Today Ive found a quizz and have sent my solution per e-mail today.
Even when Ive sent it already, I would like to ask you, if I have overseen
anything and what do you say about the solution? I needed more than 1 hour
Another question please: Once Ive learned the correct math at gammonline.com
for this type of positions, but meanwhile I forgot the straight way and
calculate it too much complicate(Still I hope the nubers are correct). Could
anyone please tell me me the easier (and common) way to calculate such
And here is the quizz and my solution:
How in the world could it possibly be
that he was favored more than me?
A bear-off position, with two men each,
but a lesson to me, it surely did teach.
He doubled me, it could not be believed.
I scooped up the cube and beavered with greed.
I said, "I'm a favorite to get off in one roll.
You sir are not. Now you'll pay the toll."
He cast his dice and rolled a failing four pips.
I smiled and said, "Now you're in quite a fix."
He said, "We were tied in the pip count;
it was right to double."
I said, "Now I'll send back the recube
for all of your trouble."
It surprised me to see his calm disposition-
to take an 8-cube in the resulting position.
With one mighty shake, as hard as I could,
out came a double, but it was no good.
"I was favored when you cubed me!"
My temperature was hot.
He looked at me, then shook his head
and countered, "You were not."
Almost two-thirds of the time,
he would win, it was shown,
because often my dice
wouldn't even be thrown.
It was then and there, I was forced to relent
that my wins totaled well under 40 percent.
The correct solution(s):
He has 2 checkers on his 3-point.
You have 1 checker on your 5-point and 1 checker on your 1-point
You have 1 checker on your 4-point and 1 checker on your 2-point.
Lets examine it:
You say, after He failed to bear off:'With one mighty shake, as hard as I
out came a double, but it was no good.'
That means in first order, that there must be at least 5 pips left.
There are 2 possible positions where are 5 pips left:
1 ckecker on 4-point and 1 ckecker on 1-point (29/36 probability to bear off
in 1 roll) and
1 ckecker on 3-point and 1 checker on 2-point (25/36 probaility to bear off
in 1 roll).
He said:'We were tied in the pip count;
it was right to double.'
You said:'I'm a favorite to get off in one roll.
You sir are not. Now you'll pay the toll.'
In both positions, where you are 5 pips away, you are favorite to bear off
in 1 roll, so you both
must have been more than 5 pips away.
Lets make the 2nd step:
How about the positions, where you two are 6 pips away?
In this case there is just 1 possible position, where you are not favorite
to bear off in 1 roll:
When both ckeckers are on the 3-point (17/36 probability).
The other 2 possible positions are equal:
If 1 ckecker is on 5-point and 1 checker on 1-point or if 1 checker is on
4-point and 1 checker is
on 2-point stays the same. Both have a 23/36 probability to to bear off in 1
roll and all 3 positions
have a 100% chance to bear off in 2 rolls.
What about the other positions when both are 7 pips or more away?
In this case there is just 1 possible position, where you are favorite to
bear off in 1 roll:
1 checker on 5-point and 1 checker on 2-point (19/36 probability).
His checkers must have been on 6- and 1-point (15/36) or on 4- and 3-point
(At the other positions, where are both 8 pips or more away the fact, that
one is favorite to bear off
in 1 roll isnt possible anymore).
That leads to following, what You said:'Almost two-thirds of the time,
he would win, it was shown,
because often my dice
wouldn't even be thrown.'
Well, the point here is, that in both cases, where you both are 7 away He is
MORE than 2/3 favorite, and not
just 'Almost two-thirds of the time'!
The Take was correct, but the Beaver was too much.
After He Doubled and you owned the Cube and He failed to bear off, then it
was correct to Re-Double,
but of course it still was a Take (He had still 36.11% (13/36) to win the
In both correct solutions (see above) he was 66,3% favorite...as you said:
'Almost two-thirds of the time'.!
P.S.: Sorry for my bad english
That's a cute little puzzle.
With problems like these, I like to draw a tree showing the
possible outcomes. In the diagram below, P1 is the first position
before O rolls. P2 is the position that results after O rolls and
fails to bear off. There are three possible outcomes shown at the
bottom of the tree. The numbers along the branches show the
number of times out of 36 that that branch will be followed.
17 / \ 19
O wins P2
13 / \ 23
O wins X wins
From position P1, 17 rolls win immediately for X; the other 19
rolls lead to position P2. From position P2, 23 rolls win
immediately for X; the other 13 rolls lead to a win for O.
It is easy to see that there is only one way for X to win: O has
to fail to bear off (19/36 probability), then X has to succeed in
bearing off (23/36 probability). These are independent events,
so their combined probability is: 19/36 * 23/36 = 437/1296 = 33.7%
If X wins 33.7% of the time, then O wins 66.3% of the time.
Followup quiz: What is O's equity at P1? In other words, how much
can O expect to win on average from this position, assuming correct
play by both sides?