Probability of getting a dice value

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Discussion Overview

The discussion revolves around the probability of rolling a specific value (4) with a set of ten dice, exploring different scenarios based on knowledge of the dice values. Participants examine the implications of known versus unknown quantities of specific outcomes and the nature of probability in repeated trials.

Discussion Character

  • Debate/contested
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • Some participants propose that if there are ten unknown dice, the probability of rolling a 4 is 1/6, assuming no prior knowledge of the dice.
  • Others argue that if it is known there are two dice showing a 4, the probability of rolling a 4 becomes 2/10, or 1/5.
  • Several participants assert that both probabilities can be correct depending on the context of the trials being considered.
  • There is a discussion about the distinction between "before-the-fact" probabilities and "momentaneous" or Bayesian probabilities, with some suggesting that prior knowledge affects probability calculations.
  • Some participants express uncertainty about how knowledge of the dice affects the probability for different individuals involved in the trial.
  • There are mentions of the importance of whether trials are conducted with or without replacement, although some participants note that this may not be relevant to the original question of picking one die.
  • One participant highlights that the theoretical probability of rolling a 4 is 1/6 in the limit of infinite trials, while acknowledging that actual outcomes may fluctuate in finite trials.
  • There is a suggestion that the understanding of probabilities may depend on whether the repeated trials are defined as "with replacement" or "without replacement," leading to different implications for subsequent draws.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the interpretation of probabilities in this scenario. There are competing views on how knowledge of the dice affects probability and whether the probabilities change based on prior knowledge or outcomes.

Contextual Notes

Participants note that the interpretation of probabilities can vary based on assumptions about the number of specific outcomes and the nature of the trials (e.g., with or without replacement). There is also ambiguity regarding the definitions of "exactly 2" versus "at least 2" dice showing a specific value.

rabbed
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If I have ten dice and pick one at random,
given that I don't know anything about them the probability of getting a dice of value 4 would be 1/6?
Given that I know that there is two dice of value 4, the probability of getting a 4 would be 2/10 = 1/5?
Which would be correct in repeated trials?
 
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They are both correct in repeated trials. For the second part one must assume that the only acceptable trials are those with two 4s and that other trials do not count. Any "given" condition means that trials satisfying those conditions are the only ones that count.
 
FactChecker said:
They are both correct in repeated trials. For the second part one must assume that the only acceptable trials are those with two 4s and that other trials do not count. Any "given" condition means that trials satisfying those conditions are the only ones that count.
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Right.
And if you have 10 unknown dice it might not be a 1/6 chance per trial, but in the long run it will be. Allright
 
rabbed said:
Right.
And if you have 10 unknown dice it might not be a 1/6 chance per trial, but in the long run it will be. Allright
No, it WILL be a 1/6 chance every time. Why would it not be?
 
Because behind the 10 unknown, there might be two 4’s hiding
 
rabbed said:
Because behind the 10 unknown, there might be two 4’s hiding
There might be ANYTHING. So what? We are talking about the probabilities. You are confusing actual outcome with probability. Based on that logic, you would expect a fair dice to roll a 4 exactly once every 6 rolls. That's not the way it works.
 
rabbed said:
Because behind the 10 unknown, there might be two 4’s hiding
The "given two 4s" does not have any effect on the first part. The 1/6 probability covers all conditions.
 
But if there actually is two 4s behind the unknowns, the probability would be 2/10 right?
Hm
 
  • #10
The 1/6 only applies in the limit of infinite trials, i think.
But in each trial there might be fluctuations or whatever to call it
 
  • #11
rabbed said:
The 1/6 only applies in the limit of infinite trials, i think.
But in each trial there might be fluctuations or whatever to call it
There are different results of a trial, but if the trials are done correctly the before-the-fact probability is always 1/6. In school problems, you should assume that the trials are done correctly unless the problem specifies otherwise.
 
