Hypothesis Test For Independence in Trials?

Clash Royale CLAN TAG#URR8PPP

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Problem

Let's say you have virtual cases in a video game, and each case has one of four distinct items inside. Let's say the items are:

• $A$ with general probability $0.4$


• $B$ with general probability $0.3$


• $C$ with general probability $0.2$


• $D$ with general probability $0.1$

However, there is rumor that these cases are programmed so that you are less likely to obtain duplicate items, as in, when opening a lot of cases, if you open $B$ in one case, for example, the probability of opening $B$ on the very next case is less than what it was originally, $0.3$

I was wondering how I would be able to test that rumor using a hypothesis test.

My guess / Attempt

Seeing as how $A$ is the most common item, I would use that one for my hypothesis test and test

$$H_0: P(A_k) = P(A_k | A_k-1)$$

That is, the probability of opening $A$ on the $k^th$ case is the same as the probability of opening $A$ on the $k^th$ case given we just opened $A$ the case before

Is this a good / viable way to test what I want? And how would I proceed from here after gathering data? Would it be similar to testing for a population proportion?

Thanks for any help given

probability hypothesis-testing

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edited Jul 24 at 13:59

asked Jul 24 at 13:52

WaveX

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up vote
3
down vote

favorite

Problem

Let's say you have virtual cases in a video game, and each case has one of four distinct items inside. Let's say the items are:

• $A$ with general probability $0.4$


• $B$ with general probability $0.3$


• $C$ with general probability $0.2$


• $D$ with general probability $0.1$

However, there is rumor that these cases are programmed so that you are less likely to obtain duplicate items, as in, when opening a lot of cases, if you open $B$ in one case, for example, the probability of opening $B$ on the very next case is less than what it was originally, $0.3$

I was wondering how I would be able to test that rumor using a hypothesis test.

My guess / Attempt

Seeing as how $A$ is the most common item, I would use that one for my hypothesis test and test

$$H_0: P(A_k) = P(A_k | A_k-1)$$

That is, the probability of opening $A$ on the $k^th$ case is the same as the probability of opening $A$ on the $k^th$ case given we just opened $A$ the case before

Is this a good / viable way to test what I want? And how would I proceed from here after gathering data? Would it be similar to testing for a population proportion?

Thanks for any help given

probability hypothesis-testing

share|cite|improve this question

edited Jul 24 at 13:59

asked Jul 24 at 13:52

WaveX

1,8111616

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

Problem

Let's say you have virtual cases in a video game, and each case has one of four distinct items inside. Let's say the items are:

• $A$ with general probability $0.4$


• $B$ with general probability $0.3$


• $C$ with general probability $0.2$


• $D$ with general probability $0.1$

However, there is rumor that these cases are programmed so that you are less likely to obtain duplicate items, as in, when opening a lot of cases, if you open $B$ in one case, for example, the probability of opening $B$ on the very next case is less than what it was originally, $0.3$

I was wondering how I would be able to test that rumor using a hypothesis test.

My guess / Attempt

Seeing as how $A$ is the most common item, I would use that one for my hypothesis test and test

$$H_0: P(A_k) = P(A_k | A_k-1)$$

That is, the probability of opening $A$ on the $k^th$ case is the same as the probability of opening $A$ on the $k^th$ case given we just opened $A$ the case before

Is this a good / viable way to test what I want? And how would I proceed from here after gathering data? Would it be similar to testing for a population proportion?

Thanks for any help given

probability hypothesis-testing

share|cite|improve this question

edited Jul 24 at 13:59

asked Jul 24 at 13:52

WaveX

1,8111616

Problem

Let's say you have virtual cases in a video game, and each case has one of four distinct items inside. Let's say the items are:

• $A$ with general probability $0.4$


• $B$ with general probability $0.3$


• $C$ with general probability $0.2$


• $D$ with general probability $0.1$

However, there is rumor that these cases are programmed so that you are less likely to obtain duplicate items, as in, when opening a lot of cases, if you open $B$ in one case, for example, the probability of opening $B$ on the very next case is less than what it was originally, $0.3$

I was wondering how I would be able to test that rumor using a hypothesis test.

My guess / Attempt

Seeing as how $A$ is the most common item, I would use that one for my hypothesis test and test

$$H_0: P(A_k) = P(A_k | A_k-1)$$

That is, the probability of opening $A$ on the $k^th$ case is the same as the probability of opening $A$ on the $k^th$ case given we just opened $A$ the case before

Is this a good / viable way to test what I want? And how would I proceed from here after gathering data? Would it be similar to testing for a population proportion?

