# Another tricky math/probability question

**URL:** <https://boards.straightdope.com/t/another-tricky-math-probability-question/750420>\
**Category:** Factual Questions\
**Created:** [March 28, 2016, 12:50am UTC](https://boards.straightdope.com/t/another-tricky-math-probability-question/750420 "2016-03-28T00:50:25Z")\
**Posts on this page:** 1\
**Page:** 2

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**Author:** ![bldysabba](https://avatars.discourse-cdn.com/v4/letter/b/ecccb3/32.png) [@bldysabba](https://boards.straightdope.com/u/bldysabba)\
**Post date:** [March 28, 2016, 9:17am UTC](https://boards.straightdope.com/t/another-tricky-math-probability-question/750420/21 "2016-03-28T09:17:21Z")

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Here was my reasoning: Expectation where X is one draw is is 0.5. The expected value of two draws is 0.5+0.5 = 1. Since the casino will draw until the sum is **greater** than 1, three draws are more likely than not, and you should play.

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