Hi Dane, if I understand your question correctly, here is my solution (note that I ignore the zero because it complicates things):

The probability of each DS hitting in 6 spins is 1.54% (see example 4 on

this page for the formula):

In the case of 3 adjacent, if we label the DS's 1-6 we could either have:

1,2,3 or 2,3,4 or 3,4,5 or 4,5,6. Each of these has probability (1/6)

^{3} and since they are mutually exclusive we just add them to get the total probability of getting 3 DS's adjacent in 3 spins:

4 x (1/6)

^{3} = 1.85%

So it's more likely that you will get 3 adjacent in 3 spins than all 6 hitting in 6 spins, but not by much.

**Edit : **I just noticed you said "at least" 3 adjacent, so the answer isn't quite correct, but calculating the exact probability will just increase the probability of 3 or more adjacent, so it still beats the "all hitting in 6 spins" scenario.