I have a question about Bill's blog post on 9/15 linked below.

Bill says: "So taking with a 10% chance of winning the game (cubeless) will also give him 5% match winning chances, and therefore the top end of the doubling window is reached when White is a 90-to-10 favorite in the game."

Where did the 10% chance come from?


https://thegammonpress.com/sep-15-20...in-match-play/

OK, here's the idea.

White is leading 11-3 in a match to 15. He's reached a racing position where he owns a 2-cube and is a big favorite. He wants to know if he's a big enough favorite to double, so he'll start by figuring out if his opponent (Black) has a take or not.

First step -- what are the chances if White doubles to 4 and Black drops? In that case White leads 13-3 to 15, and various match equity tables pretty much agree that White's winning chances are then 95%. So if Black drops he's only 5% to win the match.

Second step -- what if Black takes the cube instead? Since White will now win the match if he wins this game, Black will immediately redouble to 8 to maximize his own chances. If Black then wins, the score will be 11-11, and Black will now have 50% winning chances. That means he (Black) needs to win this game 10% of the time to have a net winning chance equal to his winning chance if he dropped instead. (10% * 50% = 5%, his winning chance if he dropped.)

So if White has more than 90% winning chances in the position, he should double and Black should drop. If White has less than 90% winning chances, Black should take. Then the problem becomes -- how much less than 90% does White need to have before he can double?

Hope this helps.

Bill,
Yes, thank you the 10% makes sense now! Thank you.