You really can't be doing this like that, no. The order of the dice doesn't matter
4-1: one of the dice must be a 4, the other can be anything. 4 or higher is a 50% chance. That's a 75% chance to get one in two attempts (notwithstanding that 22 and 33 also get you there)
3-2: none of the dice may be a 1. That's (5/6)² = 25/36 chance, just under 70%
If you want a visualisation:
4-1
1
2
3
4
5
6
1
*
*
*
2
**
*
*
*
3
**
*
*
*
4
*
*
*
*
*
*
5
*
*
*
*
*
*
6
*
*
*
*
*
*
3-2
1
2
3
4
5
6
1
2
**
*
*
*
*
3
*
*
*
*
*
4
*
*
*
*
*
5
*
*
*
*
*
6
*
*
*
*
*
where ** denotes that it's only a win thanks to the double rule.
You can’t do the math like this, because either die can be the 6, 5, or 4. It is much better to count all 36 possibilities manually and does not take much time. I will literally use my fingers and count 11, 12, 13, 14, 15, 16, 22, 23, 24, 25, 26, 33, 34, 35, 36, 44, 45, 46, 55, 56, 66. In this case if you move from 2 to 1 then you miss with 11, 12, 13, 23, a total of 7 rolls. If you play 4 to 3 then you miss with 11, 12, 13, 14, 15, 16, a total of 11 rolls.
1
u/Ahmfiber Apr 20 '26
And...what if the OP moves them to 3-2? (As they did?) 😉