Calculating Dice Roll Probability: Learn Basics with Frank

• B
• gtacs
In summary, if you have two dice, the probability of rolling a three is 1/6. If you add a third die, the probability of rolling a three goes up to 2/6. But if you keep adding dice, the probability of rolling a three goes up to 100%.
gtacs
TL;DR Summary
I'm trying to learn the basic concepts of calculating probability as it pertains to dice rolling.
Hello,
I am trying to learn the basic concepts of calculating probability as it pertains to dice rolling. I have searched the internet and not been able to figure it out.
If I have a regular 6 sided dice and I want to know the probability of rolling a 3, I know its 1/6 or 16.6%. This is where I get lost. If I add another dice to question. Meaning, with two, 6-sided dice what is the probability of rolling a 3 on either dice or both? My initial thought was it would go to 2/6 or 33%, but if I continue with this thought process and I keep adding dice when I get to 6 dice I'm at 100%, and that doesn't seem right to me.

I'm basically lost on this and have no idea how to even approach this.
Any help would be greatly appreciated.

Thanks,
Frank

gtacs said:
initial thought was it would go to 2/6 or 33%
If you make a table of six outcomes horizontal and six vertical, you can count the number of outcomes that satisfies your criterion. The fraction is NOT 1/3 !

gtacs said:
and that doesn't seem right to me.
You bet ! Going on to twelve throws would give 200% !
gtacs said:
how to even approach this
General tip: when stuck, try it from the other end: What is the probability of NOT rolling a 3 with 1 throw ? Square it for two throws, etc. Even with a hundred throws you are not at zero probability ! (and that's how it should be, although no one would believe you )

gtacs said:
Summary:: I'm trying to learn the basic concepts of calculating probability as it pertains to dice rolling.

Hello,
I am trying to learn the basic concepts of calculating probability as it pertains to dice rolling. I have searched the internet and not been able to figure it out.
If I have a regular 6 sided dice and I want to know the probability of rolling a 3, I know its 1/6 or 16.6%. This is where I get lost. If I add another dice to question. Meaning, with two, 6-sided dice what is the probability of rolling a 3 on either dice or both? My initial thought was it would go to 2/6 or 33%, but if I continue with this thought process and I keep adding dice when I get to 6 dice I'm at 100%, and that doesn't seem right to me.

I'm basically lost on this and have no idea how to even approach this.
Any help would be greatly appreciated.

Thanks,
Frank

Are you trying to learn this on your own?

BvU said:
If you make a table of six outcomes horizontal and six vertical, you can count the number of outcomes that satisfies your criterion. The fraction is NOT 1/3 !
So, I made a 6x6 grid. I counted the number 3s in the column and row. I counted 12 out of 36 possibilities. Which leads me right back to 12/36 or 1/3?

Yeah, I learning this on my own, not for a class or anything. A probability question came up at work and I realized I didn't know how to calculate it, and was just curious.

gtacs said:
So, I made a 6x6 grid. I counted the number 3s in the column and row. I counted 12 out of 36 possibilities. Which leads me right back to 12/36 or 1/3?

Yeah, I learning this on my own, not for a class or anything. A probability question came up at work and I realized I didn't know how to calculate it, and was just curious.

Are you going to study it more of just this one problem?

PeroK said:
Are you trying to learn this on your own?
Yeah, just messing around. a similar question came up at work, and I realized I didn't know how to calculate this...

PeroK said:
Are you going to study it more of just this one problem?
Yes, I'd like to learn more. Like how to approach these problems...

gtacs said:
Yes, I'd like to learn more. Like how to approach these problems...

You could look online for a pdf or videos on elementary probability theory.

For example, suppose I throw the two dice five times and get:

5+1, 3+3, 5+4, 3+2, 6+4

You have to decide whether you are counting the total number of threes you get. In this case 3/10 or 30%. Or, how often you get at least one three. In this case 2/5 = 40%.

You could try this for all 36 equally likely outcomes: 1+1, 1+2, ... 6+5, 6+6. See what you get.

gtacs said:
So, I made a 6x6 grid. I counted the number 3s in the column and row. I counted 12 out of 36 possibilities. Which leads me right back to 12/36 or 1/3?

See my underline. You don't want to count the number of threes; you want to count how many squares on your grid correspond to throwing one (or more) threes.

Note that the total number of dice rolled in your grid is 72: 36 experiments, each with two dice.

What you actually calculated was 12/72 = 1/6, which is just the probability of any single die being a three.

gtacs said:
I counted the number 3s
That is indeed 12. But you should count the number of outcomes that satisfy your criteria: 1 or 2 threes. That is NOT 12 !

Dale
gtacs said:
Summary:: I'm trying to learn the basic concepts of calculating probability as it pertains to dice rolling.

Hello,
My initial thought was it would go to 2/6 or 33%, but if I continue with this thought process and I keep adding dice when I get to 6 dice I'm at 100%, and that doesn't seem right to me.

