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Probability and Cumulative Dice Sums

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Solving a Bar Bet Using at Most Three Different Operations

A bar bet as presented in the YouTube video The HARDEST Puzzle Yet! involves starting with three of the same number from 0 through 9, then adding mathematical operations that result in an evaluation of 6. For example, if we start with three of the number 6, one solution could be 6+6−6=6. I'll demonstrate a method for solving this bar bet puzzle starting with three of any number, say N, which involves using at most three different mathematical operations (although some of these may be used many, many times).

If 0≤N≤2 we have the solutions
(0!+0!+0!)!=6(1!+1!+1!)!=62+2+2=6. If N≥3, concatenate the three numbers. Repeatedly applying the square-root operation we'll eventually end up with a result x with 3≤x<9.  If we now take the greatest integer ⌊x⌋ we have an integer n with 3≤n≤8. If we can exhibit solutions for each of these cases that use only square-roots, greatest integers and one other operation, we'll be done. Using factorial for the third operation, some possibilities are
3!=6⌊√√⌊√√√√√(4!)!⌋!⌋!=6⌊√√5!⌋!=66=6⌊√√√⌊√√7!⌋!⌋!=6⌊√√√8!⌋!=6.
I've added Wolfram Alpha links so you can verify that these do indeed evaluate to 6.

As an illustrative example, when N=1337 we have the solution ⌊√√⌊√√√√√(⌊√√√√133713371337⌋!)!⌋!⌋!=6.

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