Hello all I am a pretty new player and going through some books. I am reading Walter Trice's "Backgammon Boot Camp" and just starting to get familiar with all of the formulas and such. One thing I have had a hard time processing is wastage. Can anyone explain that to me and this....."White has a literal pipcount of 26 (one checker on the 4-point, two on the 5 and two on the 6) with only five checkers his wastage is about 6 pips." I am not sure how that was calculated. Any help would be very much appreciated.

Thanks,

Rich

You can't really calculate it, I think (unless you're a math savant). Except for very simple positions (like n-roll positions) you have to rely on experience and reference positions. If you have GNUBG, then it has a nice feature where you can see wastage and epc (current pipcount+wastage) for endgame positions.

Hi,

i had to think about your question for a while, but i think i know how it's calculated.
And it is not so difficult.

The average number of a throw is 8 1/6.

So in average you will bear 26 pips off in 4 moves. 4x 8 1/6 is 32 2/3.

Now the wastage = 32 2/3 - 26 = 6 2/3. That might explain the 6 you mention, maybe the writer calculated with an average of 8 instead of 8 1/6.

greetings k.

I'm fairly certain the numbers he mentioned are found by computer simulation. Gnubg says the wastage in the mentioned position is 5.89 btw.

The general idea is that if your pip count is, say, 30 pips, but on average you will roll 35 pips before you bear off all your checkers, then your wastage is 5 pips. You'd calculate it by doing rollouts of many bearoffs and noting the wastage after each one, then computing the average over thousands of bearoffs. It's likely there's an analytic way to calculate wastage without resorting to simulations, but the simulation method is certainly the easiest to understand.

I think it was called hyperpips in the old books

Like OP, I have a problem with wastage but it is a slighty different one.

I have let us say 5 checkers on my one point as I am bearing off. I am off in 3 rolls which gives me 8x3=24 but I only need 5 so I have a huge wastage of 19 pips.

My opponent has 5 checkers spread a checker a point over his first five points giving him a pip count of 15 and he too will need 3 rolls of 8 pips, so his wastage calculation is 24-15 = 9.

His wastage is not as high as mine but surely my position is superior to his.

I truly do not understand this.

Could someone please explain?

His wastage is lower yes but his pipcount is higher. So for you 5 + wastage = your efficient pipcount (epc). For him 15 + wastage = his epc.

Your position is very easy to calculate, because n-roll positions follow a very exact formula, where epc = 7n+1, so your epc is 22 (and your wastage is thus 17 pips). His position would be more difficult, but gnu tells me his wastage is 8,35 for an epc of 23,35. Add to this that you're on roll and it should be clear that you're a big favorite.

Of course over the board all you wouldn't need all this to figure out that this is a pass.

You can't really calculate it like this because if you roll doubles you will get off in two rolls.

Thanks for your post. How do you get to the number 22?

The rolling of doubles cancels out yes?

7*number of rolls (without doubles) + 1. I'm not enough of a math geek to tell you how to get to this formula I just know it's right.

Here's the simple math behind a two roll position ((1/6)*1)+(5/6)*2)*8,16666 = 15 (which is the same as 7*2+1)

Not exactly. Your opponent won't save a roll with 33, 22 or 11 (at least on the first roll, sometimes they work on the second roll).

My point just was, that your adjusted pipcount of 24 was a bit too high, because it didn't take doubles into account.