On Sun, 8 Mar 2020 07:28:36 +0000, Timothy Collinson - timothy.collinson at
port.ac.uk (via tml list) <xxxxxx@simplelists.com> wrote:

>                   And yes, math(s) challenged as I am, I'm still working
>out the odds of Dhe.

Rolling two fair dice, there are eleven rolls that can affect your bet. Of
those eleven rolls, only one represents a win; ten represent a loss. So, to
break even, you should expect a payoff on the winning roll of 10:1. But
somebody has to pay for all those flashing lights, slot machines,
felt-covered tables, et multae ceterae - and that's the house percentage,
so they only pay 9:1, for a whopping 10% advantage to the house (the house
advantage for roulette with an American wheel (0 and 00 both) is only just
over half that).

"What?" you say, "Eleven rolls? Not six?"

No, eleven. When you roll two dice, of two different colors, it becomes
obvious; consider the table below - the black die is across the top, the
red die is down the left:

          B L A C K   D I E
      1    2    3    4    5    6
   +------------------------------
R 1| 1 1  1 2  1 3  1 4  1 5  1 6
E 2| 2 1  2 2  2 3  2 4  2 5  2 6
D 3| 3 1  3 2  3 3  3 4  3 5  3 6
D 4| 4 1  4 2  4 3  4 4  4 5  4 6
I 5| 5 1  5 2  5 3  5 4  5 5  5 6
E 6| 6 1  6 2  6 3  6 4  6 5  6 6

So, if you've bet on "5", the rolls that can affect your bet are the
red-die row for 5 (six rolls), or the black-die column for 5 (six rolls),
both of which include one roll where both the red die and the black die
show 5, so 6 red 5s plus 6 black 5s, minus the duplicate where both are 5,
is 11. (And the same calculation holds for whatever number you bet on. If
you're playing with fair dice of other-than-six-sides, you can do the same
calculation, but the odds will be different, because the number of possible
bet-affecting rolls will be different - work it out to prove to yourself
that for any _fair_ dice, including non-platonic polyhedrals, of side (n),
the "break-even" payoff for the doubles should be (2n-2):1)

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