# Thread: Cumulative Chance and Probability of Outcome

1. ## Cumulative Chance and Probability of Outcome

I use (simple) statistics/probability fairly often with a hobby of mine. I have been using Excel to calculate the answers I want, but because I don't know the proper formula, it's a tedious, messy method. I'd appreciate anyone that can explain the formula, preferably with the proper signs, for the following types of situation. In any examples provided, please list the variable used before the equation for the example.

Whenever X happens, there is a cumulative Y% chance that Z happens.
I want to be able to calculate the probability that Y has occurred if I provide a value for X.

and

Whenever W happens, there is a cumulative X% chance that Y happens. Each time Y happens, there is a cumulative Z% increase to the chance of it happening again.
I want to be able to calculate the average number of times X must occur for Y to be 100% or higher.

I am also interested in the following probability formulas. For the following three situations, you score 1 point for rolling X or higher on a die with Y number of sides and roll Z number of dice, all with Y number of sides, each time. The dice are perfectly balanced and the sides are numbered 1 to X. I want to know the likelihood of getting any given number of points on a single roll.

You score 1 point for each dice that rolls X or higher.

You score 1 point for each dice that rolls X or higher. If you roll a 1 on the die, you lose a point. If you roll the max number on the dice, you gain 2 points.

You score 1 point for each dice that rolls X or higher. If you roll a 1 on the die, you lose a point. If you roll the max number on the dice, you gain 1 point and roll that dice again.

Thank you for taking your time to help me with my hobby, your aid is much appreciated.

2. It's very hard to follow what you are saying, because you are not using letters consistently.

Originally Posted by SzassT
Whenever X happens, there is a cumulative Y% chance that Z happens.
I want to be able to calculate the probability that Y has occurred if I provide a value for X.
First X appears to be an event, and then later it has a value. First Y appears to be a percentage (probability), and then it is an event!

Whenever W happens, there is a cumulative X% chance that Y happens. Each time Y happens, there is a cumulative Z% increase to the chance of it happening again.
I want to be able to calculate the average number of times X must occur for Y to be 100% or higher.
First X appears to be a probability, and then later it becomes an event. First Y appears to be an event, and then it is a probability! I'm also not sure just what you mean by a cumulative chance (probability), and a cumulative increase in the chance.

I am also interested in the following probability formulas. For the following three situations, you score 1 point for rolling X or higher on a die with Y number of sides and roll Z number of dice, all with Y number of sides, each time. The dice are perfectly balanced and the sides are numbered 1 to X. I want to know the likelihood of getting any given number of points on a single roll.

You score 1 point for each dice that rolls X or higher.

You score 1 point for each dice that rolls X or higher. If you roll a 1 on the die, you lose a point. If you roll the max number on the dice, you gain 2 points.

You score 1 point for each dice that rolls X or higher. If you roll a 1 on the die, you lose a point. If you roll the max number on the dice, you gain 1 point and roll that dice again.
The same problem occurs here, where X is first a particular value that is rolled, and then seems to change into the maximum number (the number of sides).

Perhaps you can explain it better if you avoid using variables in the first two questions, and just use words to describe a specific thing you want to figure out. Once we get that worked out, we can look at the dice questions.

3. Originally Posted by Dr.Peterson
It's very hard to follow what you are saying, because you are not using letters consistently.

First X appears to be an event, and then later it has a value. First Y appears to be a percentage (probability), and then it is an event!

First X appears to be a probability, and then later it becomes an event. First Y appears to be an event, and then it is a probability! I'm also not sure just what you mean by a cumulative chance (probability), and a cumulative increase in the chance.

The same problem occurs here, where X is first a particular value that is rolled, and then seems to change into the maximum number (the number of sides).

Perhaps you can explain it better if you avoid using variables in the first two questions, and just use words to describe a specific thing you want to figure out. Once we get that worked out, we can look at the dice questions.
First, thank you for your prompt reply. I am currently working and will be until early tomorrow morning. I will attempt to provide further details and clearer questions as soon as I am able.