Thanks for any help given

probability hypothesis-testing

share|cite|improve this question

edited Jul 24 at 13:59

asked Jul 24 at 13:52

WaveX

1,8111616

share|cite|improve this question

share|cite|improve this question

edited Jul 24 at 13:59

edited Jul 24 at 13:59

edited Jul 24 at 13:59

asked Jul 24 at 13:52

WaveX

1,8111616

asked Jul 24 at 13:52

WaveX

1,8111616

asked Jul 24 at 13:52

WaveX

1,8111616

1,8111616

add a comment | 

add a comment | 

1 Answer
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0
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Some inspiration from the answer posted at Stat.Stackexchange

The Runs Test was suggested, and after doing some research I learned about the extension of the runs test to $k$ categories instead of being binary (we are not restricted to "+" and "-" )

The expected number of runs is :

$$mu_r = fracn(n+1) - sumn_i^2n$$
For $i$ going from $1$ to $k$ and $n_i$ is the number of times category $i$ appears in the sequence.

The variance is $$sigma_r^2 = fracsumn_i^2 left( sumn_i^2+ n(n+1) right) -2nsumn_i^3 - n^3n^2(n-1)$$

The source of this formula was found at NCSS Analysis Of Runs (pdf)

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answered Jul 24 at 21:39

WaveX

1,8111616

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote

Some inspiration from the answer posted at Stat.Stackexchange

The Runs Test was suggested, and after doing some research I learned about the extension of the runs test to $k$ categories instead of being binary (we are not restricted to "+" and "-" )

The expected number of runs is :

$$mu_r = fracn(n+1) - sumn_i^2n$$
For $i$ going from $1$ to $k$ and $n_i$ is the number of times category $i$ appears in the sequence.

The variance is $$sigma_r^2 = fracsumn_i^2 left( sumn_i^2+ n(n+1) right) -2nsumn_i^3 - n^3n^2(n-1)$$

The source of this formula was found at NCSS Analysis Of Runs (pdf)

share|cite|improve this answer

answered Jul 24 at 21:39

WaveX

1,8111616

add a comment | 

up vote
0
down vote

Some inspiration from the answer posted at Stat.Stackexchange

The Runs Test was suggested, and after doing some research I learned about the extension of the runs test to $k$ categories instead of being binary (we are not restricted to "+" and "-" )

The expected number of runs is :

$$mu_r = fracn(n+1) - sumn_i^2n$$
For $i$ going from $1$ to $k$ and $n_i$ is the number of times category $i$ appears in the sequence.

The variance is $$sigma_r^2 = fracsumn_i^2 left( sumn_i^2+ n(n+1) right) -2nsumn_i^3 - n^3n^2(n-1)$$

The source of this formula was found at NCSS Analysis Of Runs (pdf)

share|cite|improve this answer

answered Jul 24 at 21:39

WaveX

1,8111616

add a comment | 

up vote
0
down vote

up vote
0
down vote

Some inspiration from the answer posted at Stat.Stackexchange

The Runs Test was suggested, and after doing some research I learned about the extension of the runs test to $k$ categories instead of being binary (we are not restricted to "+" and "-" )

The expected number of runs is :

$$mu_r = fracn(n+1) - sumn_i^2n$$
For $i$ going from $1$ to $k$ and $n_i$ is the number of times category $i$ appears in the sequence.

The variance is $$sigma_r^2 = fracsumn_i^2 left( sumn_i^2+ n(n+1) right) -2nsumn_i^3 - n^3n^2(n-1)$$

The source of this formula was found at NCSS Analysis Of Runs (pdf)

share|cite|improve this answer

answered Jul 24 at 21:39

WaveX

1,8111616

Some inspiration from the answer posted at Stat.Stackexchange

The Runs Test was suggested, and after doing some research I learned about the extension of the runs test to $k$ categories instead of being binary (we are not restricted to "+" and "-" )

The expected number of runs is :

$$mu_r = fracn(n+1) - sumn_i^2n$$
For $i$ going from $1$ to $k$ and $n_i$ is the number of times category $i$ appears in the sequence.

The variance is $$sigma_r^2 = fracsumn_i^2 left( sumn_i^2+ n(n+1) right) -2nsumn_i^3 - n^3n^2(n-1)$$

The source of this formula was found at NCSS Analysis Of Runs (pdf)

share|cite|improve this answer

answered Jul 24 at 21:39

WaveX

1,8111616

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 21:39

WaveX

1,8111616

answered Jul 24 at 21:39

WaveX

1,8111616

answered Jul 24 at 21:39

WaveX

1,8111616

1,8111616

add a comment | 

add a comment | 

 

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