dice probability question

what are the chances of rolling two numbers:

greater than a 3
greater than a 4
greater than a 5
greater than a 6
greater than a 7

if you roll the same number during one of the rolls, its just a reroll. so if youre trying to roll two numbers greater than a 3 and you happen to roll a 3, you just reroll the dice as if it never happened.

Quote: rudeboyoi

what are the chances of rolling two numbers:

greater than a 3
greater than a 4
greater than a 5
greater than a 6
greater than a 7

if you roll the same number during one of the rolls, its just a reroll. so if youre trying to roll two numbers greater than a 3 and you happen to roll a 3, you just reroll the dice as if it never happened.

I'll assume you are rolling 2 dice twice.

3 = 94.204152249135 %
4 = 82.644628099174 %
5 = 66.015625 %
6 = 45.889698231009 %
7 = 25 %
8 = 10.405827263267 %
9 = 3.515625 %
10 = 0.8264462809917 %
11 = 0.0865051903114 %

thank you. thats what i was looking for.

now for a slightly more complex question.

what are the probabilities for rolling a number greater than what you rolled initially, then rolling a number less than what you rolled.

so if youre trying to roll a number greater than a 4, you have to roll a 5-12. then depending on what you rolled, lets say you rolled a 6, then you have to roll a number less than 6.

is there a quick way to go about this besides.

figuring out if you are trying to roll a number greater than a 4, p(5)*p(2 to 4) + p(6)*p(2 to 5) + p(7)*p(2-6)+... and so forth.

You may find this chart helpfull. It gives the exact number of ways to roll one number then another. For example there are 15 ways to roll a 6 then a 4. There are 1296 total ways to roll 2 pairs of dice.

roll	2	3	4	5	6	7	8	9	10	11	12
2	1	2	3	4	5	6	5	4	3	2	1
3	2	4	6	8	10	12	10	8	6	4	2
4	3	6	9	12	15	18	15	12	9	6	3
5	4	8	12	16	20	24	20	16	12	8	4
6	5	10	15	20	25	30	25	20	15	10	5
7	6	12	18	24	30	36	30	24	18	12	6
8	5	10	15	20	25	30	25	20	15	10	5
9	4	8	12	16	20	24	20	16	12	8	4
10	3	6	9	12	15	18	15	12	9	6	3
11	2	4	6	8	10	12	10	8	6	4	2
12	1	2	3	4	5	6	5	4	3	2	1

thank you. this is extremely useful. id rather see it portrayed as combinations than probabilities anyways.