### Tables to Improve Your Game

Lesson of the Week—Tables to Improve Your Game

This week’s lesson is about tables.  In the past week Perry has been teaching these tables, and the importance of them, to our students.

The Hitting table below tells us how likely you are to hit a shot.

Hitting Table

 Number away 1 2 3 4 5 6 7 8 9 10 11 12 Number of hits 11 12 14 15 15 17 6 6 5 3 2 3

If your opponent has a blot that is 5 pips away, you will hit that blot 15 out of 36 times, provided, of course, there are no points made by your opponent between you and the bot.

So, if you have a checker 5 away and another 8 away, you now know that you can hit 15 plus 6, or 21 times, however, you cannot count 5-3 twice, you the total is 19.

Having these numbers off the top of your head, instead of having to calculate them, can save you a lot of time over the board as well as assure that you are using the right numbers.  This will often help you determine whether to, or where, to leave shots yourself, and of course, it will also be useful knowledge in your cube decisions.

Here’s another table that you should know.  It is the odds of coming in off the bar.

Coming In Table

 Open Points 1 2 3 4 5 6 Ways to come in 11 20 27 32 35 36

So, if you opponent has 4 points made in his inner board, that means he has two points open.   Your odds of coming in on the first rolls are 20/36.

There are many other tables and numbers we should all learn over time that I will not go into now.  For example, if you have 3 builders to make a point, there are 9 rolls out of 36 that will make the point (provided your opponent doesn’t have any points made that block rolls).

Here’s a number I learned just today from Bob Koca (Herb Roman also said this was covered in a book by Magriel).     If you have a checker on your opponent’s ace point, and he has a 5-prime from his 3 through 7, how many rolls will it take for you to get out (roll the 1-6 or roll a 1 and then a 6).

What do you think?   Scroll down for Bob's answer:

Here are a couple ways to calculate it.

1)  Suppose the question was just how long to get an ace. An ace occurs 11/36 of the time and the expected number
of rolls is the reciprocal 36/11. For more on why let me direct you to read about geometric distributions.
Similarly it will take 36/11 turns on average to get the 6.  So that is a total of 72/11.  However you might get lucky
and get the 6 with the same roll as the 1.  That will happen 2/11 of the time (2 of the 11 rolls with an ace also have the
six).  This will save the 36/11 rolls needed for the 6.  Thus, overall we have 72/11 - (2/11)(36/11) which is about 5.95

2)  Another way is to set up a recursive equation. Let's call E the answer to our question. There will be definitely at least
one roll. 2/36 of the time we got a 16 and need 0 more rolls. 9/36 of the time we got a 1 without a 6 and thus need
the 36/11 more on average to get the 6.  25/36 of the time we get no ace and are back where we started, remember that
we gave a name of E to this amount.

Thus we have the equation  E =  1  + (2/36)(0) + (9/36)(36/11) + (25/36)E
Multiplying out the fraction, adding the 1, and subtracting (25/36)E from each side results in
(11/36)E = 20/11.
Dividing each side by (11/36) we obtain  E = (20 * 36)/ (11 *11) which is about 5.95   (exact value is 5 and 115/121)

Now that I see the math, it's a wonder I never stumbled upon the answer myself.  Maybe because I am not good at multiplying???

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