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Throw Until Match
You are rolling a normal six-sided die and you're recording the outcomes. You keep rolling, until you throw any number for the second time. What are the expected number of rolls before this happens?
Solution
We can define 
as being the expected number of rolls to finish after having rolled 
distinct outcomes. We know that 
, as it will only take one more roll to get a match (whatever it is!):
, there's a 1/6 chance of needing another roll, and a 5/6 chance that this roll finishes the game. We already know . Let's work this out:
as being the expected number of rolls to finish after having rolled distinct outcomes. We know that , as it will only take one more roll to get a match (whatever it is!):
means you finished six rolls. The fact that you reached the seventh roll means that the first six rolls were all distinct numbers. Since a six-sided die only has six distinct outcomes, it means the seventh throw will always finish this game.
, there's a 1/6 chance of needing another roll, and a 5/6 chance that this roll finishes the game. We already know . Let's work this out: (1) 
(2) 
(3) 
(4) 
(5) 
(6) 
(7) 
(8) 
(9) 
(10) 
(11) 
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