4. Whenever the button on a machine is pressed, there is a cumulative 3% chance that a light turns on. So, the first time the button is pressed there is a 3% chance the light activates, on the second button press there is a 6% chance the light activates, on the third button press there is a 9% chance the light activates.
I want to be able to calculate the probability that the light is activated after any given number of button presses.
and

Whenever a button is pressed, there is a cumulative 3% chance that a light flickers. Each time the light flickers, there is a cumulative 10% increase to the chance of it happening again. So, the first time the button is pressed there is a 3% chance the light flickers, on the second button press there is a 6% chance the light flickers, on the third button press there is a 9% chance the light flickers, on the fourth button press there is a 12% chance the light flickers and it does, so on the fifth button press there is a 25% chance the light flickers (12% chance + 3% cumulative chance from pressing + 10% cumulative chance from flickering), on the sixth button press there is a 28% chance the light flickers and it does, so on the seventh button press there is a 41% chance the light flickers (25% chance + 3% cumulative chance from pressing + 10% cumulative chance from flickering)
What is the most common number of button presses required to get the chance of the light flickering to at least 100%?

Examples for the dice roll question:

On 1d4 (1 dice with 4 sides), what are the chances of getting 1 point if you must roll a 3 or higher to get one point?
On 4d4 (4 dice with 4 sides), what are the chances of getting 1 point if you must roll a 2 or higher to get one point?
On 7d10 (7 dice with 10 sides), what are the chances of getting 5 point if you must roll a 6 or higher to get one point?
On 47d12 (47 dice with 12 sides), what are the chances of getting 14 points if you must roll 11 or higher to get one point?

I hope these help clarify the formulas that I am wanting. I appreciate your time and help.

5. Originally Posted by SzassT
Whenever the button on a machine is pressed, there is a cumulative 3% chance that a light turns on. So, the first time the button is pressed there is a 3% chance the light activates, on the second button press there is a 6% chance the light activates, on the third button press there is a 9% chance the light activates.
I want to be able to calculate the probability that the light is activated after any given number of button presses.
I'll just try this first one; I don't have time to work on all of them at once.

I wouldn't call this "cumulative", exactly. I think what you are saying is that the probability of success on the nth try is 3n%. You want the probability that you will have a success WITHIN the first n tries (that is, on at least one of tries 1, 2, 3, ..., n. (And if it does succeed on one of those, then you wouldn't have to actually make any further attempts.)

The standard way to approach an "at least one" problem is to calculate 1 minus the probability of "none". That is,
P(at least one success in n) = 1 - P(try 1 fails AND try 2 fails AND ... AND try n fails).

See what you can do with that idea. What is the probability of failure on the nth try? How will you combine these.

I don't know yet whether there is actually a simple closed-form formula for this, but it may allow you to make a more efficient spreadsheet for it.

(By the way, I assume you are aware that what we generally do is to help you solve your own problem, not to solve a whole problem for you. So we'll need to see what you are able to do.)

6. Originally Posted by Dr.Peterson
I'll just try this first one; I don't have time to work on all of them at once.

I wouldn't call this "cumulative", exactly. I think what you are saying is that the probability of success on the nth try is 3n%. You want the probability that you will have a success WITHIN the first n tries (that is, on at least one of tries 1, 2, 3, ..., n. (And if it does succeed on one of those, then you wouldn't have to actually make any further attempts.)

The standard way to approach an "at least one" problem is to calculate 1 minus the probability of "none". That is,
P(at least one success in n) = 1 - P(try 1 fails AND try 2 fails AND ... AND try n fails).
So, you're suggesting using Binomial Distribution? If that is the case, the formula I'd use for that in Excel asks for "Number of Successes in Trial" (I'm assuming this would be P), Number of Independent Trials (I'm assuming this would be n), Probability of Success on each trial (I'm assuming this would be 3n%). Am I following correctly?

Originally Posted by Dr.Peterson
(By the way, I assume you are aware that what we generally do is to help you solve your own problem, not to solve a whole problem for you. So we'll need to see what you are able to do.)
I am not looking for a problem to be solved, I'm looking for a formula to use to solve problems in the future. If what you are saying is that this community does not provide formulas, I was not aware of that. The examples provided do not use the same numbers I actually need, I wouldn't learn anything if I asked for the answer. The examples I provided require the same questions answered (ie, the same formulas to solve) as what I am actually seeking. So, if you're suggesting I use Binomial Distribution, then that is a perfectly acceptable answer for me, if you are suggesting I use something else and I'm not understanding, explaining what I'm not understanding and pointing me in the correct direction is also perfectly acceptable for me. The only issue I see coming from this is that unless I know the answer to the problem, I won't know if I'm getting the correct answer out of Excel, so if the community is willing to walk me through all the steps and then wait for me to reply with the assumed answer and verify it, there shouldn't be a problem.

