In "The Numbers Game," a state lottery, four numbers are drawn with replacement from an urn containing the digits 0-9, inclusive. Find the probability of a ticket holder having a winning ticket consisting of two specified, consecutive digits in exact order (the first two digits, the middle two digits, or the last two digits).

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What's the total number of possible drawings? Since there are 4 numbers and 10 possible digits for each, you have 10^4 total drawings. Case 1: First two digits are consecutive (I'm assuming consecutive can mean 1,2 or 1,0). The first number has 10 possibilties, the second number must be wither one up or one down, so two possibilities. The third and fourth number still have 10 possibilites ( unless no other pairs can be consecutive, in which case the third number has 9 possibilities and the fourth has 8 possibilites.) The cases for the middle two and last two will be similar. Add up these possibilities and divide by the total number of possible drawings. This is the rough idea, you might have to consider edge cases and whatnot since the problem was ill-defined.

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