**When a double vtrac falls, (I'll try to use examples )..you can only play so many ways:**

**For example: say a double vtrac falls 533. (Tn's yesterday) and you have at least 2 due vtracs..in this case, we had the 2 (out 4), the 4(out 2), and the 1(0,5). The "linking" vtrac (from what fell previously) would be one of three choices: the 5 vtrac to come back, the 3 vtrac to come back or the 33 vtrac pair to come back. **

**Vtrac set up would look like this:**

**2 4 1****3****5 3 3 <-----to come back the 5, the 3 or the 33.**

**In this case, the 33 vtrac came back with the 2 vtrac which was out the longest (4 draws. The clue was the due digit ..a 6.**

**Example 2**

**Let me use ky mid's draws for the next example:**

**Their vtrac on Thursday mid was 511 (905). Their due vtracs were 2(1,6) out 3, 4(3,8) out 2 and 3(2,7).**

**their vtrac set up would have looked like this for this morning:**

**2 4 3****1, 5****5 1 1 <-------- to come back would have been the 5, the 1 or the 11 pair.**

**In this case they brought the 1 vtrac with the 2 and the 3 vtrac. Normally if I pick one from the bottom, then I would have had vtrac pairs 24,23 and 43 from the top. My due digit was a 6 so i could have eliminated the 43 vtrac pair. so now I have 24 and 23 vtrac pairs left. That gives me vtracs 124 and 123. That is 16 numbers. I normally will then try to use the root sums to narrow down my choices and/or the description list of what is due: **

**They had:**

**2 carried <--to get this I would have had to use 2 vtracs from the bottom****all even <--in both vtracs 123 and 124, there are only 2 even numbers****o carried-<this mea....**

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