If this question doesn't belong here, PLEASE let me know and I will delete it. Not sure where else to post it.

Ran into a new "shake of the day" variant at a bar I visited over the weekend. It starts with a very large cup and in it are (12) standard size dice, (1) large red die and (1) large green die. Large being maybe 1" x 1".

For your first flop, you roll all (14) dice. Whatever the red die ends up being is the number you're shooting for and whatever the green die ends up being is how many rolls you get to get all (12) of the smaller dice to show what's on the red one. Obviously, the red die doesn't really matter because whatever shows is totally random and you want the green die to be a six. Also, after the first flop, if any of the small dice match the red die, they stay out of the cup and count as one (or some) of the twelve.

There were seven of us in the group and we each played 3 times and none of us were able to get the 12 small dice to match the red die. (The best we did was needing a three of kind on the final flop).

SO, the question is...........

What is the probability of getting 12 dice to show the same number when you get 6 shakes to do it when you can pull the matching numbers after each shake?

And really, if you count the first shake with all (14) dice and a few of those match the red die, a person would get seven shakes.

Just curious as I am stumped as to what the odds might be.