Probability of getting a dice value

[rabbed]

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?

 

[FactChecker]

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.

 

[phinds]

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.

 

[rabbed]

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

 

[phinds]

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?

 

[rabbed]

Because behind the 10 unknown, there might be two 4’s hiding

 

[phinds]

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.

 

[FactChecker]

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.

 

[rabbed]

But if there actually is two 4s behind the unknowns, the probability would be 2/10 right?
Hm

 

[rabbed]

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

 

[FactChecker]

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.

 

[rabbed]

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.

 

[rabbed]

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

 

[FactChecker]

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.)

 

[rabbed]

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?

 

[FactChecker]

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.

 

[rabbed]

So I think we agree then?

 

[FactChecker]

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.

 

[phinds]

I'm not sure. I certainly don't agree with your last post and am doubtful about earlier posts.

 

[rabbed]

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?

 

[phinds]

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.

 

[rabbed]

But if we change ”depending on outcome knowledge?” to ”depending on prior knowledge?”, do we agree?

 

[Ray Vickson]

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.

 

[FactChecker]

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.

 

[Ray Vickson]

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.

 

[FactChecker]

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.

 

[phinds]

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.

 

[Ray Vickson]

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.

 

[FactChecker]

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.

 

[phinds]

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 )

 

[rabbed]

Hi

I’m reading a book about entropy where the author goes from tossing 10 die, calculating the sum of these, to analyzing the evolution of this process by (at each trial) having the outcomes of the previous trial (first configuration being all dice having lowest value), showing that the probabilities of contributing to a higher or equal sum are higher than contributing to a lower sum, until equilibrium is reached.

I imagined that in the first case all ten die were tossed (one random choice per dice), but in the other case it’s like first a random choice of picking up a dice (from the previous outcomes) and then another random choice of tossing it to get the probabilities of affecting the sum.

In the book, each dice only has 0 or 1 as values (equal amount of both).

Hope it makes sense.

 

[FactChecker]

I imagined that in the first case all ten die were tossed (one random choice per dice), but in the other case it’s like first a random choice of picking up a dice (from the previous outcomes) and then another random choice of tossing it to get the probabilities of affecting the sum.

By "other case" do you mean the case of two 4s? The way something like that is interpreted is like this: Roll 10 die. If there are not two 4s, roll them all again. Continue till there are two 4s. Then select a die at random and see if you got one of the 4s. That is one trial. There are other ways to approach that, but it is dangerous to assume that another way will really give an identical probability. These things can be very deceptive.

In the book, each dice only has 0 or 1 as values (equal amount of both).

Hope it makes sense.

I don't really understand the situation, but I hope that it is not important for getting the correct answer to your original question.

 

[rabbed]

Sort of. The 10-die configurations/outcomes are different each trial, so it might be two 4s at one trial and five 4s in some other trial I guess.

 

[FactChecker]

Sort of. The 10-die configurations/outcomes are different each trial, so it might be two 4s at one trial and five 4s in some other trial I guess.

Did you ever answer @Ray Vickson 's question of whether you mean exactly two 4s or at least two 4s? In either case, continue rolling sets of 10 die till you get a result that matches the condition. Then select your one die and see if it is a 4. That would be one trial.
The answer of probability = 2/10 is for the condition of exactly two 4s.

 

[.Scott]

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?

There are some unspecified conditions here.
Let's say we are doing repeated tests following these procedures:
1) We toss 10 dice.
2) Friend checks the dice and rearranges them.
3) Each of us picks a die - perhaps the same one.

So, on average, both we and our friend will pick a 4 with P=1/6.
In each case, our friend will know more than us, so he will be able to divide these tests into more specific populations: those with 0, 1, 2, ... or 10 die of value 4. The chance of picking a 4 in each of those populations is 0, 10%, 20%, ... 100% respectively.

There is no difference in physical results. But we are using these probabilities to predict a result and more knowledge can result in a better prediction.