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Math novice here. With a 10-sided die, the probably of rolling '1' is 10%. I'm tempted to think the probability of rolling '1' with two consecutive rolls is 20%. Would I be correct?

Not sure if I need to factor in the first roll i.e. 10% + (10% - probability of NOT rolling 1 in the first roll). Or am I overthinking this?

CLARIFICATION: I mean the probability of rolling a 1, then another 1.

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You need to clarify your question. Do you mean the probability of rolling a $1$ at least once, exactly once, or both times? –  Isaac Solomon Jan 16 '13 at 5:38
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Getting two $1$'s in a row must be less probable than getting the first one. You have to start with the first one, then can fail on the second roll. B.D has the correct answer, but this might help with intuition. –  Ross Millikan Jan 16 '13 at 5:45
    
@RossMillikan makes an excellent point. If you are trying to sync this intuition up with the array visualization I posted below, the single $1$ probability corresponds to the top-left square among just the squares in the top row; the two consecutive $1$s probability corresponds to that same square among all $100$ squares. –  Benjamin Dickman Jan 16 '13 at 6:03
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up vote 3 down vote accepted

Think of a $10 \times 10$ array of squares, where each square represents a possible roll. For example, we could say the square in row $a$ column $b$ corresponds to rolling an $a$ first, and then rolling a $b$.

With these $100$ different possibilities, only one of them - the one in the top left hand corner - corresponds to rolling two consecutive $1$s.

Therefore, the probability is $1/100 = 1\%$.

(More generally, you want to be multiplying the probabilities of independent events rather than adding them. This sometimes goes by the name of the "multiplication rule.")

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What a wonderful way of visualizing it. Thank you very much. –  hoipolloi Jan 16 '13 at 5:46
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You could also visualize it by drawing a tree: 10 different points for each of the possible rolls, then each of those points gets 10 possible points for the subsequent roll. Again, you will end up with 100 possible two roll scenarios, of which only one is 1 followed by 1. This "tree approach" might make it clearer why you are multiplying probabilities instead of adding them. –  Benjamin Dickman Jan 16 '13 at 5:48
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