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I've seen the following question asked and answered many times on various forums:

Knowing that the odds of rolling a six with one die are one in six, what are the odds of rolling a six in two, three, four (etc.) consecutive rolls? 1/6 + 1/6 + ... wouldn't make sense as you'd have a 100% chance of rolling a six in six rolls.

The answer I've seen is that the odds of rolling a six after n rolls are 1 - (5/6)^n.

Now my question is this - lets say that instead of rolling for a six, we pick a number out of a fixed sequence of six unique numbers after each roll. So for example, lets say our sequence is 1, 2, 3, 4, 5, 6. At the first roll we're looking to roll a one, the second roll we're looking for a two, then a three etc. until finally at the seventh roll we go back to one. What are the chances of matching any of the numbers in the sequence in this way after n rolls?

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Yes, if I understand your question correctly. In each roll there is one out of six numbers you consider a match, whether it is a six or any other number. Thus, $$P\{\text{At least one match in n rolls}\}=1-P\{\text{No match in n rolls}\}=1-P\{\text{No match in roll 1, no match in roll 2...}\}=1-(5/6)^n,$$ where the last step follows from independence of the rolls.

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  • $\begingroup$ @RoyG, great! If you consider the question answered, you can indicate you are satisfied by clicking the accept answer button next to the answer, or you can wait and also see what other answers may come. When you find an answer useful, you can also use the up-vote next to it to indicate so. Cheers! $\endgroup$ – ekvall Jan 27 '14 at 9:49
  • $\begingroup$ marked as answered and tried to up-vote but unfortunately it said I need at least a 15 reputation. $\endgroup$ – RKG Jan 27 '14 at 9:54
  • $\begingroup$ Yep, just FYI for future questions when your rep is higher. $\endgroup$ – ekvall Jan 27 '14 at 9:58
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    $\begingroup$ It's okay Karl, I upvoted for him. And he's a bit closer to 15 rep as well. $\endgroup$ – Glen_b Jan 27 '14 at 10:00

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