- #1

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say you throw a dice, probability of 'x' occurring is 1/6

Say you throw two dice, probability of 7 occurring is more than others.

But how do you prove this experimentally? Throw a dice? how many times?

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- #1

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say you throw a dice, probability of 'x' occurring is 1/6

Say you throw two dice, probability of 7 occurring is more than others.

But how do you prove this experimentally? Throw a dice? how many times?

- #2

Bacle2

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I don't think you can find an actual proof, but there are tests like the chi-squared, and

I think other goodness-of-fit tests to study how well your probability model fits the

actual data.

- #3

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Let's assume the die is fair.

I don't think you can find an actual proof, but there are tests like the chi-squared, and

I think other goodness-of-fit tests to study how well your probability model fits the

actual data.

Going by the probability distribution function- probability of the sum being 7 is maximum. Under what conditions is this true?

- #4

chiro

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You can't actually prove that a sample comes from a population if the sample contains a finite amount of data (and it will if it's a sample like the one you have: the only exception is when the population is finite like in the example of surveys like a nationwide census).

What you will typically do is to construct an estimator that corresponds to the probabilities of a fair-dice. You could use for example a chi-square goodness of fit test to see if the observed distribution is statistically significantly the same as a discrete uniform distribution with some particular rejection region corresponding to a p-value (probability cut-off value).

You will never be able to prove anything: we can't actually do this in statistics. The best we can do is to show enough evidence that something appears to hold enough water for it to be taken seriously.

To understand this, think of the following situation:

Lets say we have a sample of a million data points from a normal distribution and we need to get an estimator for the mean.

Now think about the following situation: the values we get from our sample give us a mean of -100 and we get a tight estimator that centres around this (with small variance in the estimator because we have so many samples).

Then we get another million data points and we find the mean of these new points: we get a sample mean of +1000, and if we want to estimate the population mean from this using classical methods, we would get our estimator to have a small variance and a mean of +1000.

This kind of example demonstrates why we can't really prove something using statistics and it also shows the importance of getting a sample that really is representative: in the above example both samples were not really good representations of the entire process but highly biased ones.

The question you will need to ask yourself is how you can obtain sample data with the least amount of bias possible and only then decide how to test your hypothesis of a fair die or not a fair die.

Sampling considerations and bias considerations are not just mathematical, but they are experimental in a physical sense and things like using say different die manufacturers, and other considerations will help you eliminate sources of biases. The key thing is to have enough of a reference point so that you can be as comfortable as possible with having removed the more significant sources of bias. Everything from the dice used, to how its thrown, recorded, and so on are important.

- #5

HallsofIvy

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If you

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One die. Two dice. How many times you throw depends on how accurate you want the result to be.

By calculation, the chance of rolling '7' on 2d6 is 1/6.

There several ways to do this. The easiest as stated is to count the combinations.

The generic way to add probabilities is:

If two events both need to occur, multiply them. If either one or the other, add them.

So you can roll a '1' on the first (1/6) die AND a '6' on the second die (1/6) = 1/6*1/6 = 1/36

OR (+)

A '2' on the first die AND a '5' on the second die (1/6 * 1/6)

OR (+)

A '3' on the first die AND a '4' on the second die (1/6 * 1/6)

... continue to 6 and 1

You get 1/36 + 1/36 + 1/36 ... (6 times in total) for 6/36 which is 1/6.

1,2 and 2,1 are both 3 but are different combinations so the chance to roll '3' is 2/36 or 1/18. However the chance to get '2' or '12' is 1/36 each because 1,1 is the same as 1,1.

By calculation, the chance of rolling '7' on 2d6 is 1/6.

There several ways to do this. The easiest as stated is to count the combinations.

The generic way to add probabilities is:

If two events both need to occur, multiply them. If either one or the other, add them.

So you can roll a '1' on the first (1/6) die AND a '6' on the second die (1/6) = 1/6*1/6 = 1/36

OR (+)

A '2' on the first die AND a '5' on the second die (1/6 * 1/6)

OR (+)

A '3' on the first die AND a '4' on the second die (1/6 * 1/6)

... continue to 6 and 1

You get 1/36 + 1/36 + 1/36 ... (6 times in total) for 6/36 which is 1/6.

1,2 and 2,1 are both 3 but are different combinations so the chance to roll '3' is 2/36 or 1/18. However the chance to get '2' or '12' is 1/36 each because 1,1 is the same as 1,1.

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- #7

DaveC426913

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You're conflating two things. The probability of the sum being 7 as maximum is true before any dice are thrown, and after a million dice are thrown. Your rolls do not affect the probability.Let's assume the die is fair.

Going by the probability distribution function- probability of the sum being 7 is maximum. Under what conditions is this true?

But your experiment will demonstrate

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