So, I did a quick bit of math. You can pull one piece every hour for two weeks, but sleeping, not playing, missing days etc, meant that I effectively pulled ~100 pieces. The chance of missing a certain piece is 7/8, so missing a certain piece 100 times in a row is (7/8)^100, which is roughly a little bit more than one in a million.
The chance of failure would be 8 times that much because there are 8 pieces you could miss, which comes to about 12.7 per million.
Yes, but they're not independent. Given that you missed piece X, the probability that any one piece it's piece Y it's 1/7, not 1/8, so the chance of missing both X and Y it's lower than the product of their individual probabilities.
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So, I did a quick bit of math. You can pull one piece every hour for two weeks, but sleeping, not playing, missing days etc, meant that I effectively pulled ~100 pieces. The chance of missing a certain piece is 7/8, so missing a certain piece 100 times in a row is (7/8)^100, which is roughly a little bit more than one in a million.
The chance of failure would be 8 times that much because there are 8 pieces you could miss, which comes to about 12.7 per million.
You couldn't possibly miss all 8 pieces. As soon as you make one attempt you'll get one piece.
Henrik's math calculated the chance of missing the "I love you" piece each time. The chance of missing any one other piece each time is the same.
Yes, but they're not independent. Given that you missed piece X, the probability that any one piece it's piece Y it's 1/7, not 1/8, so the chance of missing both X and Y it's lower than the product of their individual probabilities.
'missing each of 8 pieces individually' is not 'missing a certain piece'.