Dice roll probability

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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
 

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  • #2
BvU
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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 !
 
  • #3
BvU
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and that doesn't seem right to me.
You bet ! Going on to twelve throws would give 200% !
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 :smile:)
 
  • #4
PeroK
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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?
 
  • #5
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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.
 
  • #6
PeroK
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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?
 
  • #7
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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...
 
  • #8
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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...
 
  • #9
PeroK
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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.
 
  • #10
PeroK
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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.
 
  • #11
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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 !
 
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  • #12
PeroK
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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).
 
  • #13
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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.
 
  • #14
PeroK
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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!
 
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  • #15
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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 :smile:)

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.
 
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  • #16
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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:

https://www.physicsforums.com/threads/math-challenge-october-2019.978296/
 
  • #17
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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.
 

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