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Forum Archive :
Probability and Statistics
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> X to play 6-3:
>
> 13 14 15 16 17 18 19 20 21 22 23 24
> +---+---+---+---+---+---+---+---+---+---+---+---+---+
> | O | | O O O O O |
> | | | O O O O O |
> | | | O |
> | | | |
> | | | |
> | | | |
> | X | | |
> | X | | X |
> | X X | | X O X X X |
> | O X X | | X O X X X |
> +---+---+---+---+---+---+---+---+---+---+---+---+---+
> 12 11 10 9 8 7 6 5 4 3 2 1
>
> Several additional shot numbers appear if you strip the 6 point,
> so 8/2, 4/1 seems clear.
Counting shots in positions like this is one of the most vexing problems
for me to solve. I'm sure that there are as many techniques as there are
ways to count pips. I just mentally make the plays under consideration,
eliminate all of the combinations that can be played entirely behind the
anchor with 1's, 2's, and 3's, and then hope that I don't run out of
fingers. Invariably at some point my brain gets full and I miss something,
or as I did in this case simply give up and go with "mo spares mo betta".
Does anyone have a "shorthand" method that they'd be willing to share?
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Raccoon writes:
# of die # of dice rolls
rolls that hit including
that hit combinations
-------- ------------------
1 11
2 12
3 14
4 15
5 15
6 17
It's easy to memorize column 2 and to add two numbers in column 2 and
subtract duplicates. For example, if 5s and 6s hit, then 15+17 = 32, less
the duplicated 6-5 and 5-1 (4 numbers), so 28 numbers hit. If 5s and 3s
hit, then 14+15 = 29, less the duplicated 5-3 and 3-2 (4 numbers), so 25
numbers hit.
# of dice rolls
# of die that include one
rolls that or more of those # of dice rolls
you can play numbers that don't
----------- ---------------- -----------------
1 11 and 11 = 36 - 5*5
2 20 and 20 = 36 - 4*4
3 27 and 27 = 36 - 3*3
4 32 and 32 = 36 - 2*2
5 35 and 35 = 36 - 1*1
6 36 and 36 = 36 - 6*6
Again, column two is easy to memorize, with the help of column three. For
example, if you can only play direct 1s safely, then of the 6 die rolls, 1
is that number you can play, 5 aren't. With two dice, 5*5 is 25. And 36-25
= 11. So 11/36 rolls play safely. Or for example, if you can only play
directs 1s and 2s safely, then of the 5 die rolls, 2 are numbers you can
play, 4 aren't. With two dice, 4*4 is 16. And 36-16 = 20. So 20/36 rolls
play safely.
Similarly, if you have one checker on the bar against a 4 point board, then
4*4= 16 numbers don't contain a number corresponding to one of the two
vacant points, thus 36-16=20 do.
Rewriting column 3 in the above table:
Dice rolls that
Die rolls contain only
that play those numbers
--------- ---------------
1 of 6 1*1 = 1
2 of 6 2*2 = 4
3 of 6 3*3 = 9
4 of 6 4*4 = 16
5 of 6 5*5 = 25
all 6 6*6 = 36
In the OLM position, all numbers containing a 1, 2 or 3 play safely.
Numbers contains only 4s, 5s, or 6s might not. Since 3*3=9 dice rolls
contain only 4s, 5s, or 6s, 36-9=27 numbers contain a 1, 2 or 3 and play
safe. Of the 9 numbers that might be bad, 6-6 5-5 and 4-4 obviously play
safely after any play. If we play 6/3 now, we *could* play 5-4 from the 6
point. So all that's left to worry about are 6-4 and 6-5. So 4/36 numbers
are really bad, forcing a shot, plus 2/36 for the merely ugly 5-4.
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Chase writes:
I've been meaning to write this up more formally, but since you brought the
subject up, here is a quick and dirty explanation of a couple of aids I've
come up with to simplify all this stuff. If I ever get around to writing it
up properly, I'll post a link to the article. First a couple of
definitions:
1. Extended hit: An extended hit is a hit that comes on the third or fourth
play of a doublet.
For example, if the target is 12 pips away, 66 hits "naturally" because it
can reach the target using just two of the four available 6s. 44 and 33
require the third and fourth play of the doublet, respectively, and
therefore are extended hitters.
2. Two-folder: a pair of numbers such that one is twice the other.
Example: 63, 42, and 21 are two-folders.
With that in mind, you can quickly derive the number of "hitters" as
follows:
For direct shots (i.e. withing 6 pips of target):
1) Add 10 to the distance
2) Add 1 for each extended hitter
Example. You have a direct shot at a blot 6 pips away.
1) 6 + 10 = 16
2) 16 + 1 = 17 (22 is an extended hitter at a distance of 6, coming on the
third available 2)
For indirect shots:
1) Subtract the distance from 13
2) Add 1 for each extended hitter
Example: You have an indirect shot at a blot 9 pips away.
1) 13 - 9 = 4
2) 4 + 1 = 5 (33 is an extended hitter)
For double direct shots:
1) Add the distances together
2) Add 16
3) Add 1 for each uncounted extended hitter
4) Add 1 if the distances are a two-folder
Example: You have direct 6s and direct 3s to hit the target.
1) 6 + 3 = 9
2) 9 + 16 = 25
3) 16 + 2 = 27 (11 and 22 are extended hitters)
4) 27 + 1 = 28 (The 63 pair is a two-folder)
Example 2: You have direct 4s and 2s to hit the target.
1) 4 + 2 = 6
2) 6 + 16 = 22
3) 22 + 0 = 22 (Note that 11 is an extended hitter for 4, but it shouldn't
be added since it already hits "naturally" as a 2)
4) 22 + 1 = 23 (The pair 42 is a two-folder)
Hope this wasn't so quick and dirty that it didn't make sense.
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Fabrice Liardet writes:
Thank you for sharing that, the method for double direct shots is cool and
was new to me. Here is one other counting method I developed, which I have
found especially useful. It is probably recorded in some books too.
It counts the number of ways to kill two birds with one stone, for instance
enter and hit, enter and cover, hit and cover. It counts only moves played
by two different checkers, e.g. it does not count as "enter and hit" a move
that enters a checker and hits an indirect shot with the same checker.
Let n be the total number of dices that kill at least one bird. Let a be
the number of dices that kill bird one, but not bird two. Let b be the
number of dices that kill bird two, but not bird one. Then the number of
rolls that kill both birds is n^2 - a^2 - b^2 (because it is the number of
rolls that kill at least one bird, minus the rolls that kill only one of
the birds).
E.g. if you enter with 1,2,4,5, and hit on the other side of the board with
2 or 3, then we have n = 5 (dices 1,2,3,4,5) ; a = 3 (1,4,5) ; b = 1 (3) ;
so that you have 5^2 - 3^2 - 1^2 = 25-9-1 = 15 rolls that enter and hit.
As for the original question of counting safe rolls, I don't think that
there can be such a simple method as that. I would have done it more or
less in Raccoon's way.
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