# Deadly Sixes Game: Explain the apparent probability paradox to me **URL:** \ **Category:** The Game Room\ **Created:** [April 7, 2023, 4:49pm UTC](https://boards.straightdope.com/t/deadly-sixes-game-explain-the-apparent-probability-paradox-to-me/982326 "2023-04-07T16:49:43Z")\ **Posts on this page:** 1\ **Showing post:** 121
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [April 16, 2023, 6:41pm UTC](https://boards.straightdope.com/t/deadly-sixes-game-explain-the-apparent-probability-paradox-to-me/982326/121 "2023-04-16T18:41:26Z")
My simulation gives two kinds of results - if I compute the mean number of losers divided by the mean of (winners+losers), I get 1/36 regardless of the (finite) number of rounds. The mean number of winners for 1-round, 2-round, 3-round and 4-round games are : 0.9718, 10.4356, 102.3750 and 996.1660 while the mean number of losers are 0.0282, 0.2896, 2.9552 and 27.7942. But if I compute the mean in another way (calculating the ratio of losers to losers+winners, first and averaging that ratio), I get the ratio that increases with the limit on the number of rounds: 0.0282 for one-round, 0.0513 for two rounds, 0.0756 for three, and 0.0986 for four. --- _[View the full topic](https://boards.straightdope.com/t/deadly-sixes-game-explain-the-apparent-probability-paradox-to-me/982326)._