  • #12
Sure, but by using ”before-the-fact” probabilies to get some other answer, the other answer can only be tested by repeated trials (limit towards infinite trials).
So what you’re actually doing is using infinite-trial probabilities, which on their way to infinite trials may fluctuate.
Just want that confirmed.
 
  • #13
I guess it depends on the case also..
If you just have one dice, there can be no fluctuation of probability. Hm, or if you make a normalized histogram of the outcomes i guess it will not be uniform until a lot of trials
 
  • #14
rabbed said:
Sure, but by using ”before-the-fact” probabilies to get some other answer, the other answer can only be tested by repeated trials (limit towards infinite trials).
So what you’re actually doing is using infinite-trial probabilities, which on their way to infinite trials may fluctuate.
Just want that confirmed.
In most of these problems, they will assume theoretical probabilities of 1/6 for each side of a perfect die (or 1/2 for each side of a perfect coin). The theoretical probabilities are what you are calling "infinite-trial probabilities" but they are derived theoretically, not from varying finite-trial results. (Notice that I said varying finite-trial results, not finite-trial probabilities. Results will vary, but the probabilities will not.)
 
  • #15
So what you are saying is, that in the case of 10 unknown dice.
If my friend checks and sees two 4s and I then move them around (I’m still unaware of the number of 4s), he will have a 2/10 chance to get a 4, whereas I have a 1/6 chance?
 
  • #16
rabbed said:
So what you are saying is, that in the case of 10 unknown dice.
If my friend checks and sees two 4s and I then move them around (I’m still unaware of the number of 4s), he will have a 2/10 chance to get a 4, whereas I have a 1/6 chance?
I certainly didn't mean to implied that. Your friend's knowledge does not change the physical situation and the physical situation is identical for both of you. If there are two 4s (that is a given), then you both have a 2/10 chance of getting a 4. If 2 4s is not a given pre-condition, then the probabilities for both of you are 1/6.
That being said, there are important situations when your knowledge should be used to adjust your calculation of the probabilities. If you know that there are 2 4s, then you know that the probabilities are 2/10. A person picking from the same die trial who doesn't know there are 2 4s must calculate that the probabilities are 1/6, but he would be wrong for that trial. His actual probability of getting a 4 for that trial would also be 2/10. The method for correcting probabilities based on additional information is called Bayesian probabilities.
 
  • #17
So I think we agree then?
 
  • #18
rabbed said:
So I think we agree then?
I'm not sure. I certainly don't agree with your last post and am doubtful about earlier posts.
 
  • #19
FactChecker said:
I'm not sure. I certainly don't agree with your last post and am doubtful about earlier posts.
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  • #20
Could we agree that we have one probability that is constant over all trials, no matter at which trial we calculate it (the before-the-fact/infinite-trials probability) and then other probabilities per trial (bayesian/momentaneous probability) depending on outcome knowledge?
 
  • #21
rabbed said:
Could we agree that we have one probability that is constant over all trials, no matter at which trial we calculate it (the before-the-fact/infinite-trials probability) and then other probabilities per trial (bayesian/momentaneous probability) depending on outcome knowledge?
No. Probabilities do not depend on the outcome. I think you are confusing outcome with additional prior knowledge. If you KNOW that there are two 4s at the start of the trial, that is prior knowledge, not outcome.
 
  • #22
But if we change ”depending on outcome knowledge?” to ”depending on prior knowledge?”, do we agree?
 
  • #23
rabbed said:
If I have ten dice and pick one at random,
given that I don't know anything about them the probability of getting a dice of value 4 would be 1/6?
Given that I know that there is two dice of value 4, the probability of getting a 4 would be 2/10 = 1/5?
Which would be correct in repeated trials?

That depends on whether the repeated trials are "with replacement" or "without replacement", and whether the statement that there are 2 dice with value 4 means "exactly 2" or "at least 2".

Assuming you mean "exactly 2" and "with replacement" (and assuming some form of randomization between draws, so you have no idea where the previously-drawn die is now located), then yes, the probability of drawing a 4 is 2/10 each time.

