# Thread: unique value dice probability

1. ## unique value dice probability

Okay, from my understanding, the question goes like this. You have a normal 6 sided die and a limited number of throws, x. You've already rolled y unique numbers, you have r throws left and you need to get z unique numbers. You don't know how may throws it took to get the y unique numbers you already have but you have to find the probability that the next throw is another unique number as in, not one of the y values you've obtained.

I'm terrible at probability and my idea of solving this was (6-y)/6, as in (number of values left)/total space. This failed on a few test cases and I'm pretty sure z and x have to play a part in the equation somehow.

I'm not asking for exact answers but directions, I really want to understand this and why my answer is so awe-full, the other variables add restrictions, how do I arrive at an equation that shows this? Thank you so much. Some links to similar questions would also be greatly appreciated!

2. From my perspective, it looks like your main problem isn't really that you're "bad at probability" or even bad at math. Rather, the problem is that you don't have a clearly defined problem. I'll be blunt and hope to not come off as rude, but your first paragraph in which you explained the problem you're trying to solve felt very much like a stream of consciousness, with no attempts to organize the thoughts in a cohesive manner. For instance, the goal of the problem, as I'm interpreting what you wrote, is the last sentence here:

Originally Posted by DavidMatthews
You have to find the probability that the next throw is another unique number as in, not one of the y values you've obtained.
Given this instruction, your answer that the probability is (6-y)/6 is absolutely correct. We can examine this in detail and see why this must be the case. Consider the very first dice roll, where you've not rolled any numbers so far. Obviously, any number you roll must be unique, so you have 100% probability. In this case, we have y = 0. Your answer would indicate that the probability is (6-0)/6 = 6/6 = 1 = 100%, so we're good. Now what if y = 1? You've rolled one unique number so far, that means there are 5 unique numbers out of the 6 total numbers remaining. Your answer says the probability is (6-1)/6 = 5/6, so we're good there too. Now what if y = 2? You've rolled two unique numbers so far, that means there are 4 unique numbers out of the 6 total numbers remaining. Your answer says the probability is (6-2)/6 = 4/6, so we're good there too. And so on...

However, from the fact that you then go on to say that "This [answer] failed on a few test cases," that I've misinterpreted the goal of the problem. At this point, it should be clear that before we can even begin to use any mathematical or probability principles to solve the problem, we have to know what the problem even is. So my best recommendation is that you try to collect your thoughts and better explain what it is you're trying to ask.

3. Originally Posted by DavidMatthews
This failed on a few test cases.
In what sense did your test cases fail? You are calculating a probability so how did you verify its accuracy?

4. Originally Posted by JeffM
In what sense did your test cases fail? You are calculating a probability so how did you verify its accuracy?
I'm trying to write a software program that basically uses this formula to produce an output. I have the answers it should produce and i'm trying to get it to produce those answers. When I tried this formula, it doesn't work out. That's how I was able to verify it

5. ## Thank you for your response

Originally Posted by ksdhart2
From my perspective, it looks like your main problem isn't really that you're "bad at probability" or even bad at math. Rather, the problem is that you don't have a clearly defined problem. I'll be blunt and hope to not come off as rude, but your first paragraph in which you explained the problem you're trying to solve felt very much like a stream of consciousness, with no attempts to organize the thoughts in a cohesive manner. For instance, the goal of the problem, as I'm interpreting what you wrote, is the last sentence here:

Given this instruction, your answer that the probability is (6-y)/6 is absolutely correct. We can examine this in detail and see why this must be the case. Consider the very first dice roll, where you've not rolled any numbers so far. Obviously, any number you roll must be unique, so you have 100% probability. In this case, we have y = 0. Your answer would indicate that the probability is (6-0)/6 = 6/6 = 1 = 100%, so we're good. Now what if y = 1? You've rolled one unique number so far, that means there are 5 unique numbers out of the 6 total numbers remaining. Your answer says the probability is (6-1)/6 = 5/6, so we're good there too. Now what if y = 2? You've rolled two unique numbers so far, that means there are 4 unique numbers out of the 6 total numbers remaining. Your answer says the probability is (6-2)/6 = 4/6, so we're good there too. And so on...

However, from the fact that you then go on to say that "This [answer] failed on a few test cases," that I've misinterpreted the goal of the problem. At this point, it should be clear that before we can even begin to use any mathematical or probability principles to solve the problem, we have to know what the problem even is. So my best recommendation is that you try to collect your thoughts and better explain what it is you're trying to ask.
Thank you for your response, and no you weren't being rude at all. I really didn't define the question well but I didn't want it to come across as me just throwing my work here and asking for answers. I was trying to show that I put effort into understanding the question and solving it on my own but it seems everything came out messy. I'll ask the question again. This time I'll clearly describe the problem.

6. ## Modified version of the question

I apologize for how messy my first attempt went. I'll try to ask the question better this time:
Original Problem:
I have a normal 6 sided dice and I'm trying to figure out the probability of getting "s" unique numbers in "x" throws.

Solution attempt:
The first step was to narrow this down to a simpler equation and work from there.
If there are "r" throws left and we've already gotten "y" distinct values, let F(r, y) be the probability, this way the answer to the problem is is F(x, 0)
Next step was finding the probability that the next throw would be one of the "y" previously gotten values

I tried to find the probability it wouldn't be one of the y previous values and subtract this from 1
so 1 - ((6-y)/6) = y/6

but applying this to the original problem doesn't give me the right answer. And I still feel like other values have a part to play. I was given the advise of defining it as a recurrence relation but I can't do that without the right equation first.