7. Originally Posted by SzassT
So, you're suggesting using Binomial Distribution? If that is the case, the formula I'd use for that in Excel asks for "Number of Successes in Trial" (I'm assuming this would be P), Number of Independent Trials (I'm assuming this would be n), Probability of Success on each trial (I'm assuming this would be 3n%). Am I following correctly?
Actually, I wasn't thinking of the binomial distribution, which I don't think is relevant (because you are not counting successes); but in mentioning it, you are partially answering my question about what you are able to do, which is helpful. You know about distributions and their use in Excel.

I had something much simpler in mind as the first step: with a set of independent events combined by "and", you multiply. And the probability of failure is 1 minus the probability of success, so our first step is

P(at least one success in n) = 1 - P(try 1 fails)*P(try 2 fails)*...*P(try n fails)
= 1 - (1 - .03)(1 - .06)...(1 - .03n)
= 1 - (.97)(.94)...(1 - .03n)

But I don't see a next step; this is not a type of product that I recognize as having a closed formula. Possibly someone else will.

But you could use Excel to evaluate it for any given n.

Originally Posted by SzassT
I am not looking for a problem to be solved, I'm looking for a formula to use to solve problems in the future. If what you are saying is that this community does not provide formulas, I was not aware of that. The examples provided do not use the same numbers I actually need, I wouldn't learn anything if I asked for the answer. The examples I provided require the same questions answered (ie, the same formulas to solve) as what I am actually seeking. So, if you're suggesting I use Binomial Distribution, then that is a perfectly acceptable answer for me, if you are suggesting I use something else and I'm not understanding, explaining what I'm not understanding and pointing me in the correct direction is also perfectly acceptable for me. The only issue I see coming from this is that unless I know the answer to the problem, I won't know if I'm getting the correct answer out of Excel, so if the community is willing to walk me through all the steps and then wait for me to reply with the assumed answer and verify it, there shouldn't be a problem.
My point was that in order to help you solve the problem, we need to know what part you need help with, and what we can expect you to know already. Some of us do seem to assume that we only help with school problems; but although I disagree with that, it's true, as stated in the "Read before posting" message, that we generally want to guide, not do. (And as I see it, the "problem" here is to find a formula, if there is one. Evaluating it would, hopefully, be trivial.)

As I suggested above, I don't know whether there is a proper formula at all! Not everything has a formula (which is something many people never learn in school, because they are only shown things that have known solutions, and so assume that mathematicians can do anything easily).

Does the non-closed formula I gave above look usable as something you can evaluate in Excel and use to check what you are already doing?

8. Originally Posted by Dr.Peterson
My point was that in order to help you solve the problem, we need to know what part you need help with, and what we can expect you to know already. Some of us do seem to assume that we only help with school problems; but although I disagree with that, it's true, as stated in the "Read before posting" message, that we generally want to guide, not do. (And as I see it, the "problem" here is to find a formula, if there is one. Evaluating it would, hopefully, be trivial.)
Given the amount you've helped, I would say there is no issue here with the way this forum operates. I find this form of assistance to be perfectly acceptable!

Originally Posted by Dr.Peterson
Does the non-closed formula I gave above look usable as something you can evaluate in Excel and use to check what you are already doing?
While not ideal, I was hoping for a closed formula or a small set of formulas that would essentially allow it to be closed, I can easily enough make a spreadsheet to serve the intended purpose for both of the questions I had relating to increasing likelihood of occurrence. Now, any chance of getting some assistance with the variable fair dice?

9. I still need some clarification. I have made some corrections below that you can confirm (mostly in distinguishing the plural and singular of "die", which is important in interpreting them):

Originally Posted by SzassT
I am also interested in the following probability formulas. For the following three situations, you score 1 point for rolling X or higher on a die with Y number of sides and roll Z number of dice, all with Y number of sides, each time. The dice are perfectly balanced and the sides are numbered 1 to Y. I want to know the likelihood of getting any given number of points on a single roll.

