# Puzzler of 02/18/2012: use Bayes' theorem: P(A|B) = P(B|A) \* P(A) / P(B)

**URL:** <https://community.cartalk.com/t/puzzler-of-02-18-2012-use-bayes-theorem-p-a-b-p-b-a-p-a-p-b/71022>\
**Category:** The Show\
**Created:** [February 18, 2012, 11:05pm UTC](https://community.cartalk.com/t/puzzler-of-02-18-2012-use-bayes-theorem-p-a-b-p-b-a-p-a-p-b/71022 "2012-02-18T23:05:30Z")\
**Posts on this page:** 2\
**Page:** 3

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**Author:** ![lion9car](https://avatars.discourse-cdn.com/v4/letter/l/59ef9b/32.png) [@lion9car](https://community.cartalk.com/u/lion9car)\
**Post date:** [March 14, 2012, 7:26pm UTC](https://community.cartalk.com/t/puzzler-of-02-18-2012-use-bayes-theorem-p-a-b-p-b-a-p-a-p-b/71022/41 "2012-03-14T19:26:15Z")

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_Barkydog wrote:  
No drawing cards from a bag, no nothing else, only 2 cards can satisfy the option of showing a red side! 50/50 chance._

I’ll probably regret getting involved here, but let me just make one statement. Just because two things can happen doesn’t mean there’s a 50-50 chance of each. Today you’ll either get hit by a bus or you won’t. Is there a 50-50 chance of each? That’s where you’re going wrong in your logic.

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**Author:** ![Triedaq](https://avatars.discourse-cdn.com/v4/letter/t/65b543/32.png) [@Triedaq](https://community.cartalk.com/u/Triedaq)\
**Post date:** [March 14, 2012, 7:47pm UTC](https://community.cartalk.com/t/puzzler-of-02-18-2012-use-bayes-theorem-p-a-b-p-b-a-p-a-p-b/71022/42 "2012-03-14T19:47:39Z")

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@lioncar–in fact, if someone bets even money with the stranger that emerges from the shadows, the proabiltiy the person betting with the stranger will lose is 2/3.

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