Forum Archive :
Theory
Error ratesRepeated ND errors

If you are like me you get dinged more than anything else for not doubling
when you should. Often I will miss a cube by .05 or so for several shakes
in a row. That has led me to consider the fairness of the grading of these
errors.
Say, hypothetically, you had a position that was static for 25 rolls and
for each roll you made a .05 error by not doubling. Your cumulative error
would be 1.25, but you would have only been .05 better off if you had
doubled at any time.
Now say, in another 25roll static position, it was wrong to double by .05
and you doubled on the first roll. Your cumulative error is only .05,
twenty five times less than the other position, but the true cost of the
errors was identical.
I realize in one situation you made one bad decision and in the other you
made 25 bad decisions, but the cost in MWC of the 25 was truly no more than
the cost of the one.


Neil Robins writes:
A genuine repeating error can occur when both you and the opponent have a
checker on the bar against say four point boards. The problem is that if
its a 0.05 missed double, the 0.05 is only valid on the assumption that if
the position does repeat you then double. So you need to either take the
0.05 error each time it comes up, or else work out what the initial error
would be assuming you repeat the error each time, which is clearly a lot
higher than 0.05.


Matt Reklaitis writes:
A bot's evaluation represents values related to its own play. So when
calculating the size of any individual error, inherent to that calculation
is that your future play is similar to the bots. The more your play differs
from the bot's, the more the model breaks down. Another way of saying this
is that I suspect a bot calculates low error rates more accurately than
high error rates.
As an example: Lets say a bot thinks a position if worth 73% and is a cube
but you think it is worth 72% and is just short of a cube. The bot dings
you .05 for a missed double. If you keep repeating the same mistake in
sequence, are you really losing .05 each time? Arguably no, because the way
you are playing it (assuming the bot's play is correct), you are not really
getting 73% out of the position, so you're not really giving up .05. But
the argument seems a nonstarter because a bot cannot model how a
particular human plays  all it has for error calculation are methods
based on its own play. So even if you are arguably giving up less than
.05 * N, there's no practical way for a bot to measure that.


Matt CohnGeier writes:
13 14 15 16 17 18 19 20 21 22 23 24
++++++++++++++
 O O O   X O O O O O 
 O   O O O O 
   
  X  
   
   
   
   
  O  
   
 X   X X X X  ++
 X X X   O X X X X X   2 
++++++++++++++ ++
12 11 10 9 8 7 6 5 4 3 2 1
X to roll.
Let's say the proper cube action here is double/take and your error here
in not doubling is .05. This assumes that after fan/fan you will cube.
But after fan/fan you won't cube ... so your error is compounded more than
.05 would indicate. So regarding the error as just .05 isn't sufficient.


Joe Russell writes:
If you take the position and call it a 5point match, the error in MWC for
not doubling is about .9%. If each side dances three times and now you
roll without doubling again, you have cost yourself .9% in match winning
chances by never doubling, not 2.7%, the cumulative MWC error. In an
absurd extension of this, suppose each side danced a 100 times. Do you
think that you would have cost yourself 90% MWC, or closer to (and I think
exactly) .9% by never doubling?


Matt Reklaitis writes:
In the dance 100 times case, you cost yourself 90% and the dice gave you
back 89.1%; i.e., you gain .9% from the sequence fan/fan. (Well, here, its
not all dice. Opp is giving some back by not doubling too.)


Maik Stiebler writes:
I prepared a puzzle involving a discrete random walk to post at some point
in this thread:
Players A and B play a game of StaticishRace. A game consists of tossing a
fair coin and changing the score by +1 or 1 respectively based on the
result of the coin toss. The starting score is 0. Player A wins one point
and the game ends when the score reaches 50. Player B wins one point and
the game ends when the score reaches 50. Before each coin toss, Player A
(only player A!, to keep things simple) is given the opportunity to double
the stakes, after which Player B can either take or drop (ending the game
and losing a single point). A double is allowed only once in a game.
1. What is the optimal strategy for both players?
2. What is the theoretical (assuming optimal play from both sides) equity
of the starting position?
3. Assume Player A deviates from optimal strategy by doubling if and only
if the current score is +30. What is the practical equity of the
starting position then?
4. How much does Player A's deviation from perfect play cost him on average
per game?
Now put yourself in the position of a bot that knows the optimal strategy
and observes the game, not knowing Player A's complete strategy, but noting
the wrong plays that follow from the strategy.
5. a) At which points in the game does Player A, following the nonoptimal
strategy, blunder away theoretical equity by making a wrong play?
b) How much equity does each of these wrong plays lose?
6. How often will the opportunity for Player A to make a wrong (equity
losing) play arise in a game? Compute both an average value (a) and a
distribution (b).
7. Verify that the average number of blunder opportunities (6a) times the
cost of a blunder (5b) equals the total cost of Player A's misguided
strategy (4).


