I Probability of getting a dice value

[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.

 

[rabbed]

Thanks Scott

So our friend would have an advantage over us in predicting per-trial outcomes, but everybody has the same chance in predicting the average long-term outcome?

Also about the physical probability, everybody’s chance of getting a 4 would agree with our friends probability?

 

[rabbed]

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.

For the sake of the scenario I imagined:
Our friend would see all die after they have been tossed, before we hide and rearrange them, still having the same outcomes and still with only our friend knowing the outcomes/number of 4s (we ourselves do not know). Then we pick up a dice, putting it back so our friend can pick up a dice (could be the same that we put back).

 

[FactChecker]

For the sake of the scenario I imagined:
Our friend would see all die after they have been tossed, before we hide and rearrange them, still having the same outcomes and still with only our friend knowing the outcomes/number of 4s (we ourselves do not know). Then we pick up a dice, putting it back so our friend can pick up a dice (could be the same that we put back).

That still doesn't answer the question: Were you asking about "exactly two 4s" or about "at least two 4s"? Please tell us so that we are all talking about the same problem. If it makes no difference to you, pick one ("exactly two 4s" is a simpler calculation)

 

[rabbed]

Our friend sees all die outcomes, so if it is two 4s we can assume he uses exactly two 4s.

 

[FactChecker]

Our friend sees all die outcomes, so if it is two 4s we can assume he uses exactly two 4s.

I guess I don't know what you mean by "he uses". I thought that he just knows there are two 4s and you were just asking what the revised probabilities are with that knowledge. I find your question difficult to follow.

If you are just talking about him knowing the number of 4s, whatever that is, but still proceeding with a blind draw as though he knew nothing, then his probabilities are 1/6, as though he knew nothing.

 

[rabbed]

Note that each trial starts with all die being tossed, but the next random event (within the same trial) is about picking one of the already thrown dies.
I am pretty certain now that without knowledge about the toss outcomes we would have a 1/6 chance to get a certain dice value and with knowledge we would have another (depending on the toss outcomes).
Do we finally agree? :)

 

[rabbed]

Or rather, our best guess without knowledge would be 1/6. But in reality it would correspond to our friends knowledge about the real outcomes - let's say 2/10 if it were two 4s and predict to pick a dice with value 4.

 

[FactChecker]

Or rather, our best guess without knowledge would be 1/6. But in reality it would correspond to our friends knowledge about the real outcomes - let's say 2/10 if it were two 4s and predict to pick a dice with value 4.

I agree.