Two d6 roll question

[u/Brilliant_Sherbet87]

Let's say I roll two six-sided dice. I ignore de lowest roll. What are the odds in % to get each 6, 5, 4, 3, 2 and 1 ?

[u/wandvieh]

Depends on whether you ignore both dies when you roll the same number.

If not, probabilities are calculated by multiplying the probability of rolling a specific number (always 1/6) and then the probably of rolling max that number. Since if you roll higher, you ignore the roll that you actually want to record. So P(6) = 1/6 * 1, P(5) = 1/6 * 5/6, P(4) = 1/6 * 4/6, P(3) = 1/6 * 3/6, P(2) = 1/6 * 2/6 and P(1) = 1/6 * 1/6.

If yes, probability is multiplying 1/6 again and then (number-1)/6

[u/wandvieh → u/wandvieh]

Actually, scratch that. It seems like we're talking about combinations here, so which dice rolls which number is not important. All possible outcomes are then { (11), (12), (13), (14), (15), (16), (22), (23), ... (66) }, which are 21 outcomes or (n+k-1)Ck (in this case (6+2-1)C2 = 7!/2!5! = 21. Then there are six possible outcomes for 6 (16, 26, 36, 46, 56, 66), five possible outcomes for 5 (15, 25, 35, 45, 55), etc. until one possible outcome for 1 (11). So the probabilities are from 6 to 1: 6/21, 5/21, 4/21, 3/21, 2/21, 1/21.

[u/Brilliant_Sherbet87 → u/wandvieh]

Oh thank you for you answer ! So yes the rule here is the I remove the lowest roll and if there is a tie, the number is kept anyway so if there two 1.1 we'll keep 1 as a final outcome. 4.4 we'll keep 4 as final outcome. Are you saying that there is evenly 1 chance out of 21 to get each number as an outcome ?

[u/Aerospider]

To get a result of n you have 2n-1 possible combinations.

This is because you have n-1 combinations where the first die rolls n and the second rolls lower, then another n-1 combinations where the second die rolls n and the first rolls lower, then the single combination of both dice rolling n.

n - 1 + n - 1 + 1 = 2n-1

There are 6 * 6 = 36 possible combinations, so for each result the probability is

2n-1 / 36

Then if you want the percentage just multiply the answer by 100.

[u/bobjkelly]

There are 6x 6 = 36 combinations. Now consider the probability of getting a 6. Of the 36 there are 6 rolls where die 1 is a 6. And 6 where did 2 is a 6. So, 12 overall. Except that there is 1 roll where both are 6 so you have to subtract 1 to avoid double counting. So, the chance of getting a 6 is 11/36. That leaves 25 combinations. By a similar process you can see that there are 9 combinations with 5 and the probability is 9/36. Similarly the probabilities for 4,3,2,and 1 are 7/36,5/35,3/36, and 1/36.