1. ## Re-Rolling Dice

I have a very simple question about the act of being able to re-rolling a single die if it fails to produce the correct result and how that may or may not change the probability of the outcome. I've searched the infinite number of internet pages on dice odds and haven't found a clear example.

If I roll a single die then the odds of rolling a 5 or 6 is 2/6 or 33.33%...right? I hope I got this part correct so far! Now, if I'm allowed to re-roll the die a single time if I don't roll a 5 or 6 then how does a single re-roll affect the probability? Basic instincts tell me that the die still has a 33.33% chance on the second roll, but since a second roll is dependent on the outcome of the first would that change the probability?

Doing some basic examples reveals a probability, but I don't know if its a fare example to the above questions...If I roll 36 dice, then in a perfect world of probability 12 dice would have a 5 or 6. Now, 24 of the dice did not, so I'd be allowed a single re-roll of those dice, which would result in 8 more 5 or 6s. Once all failed dice have been allowed a re-roll I'm left with 20 dice with 5 or 6s from the original 36. That's 55% of the dice...

Back to a single die. Does the above sample suggest that the simple act of re-rolling once on a failed roll would increase a dies chances by 22% (from 33% to 55%!? That seems suspiciously high. If this is true (or not) can some one please explain the math involved for calculating the probability?

Thanks so much for any clarification!

2. If I understand it correctly your question is: What is the probability that I roll a 5 or a 6 in two rolls of one dice, tossing again if the first time was not a 5 or a 6.

Indeed, the chance of rolling a 5 or a 6 the first time is 2/6 = 1/3
In the other case (2/3) you will roll again, where you once more have a 1/3 chance. Because you will need to both NOT roll a 5 or 6 on the first roll AND roll a 5 or 6 on the second roll, you'll need to multiply these chances: 2/3 * 1/3 = 2/9 (which indeed does convert to 22%)

Add these two up to get the total chance being: 1/3 + 2/9 = 5/9 = 55%

3. Originally Posted by russ_c
I have a very simple question about the act of being able to re-rolling a single die if it fails to produce the correct result and how that may or may not change the probability of the outcome. I've searched the infinite number of internet pages on dice odds and haven't found a clear example.

If I roll a single die then the odds of rolling a 5 or 6 is 2/6 or 33.33%...right? I hope I got this part correct so far! Now, if I'm allowed to re-roll the die a single time if I don't roll a 5 or 6 then how does a single re-roll affect the probability? Basic instincts tell me that the die still has a 33.33% chance on the second roll, but since a second roll is dependent on the outcome of the first would that change the probability?

Doing some basic examples reveals a probability, but I don't know if its a fare example to the above questions...If I roll 36 dice, then in a perfect world of probability 12 dice would have a 5 or 6. Now, 24 of the dice did not, so I'd be allowed a single re-roll of those dice, which would result in 8 more 5 or 6s. Once all failed dice have been allowed a re-roll I'm left with 20 dice with 5 or 6s from the original 36. That's 55% of the dice...

Back to a single die. Does the above sample suggest that the simple act of re-rolling once on a failed roll would increase a dies chances by 22% (from 33% to 55%!? That seems suspiciously high. If this is true (or not) can some one please explain the math involved for calculating the probability?

Thanks so much for any clarification!
This is very close to a notorious practice in weapon system performance acceptance.

The required performance (call it $\displaystyle p_k$) is specified over a number of scenarios. The performance is assessed by non-deterministic simulation (which is either Monte-Carlo or uses real hardware, or a combination of both).

So we run the simulation over all of the scenarios, and the results are OK for some scenarios, but not quite good enough in the others, so you rerun all the scenarios (a more economical process is to only rerun those failing to achive the required performance) repeatedly until you get an acceptable result. (To justify the process you may make trivial adjustments to the system between reruns). With enough variability in the results this is gauranteed to result in a "pass" eventually.

Perhaps the most appalling thing about this process is that many project managers think this is a legitimate process.

(that such practices are not new see Tom Korner's book "The Pleasures of Counting" for artillary shell acceptance practices dating back to befor WWI)

CB

4. Originally Posted by russ_c
I have a very simple question about the act of being able to re-rolling a single die if it fails to produce the correct result and how that may or may not change the probability of the outcome. I've searched the infinite number of internet pages on dice odds and haven't found a clear example.

If I roll a single die then the odds of rolling a 5 or 6 is 2/6 or 33.33%...right? I hope I got this part correct so far! Now, if I'm allowed to re-roll the die a single time if I don't roll a 5 or 6 then how does a single re-roll affect the probability? Basic instincts tell me that the die still has a 33.33% chance on the second roll, but since a second roll is dependent on the outcome of the first would that change the probability?

Doing some basic examples reveals a probability, but I don't know if its a fare example to the above questions...If I roll 36 dice, then in a perfect world of probability 12 dice would have a 5 or 6. Now, 24 of the dice did not, so I'd be allowed a single re-roll of those dice, which would result in 8 more 5 or 6s. Once all failed dice have been allowed a re-roll I'm left with 20 dice with 5 or 6s from the original 36. That's 55% of the dice...

Back to a single die. Does the above sample suggest that the simple act of re-rolling once on a failed roll would increase a dies chances by 22% (from 33% to 55%!? That seems suspiciously high. If this is true (or not) can some one please explain the math involved for calculating the probability?

Thanks so much for any clarification!
The probability of rolling a 5 or a 6 on the first roll is 1/3. The probability of rolling a 5 or a 6 on the second roll is 1/3. Re-rolling again because you didn't get a 5 or a 6 the first time does NOT change the probability of getting a 5 or a 6 on the second roll.

If the question you're asking is

"What's the probability of rolling a 5 or a 6 under the rule that I can roll again if I don't get a 5 or a 6 the first time"

then the answer is 5/9 as calculated by Pim.

5. Thanks everyone for clearing this up! The math you've written Pim is exactly what I had in front of me on a sheet of paper when I was thinking about this problem. I just wasn't sure if and why you would add the probability of the first and second chance together, even though ever simulated I tested pointed to the addition of the two probabilities!

Wow, the act of being allowed a re-rolling has a surprising affect on the possibility! Thanks again!