Just one more point. If you want the average number of threes per throw, then:

For one die you get 1/6
For two dice you 2/6
For six dice you get 6/6 = 1 (i.e. if you throw six dice at a time, then you get an average of 1 six per throw)
For twelve dice you get 12/6 = 2 (if you throw twelve dice at a time, you get an average of 2 sixes per throw)

In probability theory, this is called the "expected" number or "mean" number.

This is in general different from the probability of getting at least one six on any throw (of several dice).

gtacs said:
I counted the number 3s in the column and row.
You need to count the number of outcomes with a 3 and divide that by the total number of outcomes. An outcome with two 3’s is still just one outcome.

Dale said:
You need to count the number of outcomes with a 3 and divide that by the total number of outcomes. An outcome with two 3’s is still just one outcome.

I was hoping for an alternative Bayesian solution!

BvU and Dale
BvU said:
You bet ! Going on to twelve throws would give 200% !
General tip: when stuck, try it from the other end: What is the probability of NOT rolling a 3 with 1 throw ? Square it for two throws, etc. Even with a hundred throws you are not at zero probability ! (and that's how it should be, although no one would believe you )
OK Let me see if I got this...
The probability of NOT rolling a 3 is 5/6 or 83.3% (even I can calculate the chances it WILL happen from here, lol). Then, if I add another dice, I would take (5/6)^2 or 69.2%, three dice (5/6)^3 or 57.8% etc...And, that's all there is to it? So, each time you add a dice you reduce the probability of NOT rolling a 3 by multiplying the previous value by .833?
I went to khan academy and watched the video, as well as read other stuff but never got it. I would never had thought of this. Thanks to everyone for their help.

BvU and PeroK
gtacs said:
OK Let me see if I got this...
The probability of NOT rolling a 3 is 5/6 or 83.3% (even I can calculate the chances it WILL happen from here, lol). Then, if I add another dice, I would take (5/6)^2 or 69.2%, three dice (5/6)^3 or 57.8% etc...And, that's all there is to it? So, each time you add a dice you reduce the probability of NOT rolling a 3 by multiplying the previous value by .833?
I went to khan academy and watched the video, as well as read other stuff but never got it. I would never had thought of this. Thanks to everyone for their help.

Yes. You can either count the basic equally likely outcomes. Of the 36 outcomes we have:

10 outcomes with exactly one 3
1 outcome with a double 3
25 outcomes with no 3

The probability, therefore, of getting at least one 3 is ##11/36##. And the probability of not getting a 3 is ##25/36##.

This gives the same answer as the complement method. Probability of not getting a 3 is ##(5/6)(5/6) = 25/36##.

If you want more of a challenge you could try to calculate the probability of being dealt each of the poker hands, from a simople 5-card deal: royal flush, straight flush, all the way down to a pair.

There was an interesting probability problem here (question number 3) in one of the monthly maths challenges:

For any two numbers on two independent dice, the probabilities of each taken alone should be multiplied together. So, for instance, the probability of getting a 1 on one die and a 2 on the other die is 1/6 * 1/6 = 1/36. Now to calculate the probability of getting any total from two dice, you must add the probabilities of getting the combinations of numbers that would give the desired total.

1. What is the probability of rolling a specific number on a single die?

The probability of rolling a specific number on a single die is 1/6 or approximately 16.67%. This is because there are six possible outcomes (numbers 1-6) and each outcome has an equal chance of occurring.

2. How do you calculate the probability of rolling a certain sum with multiple dice?

To calculate the probability of rolling a certain sum with multiple dice, you need to first determine the total number of possible outcomes. Then, count the number of outcomes that result in the desired sum. Finally, divide the number of desired outcomes by the total number of possible outcomes to get the probability. For example, if you want to know the probability of rolling a sum of 8 with two dice, there are 36 total outcomes and 5 desired outcomes (2+6, 3+5, 4+4, 5+3, 6+2), so the probability would be 5/36 or approximately 13.89%.

3. What is the probability of rolling a higher number with two dice compared to one die?

The probability of rolling a higher number with two dice compared to one die is higher. This is because with two dice, there are more possible outcomes and thus a higher chance of rolling a higher number. For example, the probability of rolling a 6 with one die is 1/6, but the probability of rolling a 6 with two dice is 5/36.

4. How do you calculate the probability of rolling a specific sequence of numbers with multiple dice?

To calculate the probability of rolling a specific sequence of numbers with multiple dice, you need to first determine the total number of possible outcomes. Then, count the number of outcomes that result in the desired sequence. Finally, divide the number of desired outcomes by the total number of possible outcomes to get the probability. For example, if you want to know the probability of rolling a sequence of 2-3-4 with three dice, there are 216 total outcomes and 1 desired outcome (2+3+4), so the probability would be 1/216 or approximately 0.46%.

5. How does the number of dice rolled affect the probability of rolling a certain number or sequence?

The number of dice rolled directly affects the probability of rolling a certain number or sequence. As the number of dice increases, the total number of possible outcomes also increases, leading to a lower probability of rolling a specific number or sequence. For example, the probability of rolling a 6 with one die is 1/6, but the probability of rolling a 6 with two dice is 5/36 and the probability of rolling a 6 with three dice is 25/216.

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