If you do draws without replacement (that is, you do not replace a drawn die) then it becomes more interesting and more challenging. If your first draw is a non-4, that leaves 9 dice among which there are two 4's, so you chance of drawing a 4 on your second draw is 2/9. However, if your first draw is a 4, that leaves 9 dice of which exactly one is a 4, so you chances of getting a 4 on your second draw would be 1/9. Similarly, the probabilities for your third draw depend on what you got in your first two draws, etc.
 
  • #24
Ray Vickson said:
That depends on whether the repeated trials are "with replacement" or "without replacement",
He only picks one, so replacement is not relevant.
"exactly 2" or "at least 2"
Good point.
 
  • #25
FactChecker said:
He only picks one, so replacement is not relevant.
Good point.

In his first post the OP asked about repeated trials also. I was dealing with that question. So did you in post #2, where you said "They are both correct in repeated trials."

I suppose one could argue that trials are not actually "repeated" unless all (controllable) conditions are the same, so would need replacement to be truly repeated, at least in the case of a finite population.
 
  • #26
Ray Vickson said:
In his first post the OP asked about repeated trials also. I was dealing with that question. So did you in post #2, where you said "They are both correct in repeated trials."
By repeated trials, I mean that each trial is to roll 10 different die and pick one. So there is no replacement in each trial and each trial has probabilities of 1/6 or 2/10 for the cases with no condition and conditional on two 4s, respectively.
 
  • #27
FactChecker said:
By repeated trials, I mean that each trial is to roll 10 different die and pick one. So there is no replacement in each trial and each trial has probabilities of 1/6 or 2/10 for the cases with no condition and conditional on two 4s, respectively.
I agree that that is the way it should be interpreted / stated.
 
  • #28
FactChecker said:
By repeated trials, I mean that each trial is to roll 10 different die and pick one. So there is no replacement in each trial and each trial has probabilities of 1/6 or 2/10 for the cases with no condition and conditional on two 4s, respectively.

The OP never mentioned "tossing" the dice, so my interpretation was different---perhaps ridiculous, but there you go. To me he seemed to be saying something like having a fixed set of dice lying on a table, perhaps covered by an opaque screen. He reaches under the screen and picks a die. Then a repeated trial (with replacement) would consist of putting back the chosen die in exactly the same orientation it had before (so if it showed a three it would be put back as a three), but put in a different random location under the screen.

Of course, I have never before heard of such a way of dealing with dice, but there is always a first time, and---absurd or not---that was how I interpreted the OP's contribution.
 
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  • #29
Ray Vickson said:
The OP never mentioned "tossing" the dice, so my interpretation was different---perhaps ridiculous, but there you go. To me he seemed to be saying something like having a fixed set of dice lyin on a table, perhaps covered by an opaque screen. He reaches under the screen and picks a die. Then a repeated trial (with replacement) would consist of putting back the chosen die in exactly the same orientation it had before (so if it showed a three it would be put back as a three), but put in a different random location under the screen.

Of course, I have never before heard of such a way of dealing with dice, but there is always a first time, and---absurd or not---that was how I interpreted the OP's contribution.
That interpretation hadn't occurred to me. @rabbed was the first to mention repeated trials and I don't know what he had in mind.
 
  • #30
Ray Vickson said:
The OP never mentioned "tossing" the dice, so my interpretation was different---perhaps ridiculous, but there you go. To me he seemed to be saying something like having a fixed set of dice lying on a table, perhaps covered by an opaque screen. He reaches under the screen and picks a die. Then a repeated trial (with replacement) would consist of putting back the chosen die in exactly the same orientation it had before (so if it showed a three it would be put back as a three), but put in a different random location under the screen.

Of course, I have never before heard of such a way of dealing with dice, but there is always a first time, and---absurd or not---that was how I interpreted the OP's contribution.
Like factchecker, this had not occurred to me either but it is a reasonable interpretation. Another good example of why we insist that people post clearly defined conditions for their questions (and then sometimes don't point out when they don't :smile:)
 

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