Need a Probability Equation for Dice Outcomes

Condad

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Joined
Apr 13, 2021
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4
Hello,

I'm looking for assistance in devising an equation to determine the probability of outcomes on dice pools. I'm having two issues in doing it myself, because I'm not looking for the probability of the sums of the dice, but the likelihood that each group of dice will have dice that roll equal to or greater than a given threshold (4 or 8). These thresholds represent "success". Greater than 4 is 1 success. Greater than 8 is 2 successes. The dice we use are d4, d6, d8, and/or d10.

The system scales the dice. We start with 1d4, move up to 1d6, then 1d8, and 1d10. After this, we add a second die that scales. So: 1d10 & 1d4, 1d10 & 1d6, etc. The same thing happens after 2d10, all the way up to 5d10.

So, for example, in 1d10&1d4, the chance of one of them rolling => 4 (1 success) is 77.50% (70% for the 1d10 to land 4-10, 7.5% chance for the 1d4 to land 4 and the 1d10 to land 1-3) if I've done my math right. The chance of 2 successes (1d10 >= 8 OR 1d10 = {4..7} & 1d10 = 4) is 30%. I think.

It's been slow going, but I'm hoping to speed up the process.
 

Cubist

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Oct 29, 2019
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930
...The chance of 2 successes (1d10 >= 8 OR 1d10 = {4..7} & 1d10 = 4) is 30%. I think.
You made a mistake above. (30% for 1d10 >= 8 OR 10% for 1d10 = {4..7} & 1d4 = 4) which comes to 40%

...but I'm hoping to speed up the process.
You could make a spreadsheet like this...

die.png

I did this by using the spreadsheet's min/max functions. Note that having die B as one-sided is equivalent to having just one die A. I'm going to give you quite a lot of help since I suspect that this isn't a math question from your school/ college. (Are you designing a game of chance?)

Let a,b be the number of sides on each die. The chances of one or more success is 1-P(no success)
=1 - P(no success with A)*P(no success with B)
=1 - (min(3,a)/a) * (min(3,b)/b)

--

The two or more success case is a little more challenging.

The chances of two or more successes is 1-P(zero or one successes)

To help with the above, what is P(one success with a, and none with b) = ?
min(4,max(a-3,0))/a * (min(3,b)/b)

Can you continue given the info above?
 

Condad

New member
Joined
Apr 13, 2021
Messages
4
You made a mistake above. (30% for 1d10 >= 8 OR 10% for 1d10 = {4..7} & 1d4 = 4) which comes to 40%



You could make a spreadsheet like this...

View attachment 26479

I did this by using the spreadsheet's min/max functions. Note that having die B as one-sided is equivalent to having just one die A. I'm going to give you quite a lot of help since I suspect that this isn't a math question from your school/ college. (Are you designing a game of chance?)

Let a,b be the number of sides on each die. The chances of one or more success is 1-P(no success)
=1 - P(no success with A)*P(no success with B)
=1 - (min(3,a)/a) * (min(3,b)/b)

--

The two or more success case is a little more challenging.

The chances of two or more successes is 1-P(zero or one successes)

To help with the above, what is P(one success with a, and none with b) = ?
min(4,max(a-3,0))/a * (min(3,b)/b)

Can you continue given the info above?
You're correct, I'm tinkering with a game mechanic that would depend on those thresholds instead of sums, like most dice games, and I knew there must be a mathematical way to figure these odds, but having dice of different sizes and two thresholds was throwing me for a loop.

I think this is exactly what I was looking for. I'll throw together an excel sheet and see what happens. Thanks a lot, Cubist. I think that's everything I needed.
 

Condad

New member
Joined
Apr 13, 2021
Messages
4
You made a mistake above. (30% for 1d10 >= 8 OR 10% for 1d10 = {4..7} & 1d4 = 4) which comes to 40%



You could make a spreadsheet like this...

View attachment 26479

I did this by using the spreadsheet's min/max functions. Note that having die B as one-sided is equivalent to having just one die A. I'm going to give you quite a lot of help since I suspect that this isn't a math question from your school/ college. (Are you designing a game of chance?)

