Problem of the Week #74: August 29


Cash game, center cube.




(a) Should Black double? If he does, Should White take?




(b) Should Black double? If he does, Should White take?


Note: All ‘cash game’ problems assume the Jacoby Rule is in effect. That is, you can’t win a gammon unless the cube has been turned.

Black should double both games and White should drop both cubes. 5ive's are really intresting in the second game

In Problem of the Week #73, Bill Robertie provided a simple reference position and rule for this kind of problem. His basic "5-10 Rule" for "One-Man-on-the-Bar" positions gives Black a redouble when he trails by no more than 10 pips, and lets White take when his lead is at least 5. I like the restatement by Uberkuber calling this the "5-10-11 Rule." It reminds us that Black can make an initial double when trails by as much as 11 pips.

Problem 74a modifies the reference position by shifting White's gap from his six point down to the four. White's pip count has been held constant at 66 by moving his spare on the five point down to the one. The result is that White's lead is unchanged at 8 pips. Black may be hindered, however, after a hit by White. If he is hit, and then enters by rolling 42 or 41, he won't be able to jump into White's outer board. Instead, the 1 or 2 will be played inside, wasting pips in the race. Thus Black may want to be a somewhat slower to double.

The other modification in Problem 74a is that the cube has be placed back in the center. This change makes it a tad easier for Black to double. To be precise, about one pip easier. For Black, these two changes make for a wash. He should still double.

White's take, however, may be problematic. His gap on the four point is dangerous. It's not the type of gap you can fill when you roll the missing number. An adjustment should be made for the gap, perhaps adding 2 to 4 pips to White's count. Furthermore, White has also wasted pips by placing a fourth checker on his one point. It will probably be borne off inefficiently. Most racing formulas would add 2 pips to White's count for this. After these adjustments, White no longer has the 5-pip advantage he needs for a take.

My solution for Problem 74a: Double, pass.

Problem 74b is more difficult to analyze. Both players have been ******ed substantially. For Black, this means an increase of 17 pips. His his spare men on the three and four points have been moved back to the ten point, and his trailers slipped backwards an additional 2 pips each. Similarly, White's count is up by 17. His checkers from the three and five points have been shifted back to the midpoint. White now leads in the race by 7 pips, 84 to 91. We are still in the double/take range according to our rule.

Against this must be weighed the increased chance of contact, and -- especially for White -- the increased chance of losing a gammon. If we call the outer board contact a wash, granting that whoever hits first will gain a big advantage, then it is the gammon threat that turns the tide. For here, the action is mostly one-sided. Black has all five spare checkers in play, available to make his five point. Barring that, Black has good pick-and-pass chances. Problem #73, our reference which is essentially gammon-free, has morphed into a position where White might lose a gammon 15-20% of the time.

My solution for Problem 74b: Double, pass.

For the Record
I am so often wrong that I like to post my record in these messages. It's kind of a truth-in-advertising thing. Grunch: I have been answering these problems without the use of a bot, and before checking the excellent solutions of others, since Problem 28. My record at this writing is 48%.

Appreciate the analyisis Taper_Mike

Yeah, double/pass for both IMO.

In (a), the inefficient bearoff position is the problem for White. That said, it's easy to read too much into such things. I wouldn't be all that surprised if this was still a thin take, but over the board I would just let it go.

In (b) I think both the contact and the gammons now favor Black. The 10-point is a nice base of operations both for bringing his stragglers around safely, and for hitting if White enters. White may try to enter, stay there, and then get a shot when Black tries to clear the 10-point, but between pick-and-pass numbers and other safe numbers, there are plenty of games where that won't happen at all. Throw a few gammons in there, and I'm not at all tempted by the cube as White.

Regarding Problem 74b

On reflection, I think you're right about this. I've overestimated the impact of gammons, while underestimatiing Black's advantage with other types of contact. Both are important.

If White is lucky enough to come in, but unlucky enough to miss a fly shot in the process, he'll find his moves are almost all forced. He's stripped everywhere, and will have to choose between running out to meet Black's entourage, breaking his midpoint, or crashing his board.

BTW: I miscounted above. In Problem 74b, White has upped his count by 18 pips, not the 17 mentioned in my solution.

Hi,

this time i couldn't resist reading the other solutions first.
I think Taper Mike is correct with his analysis, there's only one extra variation in the b) part I would like to mention.
What if black throws D3,D6 or 63. Even those, at first sight, bad numbers for black, don't cost him that much.
White will have a little bit more control on the outfield after entering, since the black checkers are a bit disconnected.
I don't think this weighs up against the extra gammons, so it's still DP.

greetings k.

Based on the rule stated by bill in the last post double/drop both. Great analysis also mike, very thorough.

(a) It looks like last week's problem, or at least, the parameters look the same. Black is trailing the race by 8 pips (74-66), so it would be double and take.

Double / Take



(b) Black trails by 7 pips(91-84). He also blocks 5-5. But White has a point in the outfield that could prove useful. I'm not sure if it cancel each other, but I'll still go with double and take, although it's probably a thin take because White's 6-pt is open.

Double / Take

Just read the other posts.

Sigh!

