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October 1st, 2017, 06:13 PM  #1 
Newbie Joined: Oct 2017 From: USA Posts: 1 Thanks: 0  Probability question
Hi, hopefully I can explain this well. So let's say you roll for a chance to get a number from 121. Every time you get 21, you roll for a chance to get 118. You do this until you get every number in 118 (1,2,3,4,5,6,7, etc) once. How many 121 rolls would this take? And can you provide a formula of some sort? Thanks!

October 3rd, 2017, 06:06 PM  #2 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 561 Thanks: 79 
The formula for how many times it will take to get every number from 1 to 18 is the sum of 18/x for the 18 values of x. To the nearest whole number, the expected numbers of times this will take is 63. The expected rolls of the 1 to 21 where only 21 is a success is 63*21 = 1,323. I'm not guaranteeing that I understand the question, and it's important to know that the probabilities you're asking about do not have guarantees. For example, you could hypothetically roll a 10 for 100 consecutive times. A question like "how many coins does it take to guarantee at least 2 heads" may sound simple, but it doesn't have an answer.


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