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Chicago dice [1] - Each round is an independent event; the probability of winning each round is 1/36, 2/36, etc.; totaling each probability gives 36/36 = 1, clearly a wrong answer since it's quite conceivable that every round sees rotten luck. Where did I go wrong? Thank you. Imagine Reason (talk) 14:12, 28 January 2018 (UTC)[reply]
What are you looking for exactly? Do you mean the probability of winning at least one round? That would be
You didn't go wrong! You successfully calculated the expected value of the number of rounds won in one game. By realizing that the rounds were independent events, you calculated the expected value the easy way. The harder way would have been to calculate pn, the probability of wining exactly n rounds in a single game, and then sum , getting the same value 1. -- ToE20:35, 28 January 2018 (UTC)[reply]
Further study: If you play your Chicago Dice with only 1 die, giving six rounds per game numbered 1..6, then you have a 1/6 chance of winning each round and the expected number of rounds won per game is still 1. What if you play with 3 dice, giving 16 rounds, numbered 3..18? What if you play with n dice, giving 5n+1 rounds, numbered n..6n? What are the expected number of rounds won per game and why? -- ToE23:41, 29 January 2018 (UTC)[reply]
A relevant sequence is OEIS: A063260, Sextinomial (also called hexanomial) coefficient array, which "can be used to calculate the number of occurrences of a given roll of n six-sided dice". -- ToE23:52, 29 January 2018 (UTC)[reply]