Bob Koca writes:
> 1. What is the optimal strategy for both players?
Player A doubles at +25 at which point B has an optional take/pass. That
is because B is then has the required 1/4 win chance. A does not double
before then as there is no market loss.
> 2. What is the theoretical (assuming optimal play from both sides) equity
> of the starting position?
A has +1/3 equity. Going from 50 to +25 is 75 and A starts 2/3 of the way
there. 2/3  1/3 = 1/3.
> 3. Assume Player A deviates from optimal strategy by doubling if and only
> if the current score is +30. What is the practical equity of the
> starting position then?
A wins 50/80 of the games for an equity of +1/4
> 4. How much does Player A's deviation from perfect play cost him on
> average per game?
1/3  1/4 = 1/12
Now put yourself in the position of a bot that knows the optimal strategy
and observes the game, not knowing Player A's complete strategy, but noting
the wrong plays that follow from the strategy.
> 5. a) At which points in the game does Player A, following the non
> optimal strategy, blunder away theoretical equity by making a wrong
> play?
> b) How much equity does each of these wrong plays lose?
Only when the game is at exactly 25 does A lose equity. Values below +25
are a nodouble and values above +25 are an optional double. The cash is
not lost in the next sequence.
At +25, B has an optional take/pass. Let's suppose he would pass. Then
the equity lost by not doubling at +25 is 1/2 the equity lost by the game
reaching +24 (if it reaches +26 it is a cash anyways). At +24, A's
theoretical equity is 74/75  1/75 = 73/75, a loss of 2/75. So failing to
double at +25 costs 1/75 equity.
> 6. How often will the opportunity for Player A to make a wrong (equity
> losing) play arise in a game? Compute both an average value (a) and a
> distribution (b).
It happens at least once in 2/3rds of the games. Suppose that the game is
exactly +25. We need to find the probability of that state occurring again.
This is harder than the previous questions. When you are at +25 you go to
+26 half the time. From there you go to 25 before 30 with probability 4/5.
The other half the time you go to 24 and then the chance of returning to 25
before 50 is 74/75. (1/2)(4/5) + (1/2)(74/75) = 67/75 chance of a repeat
visit to 25 if start from 25. The expected number of times needed to not
repeat has a geometric distribution with p = 8/75 and has an expected value
of 1/(8/75) = 75/8. The expected total number of visits is thus (2/3)
(75/8) = 25/4.
The distribution is as follows:
Exactly 0 visits occurs with probability 1/3
Exactly 1 visit occurs with probability (2/3)(8/75)
Exactly 2 visits occurs with probability (2/3)(67/75)(8/75)
Exactly 3 visits occurs with probability (2/3)(67/75)**2(8/75)
...
Exactly n visits occurs with probability (2/3)(67/75)**(n1)(8/75)
(This was jointly done with Chris Yep).
> 7. Verify that the average number of blunder opportunities (6a) times the
> cost of a blunder (5b) equals the total cost of Player A's misguided
> strategy (4).
(25/4)(1/75) = 1/12.




Theory
 Derivation of drop points (Michael J. Zehr, Apr 1998)
 Double/take/drop rates (Gary Wong, June 1999)
 Drop rate on initial doubles (Gary Wong, July 1998)
 Error rateWhy count forced moves? (Ian Shaw+, Apr 2009)
 Error ratesRepeated ND errors (Joe Russell+, July 2009)
 Inconsistencies in how EMG equity is calculated (Jeremy Bagai, Nov 2007)
 Janowski's formulas (Joern Thyssen+, Aug 2000)
 Janowski's formulas (Stig Eide, Sept 1999)
 Jump Model for money game cube decisions (Mark Higgins+, Mar 2012)
 Number of distinct positions (Walter Trice, June 1997)
 Number of nocontact positions (Darse Billings+, Mar 2004)
 Optimal strategy? (Gary Wong, July 1998)
 Proof that backgammon terminates (Robert Koca+, May 1994)
 Solvability of backgammon (Gary Wong, June 1998)
 Undefined equity (Paul Tanenbaum+, Aug 1997)
 Underdoubling dice (Bill Taylor, Dec 1997)
 Variance reduction (Oliver Riordan, July 2003)
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