Let a,b be the number of sides on each die. The chances of one or more success is 1-P(no success)
=1 - P(no success with A)*P(no success with B)
=1 - (min(3,a)/a) * (min(3,b)/b)

--

The two or more success case is a little more challenging.

The chances of two or more successes is 1-P(zero or one successes)

To help with the above, what is P(one success with a, and none with b) = ?
min(4,max(a-3,0))/a * (min(3,b)/b)

Can you continue given the info above?
Hey Cubist, I'm back. I think I'm misunderstanding how the MIN and MAX functions are working here.

When plugging in the equation you gave in the spoiler, I'm getting 70% for 1d10+1d4, even though we both know it should be 0.775.
min(4,max(a-3,0))/a * (min(3,b)/b) was the equation you gave, and plugging everything in:
min(4,max(10-3,0))/10 * (min(3,4)/4)
Excel is telling me 70%. What am I not seeing here?
 

Cubist

Full Member
Joined
Oct 29, 2019
Messages
930
When plugging in the equation you gave in the spoiler, I'm getting 70% for 1d10+1d4, even though we both know it should be 0.775.
The expression for the 0.775 figure is not given in the spoiler, it is given higher up in the post...

Let a,b be the number of sides on each die. The chances of one or more success is 1-P(no success)
=1 - P(no success with A)*P(no success with B)
=1 - (min(3,a)/a) * (min(3,b)/b)
Plugging in a=10 and b=4 gives 1 - (min(3,10)/10) * (min(3,4)/4) = 1 - (3/10) * (3/4) = 0.775

The spoiler contains help towards obtaining an expression for the probability of 2 or more successes. It's not the final formula. I guess that you're struggling with this, and since it isn't homework then I'll give you the final answer in the double spoiler below...

The chances of two or more success is 1-P(zero or one success)
=1 - (P(no success) + P(one success with a, and none with b) + P(one success with b, and none with a))
=1 - ((min(3,a)/a) * (min(3,b)/b) + min(4,max(a-3,0))/a * (min(3,b)/b) + (min(3,a)/a) * min(4,max(b-3,0))/b)
 

Condad

New member
Joined
Apr 13, 2021
Messages
4
The expression for the 0.775 figure is not given in the spoiler, it is given higher up in the post...


Plugging in a=10 and b=4 gives 1 - (min(3,10)/10) * (min(3,4)/4) = 1 - (3/10) * (3/4) = 0.775

The spoiler contains help towards obtaining an expression for the probability of 2 or more successes. It's not the final formula. I guess that you're struggling with this, and since it isn't homework then I'll give you the final answer in the double spoiler below...

The chances of two or more success is 1-P(zero or one success)
=1 - (P(no success) + P(one success with a, and none with b) + P(one success with b, and none with a))
=1 - ((min(3,a)/a) * (min(3,b)/b) + min(4,max(a-3,0))/a * (min(3,b)/b) + (min(3,a)/a) * min(4,max(b-3,0))/b)
Hey thanks again, I really appreciate all your help.

I found another approach that's much easier for me to read. I'm using the PROB() function in excel and simply have columns dedicated to the dice and their roll probabilities. Then it's simple addition of probability. I've attached the document of my progress so far (I just made this breakthrough like five minutes ago).

On the left I have which step of the dice progression I'm on, the max possible chances, and then the chances of each number of successes for a given dice combination. On the far right (partially cut off) I have the probability pools I'm calling on for my PROB() functions. So far the math checks out, so I'm gonna keep this. It's much easier for me to comprehend and I think will save my wrists on the carpal tunnel of interchanging min()/max() functions.

Thanks for helping me understand the issue, though!
 

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Cubist

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Oct 29, 2019
Messages
930
So far the math checks out
FYI: The result in cell 7e is incorrect (ought to be 0.4 exactly) . But, I guess that the spreadsheet is still a work in progress since you've only just come up with the idea. Good luck in progressing it!

It's much easier for me to comprehend ... Thanks for helping me understand the issue, though!
Glad you've found a method that makes more sense to you. You're welcome for the help!
 
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