# Probability of getting a dice value

• I
• rabbed
In summary, both the statements "the probability of getting a dice of value 4 would be 1/6" and "the probability of getting a 4 would be 2/10 = 1/5" are correct in repeated trials, but the second statement only applies if the only acceptable trials are those with two 4s and that other trials do not count. Additionally, knowing the number of 4s present can affect the calculation of probabilities, using methods such as Bayesian probabilities. However, probabilities themselves do not depend on the outcome of a trial.
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?

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FactChecker said:
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).

rabbed said:
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)

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

rabbed said:
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.

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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? :)

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.

FactChecker
rabbed said:
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.

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