巴打係咪差唔多開估
定係太少人玩唔到
整緊
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
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.
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
整緊
巴打係咪差唔多開估
定係太少人玩唔到
整緊
唔識整
呢度係個EXCEL表
有冇巴打幫幫手
https://docs.google.com/spreadsheets/d/1RdtMihlBQWSbcFqXzJKzX9rXkIR4nmFG89JGLBIKMaw/edit#gid=2079417308
巴打係咪差唔多開估
定係太少人玩唔到
整緊
唔識整
呢度係個EXCEL表
有冇巴打幫幫手
https://docs.google.com/spreadsheets/d/1RdtMihlBQWSbcFqXzJKzX9rXkIR4nmFG89JGLBIKMaw/edit#gid=2079417308
參與人數為 221人
目測結果
第二round 勝利者為 145
上面嗰個唔係一千人咩,變咗二百人,成隻game唔同咗啦
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.
多謝ching 今日畀人笑我傻仔揀100
一早講左 個game 係玩唔同其他人撞為先 你愈有理由去分析點解一個答案係最佳既時候 果個答案就係一定係最差
你掉番轉問我叫我解釋點為63會赢 145會赢 我係解釋唔到 正正就係解釋唔到 所以佢係最好既答案
其實話簡最大(100)係best strategy既人睇到兩個TEST 最大既號碼都係大比數咁輸, 有咩感想?
其實話簡最大(100)係best strategy既人睇到兩個TEST 最大既號碼都係大比數咁輸, 有咩感想?
best strategy唔係用嚟贏架
其實話簡最大(100)係best strategy既人睇到兩個TEST 最大既號碼都係大比數咁輸, 有咩感想?
best strategy唔係用嚟贏架