Grunch: (a) double/take, (b) double/take.

a) No double/take. Black is too far behind in the race to warrant a double

b) Black should double, I think that is pretty clear. He will get a lot of shots at white and he can still close white out. White is a bit worse in this position than in position a. I think he may have to drop since it will be very difficult to escape until black has brought all his men in. After that, white is far behind because he'll also have the checkers on the 12 point to bring in.

Answer - double/drop

Looks like I don't know how to count pips. I forgot to count the checker on the bar! Yeah, position a should be double/take

a) The gap on four point would be more severe disadvantage if more checkers were on higher points. So , as with the last problem, Double/ take.

b) In the last problem Robertie stressed importance of black making the point 6 points away from gap. Here it is 5 points away. Although this turn not many numbers close out the board, black will be able to arrange checkers at minimum risk to close out / contain white. Double /pass.

Problem 74a
A - Should Black double?
Black has a good position:
- Black has a 5-point board
- White has one man on the bar.
Black can expect White to stay on the bar for three rolls.
Black is trailing by 8 pips: his pip count is 74, White’s is 66. In this position, Black will probably succeed in moving his two back men into his home board with only two rolls.

For these reasons, Black should double.

B- Should White take?
White should take a double if he has at least 25% of winning the game.

Against the 5-point board, White has 31% chance to enter.

1) White is ahead in the race with 8 pips (more than half a roll). It’s good for White: it increases his chance to win.
He has also an exceptionally good roll to win if he rolls 5-5.

2) The open point is the 5-point.
- If Black leaves one blot in his outer board, the White’s best move is probably to hit the blot or to leave his checker at the 5-point.
- If Black succeeds to move one checker into his home board, White’s best move is probably to run.

It seems that Black has numerous opportunities to hit White using single number.
This factor decreases White's chance.

3) White's distribution is not very good.
White has 4 men in his As point and an open point: the 4 point. White will probably waste rolls.
Given 1, 2 and 3, I think the chance to win is less than 25%. So White shouldn’t take.

My solution: Double, Pass.


Problem 74B
A - Should Black double?
Black has a good position:
- Black has a 5-point board
- White has one man on the bar.
Black can expect White to stay on the bar for three rolls.
Black is trailing by 7 pips: his pip count is 91, White’s is 84. In this position, Black will probably succeed in moving his two back blots into his home board with only two rolls.

More over, Black has two jokers : 5-1 and 5-5.
For these reasons, Black should double.


B- Should White take?
White should take a double if he has at least 25% of winning the game.

Against the 5-point board, White has 31% chance to enter.

1) White is ahead in the race with 7 pips (more than half a roll). It’s good for White: it increases his chance to win.
He has also an exceptionally good roll to win if he rolls 5-5.

2) The open point is the 5-point and Black has two men in the 10 point.
Timing is good for Black. Black has numerous opportunities to hit White using single number. This factor decreases White's chance.

Given 1 and 2, I think the chance to win is less than 25%. So White shouldn’t take.


My solution: Double, Pass.

Why do you say three rolls instead of two?

Here's a table lising White's chance to enter on a given roll ("P" stands for "Probability"; parenthesized expressions should be multiplied):

31% = P(enter on roll 1) = (11/36)
21% = P(enter on roll 2) = (11/36)(25/36)
15% = P(enter on roll 3) = (11/36)(25/36)(25/36)
10% = P(enter on roll 4) = (11/36)(25/36)(25/36)(25/36)

Note that White is a slight favorite to enter on one of his first two rolls.

Hi,
I don’t know if I’m wrong: 11/36= 31% ~ 33% and 1/33% = 3.

Hey, Yules,

I see what you are thinking. You have fallen victim, however, to a common fallacy. You are adding probabilities that are not independent of one another. For instance, if Black enters on the first roll, then his chance of entering on the second roll is zero, because he is already in. Only if Black fails to enter on the first roll is his chance of entering on the second equal to 11/36. The chance of entering on the second roll depends on what happens on the first.

The easy way around the problem is to construct probabilities for events that are mutually exclusive. The events and their associated probabilities are then independent of each other. Independent probabilities can be added, if desired, to find the probability that one or the other of the mutually exclusive events might occur. The other trick you need involves the multiplication of independent probabilities when you want to find out the chance that two independent events will both occur. Hence, my tables above.

For Black to enter on the second roll, he must fail to enter on the first roll. The odds of this are 25/36. Then, on the second roll, he has an 11/36 chance to enter. By multiplying these two independent probabilities, we find the chance of both failing to enter on the first roll and then entering on the second roll. The result is about 21%. Adding this to Black's independent 31% chance of entry on his first roll gives a chance of about 52% that he will enter either on the first or second roll.

I hope this hand-waving explanation is helpful!

- Mike

Mike,
I understand your table:

31% = P(enter on roll 1) = (11/36)
21% = P(enter on roll 2) = (11/36)(25/36)
15% = P(enter on roll 3) = (11/36)(25/36)(25/36)
10% = P(enter on roll 4) = (11/36)(25/36)(25/36)(25/36)

You are right: it needs two rolls to enter (31%+21%>50%)
Thank you