#1
0
(由 Microsoft 翻译) 我认为大约 25% 的时间没有 "假" 硬币。 要么那个， 要么重量差是如此之小， 可以忽略不计地有很多硬币在规模上。
(原文) Coin Weighing Game
I think that about 25% of the time there is no "fake" coin. Either that or the weight differential is so small as to be negligible with lots of coins on the scale.
I think that about 25% of the time there is no "fake" coin. Either that or the weight differential is so small as to be negligible with lots of coins on the scale.
作者 DingleBerries
2006-08-24 02:56:33
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#2
0
(由 Microsoft 翻译) 经过进一步审查 - 我只是不太擅长比赛。
(原文) Upon further review - I'm just not that good at the game.
作者 DingleBerries
2006-08-24 05:10:56
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#3
(由 Microsoft 翻译) 我们已经发布了更新。
这应该
1） 解决您提到的问题
2） 确保如果移动错误，游戏将丢失
3） 如果做出正确的移动，使结果更加随机
这应该
1） 解决您提到的问题
2） 确保如果移动错误，游戏将丢失
3） 如果做出正确的移动，使结果更加随机
(原文) We've posted an update.
This should
1) solve the problem you mentioned
2) ensure that the game will be lost if a wrong move is made
3) make the result more random if a correct move is made
This should
1) solve the problem you mentioned
2) ensure that the game will be lost if a wrong move is made
3) make the result more random if a correct move is made
作者 Novel Games
2006-08-24 17:11:31
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#4
2601
(由 Microsoft 翻译) 我同意 DingleBerries 的观点，尽管这条信息是 2006 年发布的。
(原文) I AGREE with DingleBerries, even though this message is from 2006.
作者 Alex Gordillo
2025-05-06 12:44:34
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#5
-997308
(由 Microsoft 翻译) 在 39 个硬币中，最初有 78 种可能性（39 个硬币中的一个更重或更轻）。 @Everyone 在称重结果为 <、= 或 >的情况下，您必须确保第一次称重将可能性分成不超过 27 个可能性。然后 @Everyone 您必须确保第二次权衡将可能性分成不超过 9 个可能性（如果称重结果为 <、= 或 >）。再说一次， @Everyone ，您必须确保第三个称量在称量结果为 <、= 或 >的情况下将可能性分成不超过 3 种可能性。 最后， @Everyone， 您必须确保在称量结果<的情况下，第四个称量将可能性分为不超过 1 种。 = 或 >。缩小范围后，您就可以选择答案。
(原文) In 39 coins, there are initially 78 possibilities (one of 39 coins is either heavier or lighter). @Everyone You must make sure that the first weigh splits the possibilities into no more than 27 each in cases of weigh result being <, =, or >. Then @Everyone you must make sure that the second weigh splits the possibilities into no more than 9 each in cases of weigh result being <, =, or >. Then again, @Everyone you must make sure that the third weigh splits the possibilities into no more than 3 each in cases of weigh result being <, =, or >. Then finally, @Everyone, you must make sure that the fourth weigh splits the possibilities into no more than 1 each in cases of weigh result being <, =, or >. And after it's been narrowed down then you can pick the answer.
作者 Piotr Grochowski
2025-05-06 19:57:01
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