99
佢實寫返100 佢地就收皮

假如3個人玩
然後你偷睇到A填100，B填99
咁你應該填乜

緊係填99然後比填99嘅人見到你填99，

咁佢得返100呢個選擇轉

然後揀100嗰兩條友攬炒

簡99再話比b知我寫99
佢實寫返100 佢地就收皮

或者咁講 你要另佢地一樣 你寫1都贏

玩既人愈多,應該揀既數字就愈細,但係又唔係揀0-10果d數字,點解會咁

我幾頁前打左幾條式, 可以參考一下
P(n wins)
= P(n is the largest unique number)
= P(n is unique)*P(n+1,...,100 are not unique)

Let
a_n = P(n wins) = P(n is the largest unique number)
b_n = P(n is unique), 1-b_n = P(n is not unique)
a_n = b_n*(1-b_{n+1})*...*(1-b_100)

你expect個度淨係計左b_n, 但b_n =/= P(n wins)
加新人用stat個distribution + uniformly random黎gen, 應該好過只用uniformly random?
btw, 可以model player既strategy
let p_n be the probability of selecting n
即用stat拎p_n, 再用p_n計b_n先, 跟住計a_n
然後再用返a_n 黎define新p_n
e.g. p_n = a_n/(a_1+...+a_100)
loop黎睇下會點

假如3個人玩
然後你偷睇到A填100，B填99
咁你應該填乜

緊係填99然後比填99嘅人見到你填99，

咁佢得返100呢個選擇轉

然後揀100嗰兩條友攬炒

簡99再話比b知我寫99
佢實寫返100 佢地就收皮

或者咁講 你要另佢地一樣 你寫1都贏

呢個情況主導權就係自己到
問佢地睇下開幾大條件黎換取勝利

我改咗RX-78-2段code再寫咗個simulation
link: https://pastebin.com/VwnHggRD
data主要係用之前所得嘅sample distribution模擬出來

Output:

[63, 52, 48, 62, 65, 63, 66, 63, 63, 63, 63, 65, 63, 44, 63, 52, 64, 63, 64, 63, 63, 65, 54, 50, 40, 65, 63, 48, 63, 56]
The mean of winning places for 1000 players is 59.3

[31, 27, 27, 63, 19, 18, 27, 6, 33, 52, 31, 21, 33, 46, 63, 15, 18, 31, 48, 63, 21, 31, 63, 31, 33, 52, 49, 28, 46, 14]
The mean of winning places for 2000 players is 34.6666666667

[27, 6, 14, 27, 28, nan, 12, nan, 28, 18, nan, 15, 21, 5, nan, 33, 28, 14, 11, nan, 31, nan, nan, 28, nan, 63, 33, 15, nan, 5]
The mean of winning places for 3000 players is 22.0

[11, 63, nan, nan, nan, nan, 63, nan, 8, nan, 33, nan, nan, 27, nan, 14, 21, 63, 31, nan, 0, nan, nan, 21, nan, 18, nan, 18, nan, nan]
The mean of winning places for 4000 players is 27.9285714286

[nan, nan, 8, nan, 0, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, 63, 31, 28, 31, nan, nan, nan, 11, 3, nan, nan, nan, nan, 8, nan]
The mean of winning places for 5000 players is 20.3333333333

[nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan, nan]
No winning places

[Finished in 12.3s]

我前面話「如果有一萬人玩隻game，啲人重夠唔夠膽講60-70係合理範圍」係有少少根據嘅
每一條array有30個elements，分別代表30回合裡面每一場邊個數字贏咗
3000人嗰度你已經見到有好多nan，因為嗰一場裡面無人贏，所以我set咗winning position係nan

睇返上面output你就會見到愈多參賽者個winning number就會愈低
其實4000同5000個players出嘅sample mean唔係太靠得住，因為太多nan啦，拖低咗個sample size of non-missing data
去到6000個players就直情無人贏
背後原因我諗同birthday problem/pigeonhole principle有啲關係

巴打係咪差唔多開估
定係太少人玩唔到

順帶一提，唔好諗住用simulation嘅outcome做prediction，你會死得好慘