跟读练习: Can you steal the most powerful wand in the wizarding world? - Dan Finkel - 通过YouTube学习英语口语
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The Fabled Mirzakhani Wand is the most powerful magical item ever created,
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The Fabled Mirzakhani Wand is the most powerful magical item ever created,
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and that's why the evil wizard Moldavort is planning to use it to conquer the world.
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You and Drumbledror have finally discovered its hiding place in this cave.
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The wand is hidden by a system of 100 magical stones,
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including a glowing keystone, and 100 platforms.
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If the keystone is placed on the correct platform,
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the wand will be revealed.
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If placed incorrectly, the entire cave will collapse.
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The keystone is immune to all magic,
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but the other stones aren't,
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meaning you can pick them up and cast a placement spell,
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and the platform that stone belongs on will glow.
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Place all 99 stones correctly,
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and the final platform must be the keystone's correct resting place.
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You're about to get started when one of Moldavort's henchmen arrives and irreversibly seals a random stone to a random platform.
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If you need to place a stone that belongs on a platform that's already occupied,
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your spell will make some random unoccupied platform glow instead.
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What are your odds of placing the keystone on the correct platform?
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Let's imagine we knew everything about this situation.
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With perfect knowledge, we could label the stones 1 to 100 based on the order we plan to place them,
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and label the platforms they belong on in the same way.
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We'll label the stone the henchman placed as 1,
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meaning it was supposed to go on platform 1,
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and the keystone as 100, belonging on platform 100.
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Of course, we don't know which platform is which,
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so the numbering of the platforms is actually invisible to us.
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There are three possibilities.
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One, that first stone was placed randomly onto its own platform,
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in which case you're guaranteed to succeed.
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2. It was placed on the Keystone's platform,
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and you're doomed to fail.
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But most likely, Scenario 3,
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it was placed somewhere else.
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Suppose the henchman placed Stone 1 on, say, Platform 45.
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Then you'd place Stone 2 on Platform 2,
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3 on 3, and so on until you got to Stone 45.
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Its platform being taken, a random platform would light up.
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And here, there are three possibilities.
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If it's platform 1, you'll win,
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because all of the remaining stones will go to the correct platforms.
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If platform 100 lights up,
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you lose, because the keystone spot will be taken.
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Any other platform, and you're essentially back where you started,
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just with 54 remaining stones and one on the wrong platform.
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In that scenario, let's say the spell tells us to place Stone 45 on platform 82.
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Then we'd place 46 to 81 correctly, and 82 at random.
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And here, we reach the same three possibilities.
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Pedestal 1, you win.
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Pedestal 100, you lose.
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Any other, you continue the process.
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In other words, you're playing a game where you have equal chances to win and lose,
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and some chance to delay the decisive moment.
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No matter how many times this process repeats,
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you'll inevitably either place a stone on Pedestal 1 or Pedestal 100 before you reach the keystone.
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That's all that determines whether you succeed or fail,
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and critically, the chances of those events are equal.
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This can be unintuitive, so let's imagine another similar game.
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Say, Drumbled Drawer magically generates numbers from 1 to 100.
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If it's a one, you win.
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If it's a 100, you lose.
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If it's anything else, he picks again.
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Since the odds of winning by getting 1 are the same as losing by getting a 100,
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this is a game you're just as likely to win as to lose.
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It might take a while,
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but the delays don't give an advantage to getting a 1 before 100, or vice versa.
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The same essential reasoning applies to our situation.
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You're debating whether it's worth risking a 50-50 chance of a cave-in when Drumbledore reveals his secret weapon,
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a rare Feluche Felucious potion,
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which grants extraordinary luck for a brief period of time.
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There's a 1 in 100 chance the Keystone's platform was taken by the first stone,
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and you've lost already.
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But otherwise, you've got even odds to win or lose.
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And right now, you're feeling lucky.
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背景与上下文
在这段精彩的视频中，达恩·芬克尔（Dan Finkel）探讨了一个与魔法世界相关的游戏，主要围绕着传说中的米尔扎卡尼魔杖展开。这根魔杖被邪恶巫师莫尔德沃特（Moldevort）计划用来征服世界，然而它被藏在一个由100块魔法石和100个平台组成的系统中。参与者必须通过准确的拼图将魔法石放到正确的平台上，才能揭示魔杖的位置。这部视频不仅强调了计算概率的重要性，还巧妙地运用了语言技巧，引导观众深入思考。
日常交流的五个关键短语
- “你准备好了吗？” - 用于询问他人是否准备好进入某个情况或任务。
- “如果我说……” - 用于引入一个条件或假设，引发讨论。
- “考虑到所有可能性…” - 引导听众思考多个选择或结果。
- “让我们开始吧！” - 表达开始一个新活动或挑战的兴奋。
- “再次确认…” - 提醒他人对已知信息或选择进行重新审视。
逐步跟读指南
为了提高英语发音和听力水平，您可以通过以下步骤进行跟读练习，尤其适合需要加强雅思口语练习的学生：
- 选择合适的材料： 从视频中选择一小段，确保它的难度适合您。
- 初次倾听： 只听一次完整内容，了解整体情境，培养对语音的感知。
- 逐句重复： 每听一句，暂停并尝试模仿发音。利用 shadowspeak 方法，通过音节和重音同步发声。
- 记录与回放： 使用手机或录音设备记录自己的声音，与原音比较，寻找发音差异。
- 定期复习： 每周选择不同主题进行实战练习，确保长沙时间保持练习。
通过这种方式，您可以有效提升您的英语口语能力与发音，将其运用于日常交流中，在愿望与挑战之间找到平衡，最终实现提高英语发音的目标。
什么是跟读法？
跟读法 (Shadowing) 是一种有科学依据的语言学习技巧，最初开发用于专业口译员的培训，并由多语言者Alexander Arguelles博士普及。这个方法简单而强大：您在听英语母语原声的同时立即大声重复——就像是一个延迟1-2秒紧跟说话者的影子。与被动听力或语法练习不同，跟读法强迫您的大脑和口腔肌肉同时处理并模仿真实的讲话模式。研究表明它能显着提高发音准确性，语调，节奏，连读，听力理解和口语流利度——使其成为雅思口语备考和真实英语交流最有效的方法之一。