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Feb 10, 2024, 12:47:24 PMFeb 10

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Tim posted a clip in another thread of Gus Hansen against Bob Koca.

Naturally, I watched the clip, heard the commentary and ordered the t-shirt.

The commentators (one of them) made a claim that in a gammonless

position at 3A 1A post-Crawford, the leader's take point is only 1%.

At time of writing, I haven't checked this. The purpose of this post is

to verify this in real time. (My real time anyway -- it won't be real time if

you're reading this later.)

Dropping gives a 2A 1A post-Crawford position. This free drop position

gives the leader 51.2% winning changces according to Rockwell-Kazaross.

Let x be the match-leader's game-winning probability and select x

so that the match-winning chances from dropping and taking are the same.

Such an x will be the take-point -- the point at which dropping and taking

are equally good.

Dropping leads to 51.2% but what does taking lead to?

Clearly, x will be part of the answer.

Taking leads to a match-winning-probability (for the match-leader)

of x + (1-x)/2. This is because the match winning possibilities are to

win immediately (x) or to lose and win the final game (1/x)/2.

So x + (1 - x)/2 = 51.2%.

(x + 1)/2 = 51.2%

x/2 = 1.2%

So x = 2.4%

This is absolutely miles away from the commentator's "one percent".

Bear in mind that these computations use a very low estimate for

the value of a free drop. So x might well be even larger.

The commentary was just major bullshit!

If Trump loses the presidential election, should we bring him in as a

backgammon commentator? He wouldn't be able to do any harm that

way, and if spreading bullshit is the main qualification, he might be

an ideal candidate.

Paul

Naturally, I watched the clip, heard the commentary and ordered the t-shirt.

The commentators (one of them) made a claim that in a gammonless

position at 3A 1A post-Crawford, the leader's take point is only 1%.

At time of writing, I haven't checked this. The purpose of this post is

to verify this in real time. (My real time anyway -- it won't be real time if

you're reading this later.)

Dropping gives a 2A 1A post-Crawford position. This free drop position

gives the leader 51.2% winning changces according to Rockwell-Kazaross.

Let x be the match-leader's game-winning probability and select x

so that the match-winning chances from dropping and taking are the same.

Such an x will be the take-point -- the point at which dropping and taking

are equally good.

Dropping leads to 51.2% but what does taking lead to?

Clearly, x will be part of the answer.

Taking leads to a match-winning-probability (for the match-leader)

of x + (1-x)/2. This is because the match winning possibilities are to

win immediately (x) or to lose and win the final game (1/x)/2.

So x + (1 - x)/2 = 51.2%.

(x + 1)/2 = 51.2%

x/2 = 1.2%

So x = 2.4%

This is absolutely miles away from the commentator's "one percent".

Bear in mind that these computations use a very low estimate for

the value of a free drop. So x might well be even larger.

The commentary was just major bullshit!

If Trump loses the presidential election, should we bring him in as a

backgammon commentator? He wouldn't be able to do any harm that

way, and if spreading bullshit is the main qualification, he might be

an ideal candidate.

Paul

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