# Newcomb's Paradox

**URL:** <https://www.thephilosophyforum.com/t/newcombs-paradox/298>\
**Category:** Logic & Philosophy of Mathematics\
**Created:** [March 11, 2026, 3:09am UTC](https://www.thephilosophyforum.com/t/newcombs-paradox/298 "2026-03-11T03:09:32Z")\
**Posts on this page:** 1\
**Showing post:** 535

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**Author:** ![Michael](https://avatars.discourse-cdn.com/v4/letter/m/4bbf92/32.png) [@Michael](https://www.thephilosophyforum.com/u/Michael)\
**Post date:** [March 16, 2026, 7:03pm UTC](https://www.thephilosophyforum.com/t/newcombs-paradox/298/535 "2026-03-16T19:03:28Z")

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I haven’t changed anything. It’s very simple. All of these are valid arguments:

**Argument A**

1. I will one-box
2. I will almost certainly be predicted to one-box
3. Therefore, I will almost certainly win exactly $1,000,000

**Argument B**

1. I will two-box
2. I will almost certainly be predicted to two-box
3. Therefore, I will almost certainly win exactly $1,000

**Argument C**

1. I will one-box
2. I will almost certainly be predicted to two-box
3. Therefore, I will almost certainly win exactly $0

**Argument D**

1. I will two-box
2. I will almost certainly be predicted to one-box
3. Therefore, I will almost certainly win exactly $1,001,000

But, as per the premise of the experiment, if “I will one-box” is true then “I will almost certainly be predicted to one-box” is true, and if “I will two-box” is true then “I will almost certainly be predicted to two-box“ is true, and so only Arguments A and B are applicable to the problem.

Therefore, we are left with:

1. If I will one-box then I will almost certainly win exactly $1,000,000
2. If I will two-box then I will almost certainly win exactly $1,000

This is the proof that it is most rational to one-box.

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