You score 1 point for each die that rolls X or higher.

You score 1 point for each die that rolls X or higher. If you roll a 1 on a die, you lose a point. If you roll the max number on a die, you gain 2 points.

You score 1 point for each die that rolls X or higher. If you roll a 1 on a die, you lose a point. If you roll the max number on a die, you gain 1 point and roll that die again.
There are no significant corrections needed here:

Originally Posted by SzassT
Examples for the dice roll question:

On 1d4 (1 die with 4 sides), what are the chances of getting 1 point if you must roll a 3 or higher to get one point?
On 4d4 (4 dice with 4 sides), what are the chances of getting 1 point if you must roll a 2 or higher to get one point?
On 7d10 (7 dice with 10 sides), what are the chances of getting 5 points if you must roll a 6 or higher to get one point?
On 47d12 (47 dice with 12 sides), what are the chances of getting 14 points if you must roll 11 or higher to get one point?
Taking the second of these examples, here is my understanding as I read it, which again you can verify:

You roll 4 4-sided dice, and you get 1 point for each die that shows a 2 or higher. You want to know the probability that you will get EXACTLY 1 point (not AT LEAST 1 point) -- that is, that exactly one die will show 2 or higher.

This one is just a simple case of the binomial distribution. The other examples are similar. Other cases in the original question would be more complicated, but I'm not considering them until confirmation (and even then, I hope others will share ideas!).

10. Originally Posted by Dr.Peterson
I still need some clarification. I have made some corrections below that you can confirm (mostly in distinguishing the plural and singular of "die", which is important in interpreting them)

Originally Posted by Dr.Peterson
Taking the second of these examples, here is my understanding as I read it, which again you can verify:

You roll 4 4-sided dice, and you get 1 point for each die that shows a 2 or higher. You want to know the probability that you will get EXACTLY 1 point (not AT LEAST 1 point) -- that is, that exactly one die will show 2 or higher.

This one is just a simple case of the binomial distribution. The other examples are similar. Other cases in the original question would be more complicated, but I'm not considering them until confirmation
Actually, I am interested in getting AT LEAST a set number of points, not EXACTLY, but that is completely my failure for not specifying such. Perhaps an explanation of my intended application will help in determining an appropriate way to assess the variations of the questions. I am attempting to determine which of the variations yields the most dependable (consistent) results (i.e. More dice = more likely to attain a set point requirement or higher consistently).

My assumption is that the better rule would vary depending on the target Points Attained, number of dice rolled, number to be rolled or exceeded, and number of sides per die, but I'd like to see if there is a particular method among those listed that is more consistent than the other two sets.

My method of testing was going to revolve around the following set of rules.

10d6 (10 dice with 6 sides each) are rolled. A point is gained for each 4 or better. What are the odds of rolling 3 or more points? What are the odds of having 0 points and rolling at least one 1?

10d6 (10 dice with 6 sides each) are rolled. A point is gained for each 4 or 5. A point is lost for each 1. 2 points are gained for each 6. What are the odds of rolling 3 or more points? What are the odds of rolling 0 points and losing points? What are the odds of rolling negative points?

10d6 (10 dice with 6 sides each) are rolled. A point is gained for each 4 or better. A point is lost for each 1. Each die rolls a 6 is rolled again. What are the odds of rolling 3 or more points? What are the odds of rolling 0 points and losing points? What are the odds of rolling negative points?

I hope this helps clarify things and not complicate them!

Originally Posted by Dr.Peterson
(and even then, I hope others will share ideas!).
On this note, I do feel you deserve a specifically directed thank you for helping me thus far and being the only one to do such. So, Dr. Peterson, I appreciate all of the patience and help you've provided thus far. You have been a warm welcome to this community and I greatly appreciate the help you have provided thus far. I hope that I may look to you for assistance in the future and that I may be able to provide the guidance and tutelage that you have provided myself to others in the future. I can only hope the there are more like you in this community. Your students are in excellent hands and I hope that they realize the opportunity they have been given in studying under you.

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