So I saw this problem in two different books now and both times with the same explanation I think is wrong. Since i dont know how to upload a diagram i will explain the situation, which is pretty simple.

both sides have only two checkers left, both on the 2point (23point)

now it says, that the player to move should double, which obviously is correct. but then it says, the other player should take, because the first player will win only 26/36th of the time, which is wrong.

true, the player to move will win immediately with 26 out of 36 throws. but even if he misses, he will win 10/36 of the remaining time, making him a 80% favorite to win.

I did the complete EV-calculation for this scenario and the fact, that player 2 will redouble and player 1 has to take turns it into a take with an EV of -0.95.

but the explanation really threw me off, because it was obvious to me, that player 2 had a chance smaller than 25% to win.
is the total winning percantage irrelevant if there is still the possibility to redouble? I am having trouble to fully understand the mathematics of these situations.

If the player who doubled rolls one of the 10 rolls that don't win the game immediately, there will be a strong redouble. This redouble must be taken, because there's still a 10/36 or 27.8% chance to win.

So the EV formula reads like this:

(26/36) * -2 + (10/36) * ((10/36) * -4 + (26/36) * 4)
= -1.45 + (10/36) * (-1.11 + 2.89) = -1.45 + 0.49 = -0.96

Compare this to the -1.00 if you drop (-:

PS:

Oh, I've seen you worked the EV increase out by yourself. Well, it's the possibility of the strong redouble that makes this take correct. If the cube was dead after taking, you would have to drop.

yeah, I figured that out. that problem leaves me with a general question regarding doubles and takes.
in a money game (to take match consideration out of the equation), if I can calculate my winning chances straightforward liike the example above:

how do I include the worth of the cube? like in the example above my winning chances are below 25%, but the worth of the cube makes up for that lack of winning percantage. do I really have to go through EV-calculation to know the answer or is there a quicker way?

to clarify my point:
I learned that 25% is the breakeven point for taking a double. but even without the possibility of a gammon, this is not always true, as proved in the example above. thats what really threw me off.

Well for the endgame you could probably calculate it on the fly. But it also helps learning the cube actions for these types of endgames by heart.

For example, you should simply know that in a four roll position (both have like 8 checkers on the ace point, or 4 each on ace and deuce point) the correct cube action is double/take, while in a three roll position (6 instead of 8 checkers), it is double/pass.

The reason why the four roll position is a take is exactly because you have the chance to offer a redouble that can be taken. If you catch up in a three roll position, your opponent will always drop a redouble.

For comparable reasons, a five roll position is double/take when the cube is still in the middle. If you own the cube in a five roll position, it is better to wait one move and offer the redouble in a four roll position when your opponent is in a tougher spot.

Your analysis is correct. I wrestled with including the complete explanation when I wrote the book, but decided to leave it in its very simplified (but technically incorrect form) because the book's audience was complete beginners, and I didn't want to lose them. I figured the potentially interested players who saw the real problem could figure out the math for themselves.

And so you did!

Actually the four roll position never gets to a double/take position. It's a take because you have three shots at rolling a double, which makes your raw winning chances just over 25%.

Oh dear...

I apologize. I try to only post statements I am sure of. Here, I just proved that I'm a bit rusty.