#1
1
(已翻譯) 我認為大約 25% 的時間沒有 「假」 硬幣。 要麼那個, 要麼重量差是如此之小, 可以忽略不計地有很多硬幣在規模上。
(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
讚好
回覆
#2
1
(已翻譯) 經過進一步審查 - 我只是不太擅長比賽。
(原文) Upon further review - I'm just not that good at the game.
作者 DingleBerries
2006-08-24 05:10:56
讚好
回覆
#3
(已翻譯) 我們已經發佈了更新。
這應該。
1) 解決您提到的問題。
2) 確保如果移動錯誤,遊戲將丟失。
3) 如果做出正確的移動,使結果更加隨機。
這應該。
1) 解決您提到的問題。
2) 確保如果移動錯誤,遊戲將丟失。
3) 如果做出正確的移動,使結果更加隨機。
(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
讚好
回覆
#4
2606
(已翻譯) 我同意 DingleBerries 的觀點，儘管這條資訊是 2006 年發佈的。
(原文) I AGREE with DingleBerries, even though this message is from 2006.
作者 Alex Gordillo
2025-05-06 12:44:34
讚好
回覆
#5
2679
(已翻譯) 在 39 個硬幣中，最初有 78 種可能性（39 個硬幣中的一個更重或更輕）。 @Everyone 在稱重結果為 <、= 或 >的情況下，您必須確保第一次稱重將可能性分成不超過 27 個可能性。然後 @Everyone 您必須確保第二次權衡將可能性分成不超過 9 個可能性（如果稱重結果為 <、= 或 >）。再說一次， @Everyone ，您必須確保第三個稱量在稱量結果為 <、= 或 >的情況下將可能性分成不超過 3 種可能性。 最後， @Everyone， 您必須確保在稱量結果<的情況下，第四個稱量將可能性分為不超過 1 種。 = 或 >。縮小範圍后，您就可以選擇答案。
(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
讚好
回覆