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首先唔應該揀40或以下, 因為就算加左60都高唔過直接揀100
假設加到60分, 41=>101,...,100=>160
如果參加者眾, 直接揀100會比較好, 因為好大機會個個數都有人揀, 但另一方面揀4x可能可以出奇制勝, 但輸左就得4x分
如果人少少, 應該揀貼近100既數, 搏冇人同你一樣
:^(
楚歌
自
首先唔應該揀40或以下, 因為就算加左60都高唔過直接揀100
假設加到60分, 41=>101,...,100=>160
如果參加者眾, 直接揀100會比較好, 因為好大機會個個數都有人揀, 但另一方面揀4x可能可以出奇制勝, 但輸左就得4x分
如果人少少, 應該揀貼近100既數, 搏冇人同你一樣
自
:^(
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數學物理
實際上唔係
it sounds good but it never works
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.
實際上唔係
it sounds good but it never works
都係得六個字
好明顯考你既唔係實際你填既數字而係背後既concept
呢個咪就係正確答案
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.
實際上唔係
it sounds good but it never works
好明顯考你既唔係實際你填既數字而係背後既concept
呢個咪就係正確答案
:^(
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碗Bowl
study下Nash Equilibrium先啦
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.
實際上唔係
it sounds good but it never works
好明顯考你既唔係實際你填既數字而係背後既concept
呢個咪就係正確答案
study下Nash Equilibrium先啦
好出奇呀
有問題
前設係唔係如果成堆人寫100
個堆100就輸哂 (取消資格)
咁寫99 個個人會贏
如果冇呢個前設就寫硬100
如果有
case 2 Alice 寫99未必輸
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.
實際上唔係
it sounds good but it never works
好明顯考你既唔係實際你填既數字而係背後既concept
呢個咪就係正確答案
有問題
前設係唔係如果成堆人寫100
個堆100就輸哂 (取消資格)
咁寫99 個個人會贏
如果冇呢個前設就寫硬100
如果有
case 2 Alice 寫99未必輸
假設加到60分, 41=>101,...,100=>160
如果參加者眾, 直接揀100會比較好, 因為好大機會個個數都有人揀, 但另一方面揀4x可能可以出奇制勝, 但輸左就得4x分
如果人少少, 應該揀貼近100既數, 搏冇人同你一樣