Indicator Random Variables for Card Game (2 Red in a row)
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Working through some practice problems in Introduction to Probability, Blitzstein. (I found similar problems to this one but want to make sure I understand this concept.)
You have a well-shuffed 52-card deck. On average, how many pairs of adjacent cards are there such that both cards are red?
Create indicator random variable I, where I=1 if both red, else 0
Cards could be RR, RB, BB, BR, so probability of indicator variable success is:
- P(I=0)=$frac{3}{4}$
- P(I=1)=$frac{1}{4}$
You don't have to check the 1st card alone, so there are 52-1 cards to check. Expectation is therefore:
$E(I) = sum_{i=1}^{51} P(I_i=1) = 51(0*frac{3}{4}+1frac{1}{4}) = frac{51}{4}$
Questions:
1. Is the logic correct?
2. Can someone verify and further explain why sum from 1 to 51 is correct (or incorrect if I'm wrong)
3. Is the nomenclature for what to sum correct?
(Edit: edited P(I=1) and P(I=0) per comment below)
random-variables card-games
asked Dec 1 '18 at 19:58
user603569user603569
728
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add a comment |
$begingroup$
Working through some practice problems in Introduction to Probability, Blitzstein. (I found similar problems to this one but want to make sure I understand this concept.)
You have a well-shuffed 52-card deck. On average, how many pairs of adjacent cards are there such that both cards are red?
Create indicator random variable I, where I=1 if both red, else 0
Cards could be RR, RB, BB, BR, so probability of indicator variable success is:
- P(I=0)=$frac{3}{4}$
- P(I=1)=$frac{1}{4}$
You don't have to check the 1st card alone, so there are 52-1 cards to check. Expectation is therefore:
$E(I) = sum_{i=1}^{51} P(I_i=1) = 51(0*frac{3}{4}+1frac{1}{4}) = frac{51}{4}$
Questions:
1. Is the logic correct?
2. Can someone verify and further explain why sum from 1 to 51 is correct (or incorrect if I'm wrong)
3. Is the nomenclature for what to sum correct?
(Edit: edited P(I=1) and P(I=0) per comment below)
random-variables card-games
asked Dec 1 '18 at 19:58
user603569user603569
728
$endgroup$
add a comment |
$begingroup$
Working through some practice problems in Introduction to Probability, Blitzstein. (I found similar problems to this one but want to make sure I understand this concept.)
You have a well-shuffed 52-card deck. On average, how many pairs of adjacent cards are there such that both cards are red?
Create indicator random variable I, where I=1 if both red, else 0
Cards could be RR, RB, BB, BR, so probability of indicator variable success is:
- P(I=0)=$frac{3}{4}$
- P(I=1)=$frac{1}{4}$
You don't have to check the 1st card alone, so there are 52-1 cards to check. Expectation is therefore:
$E(I) = sum_{i=1}^{51} P(I_i=1) = 51(0*frac{3}{4}+1frac{1}{4}) = frac{51}{4}$
Questions:
1. Is the logic correct?
2. Can someone verify and further explain why sum from 1 to 51 is correct (or incorrect if I'm wrong)
3. Is the nomenclature for what to sum correct?
(Edit: edited P(I=1) and P(I=0) per comment below)
random-variables card-games
asked Dec 1 '18 at 19:58
user603569user603569
728
$endgroup$
Working through some practice problems in Introduction to Probability, Blitzstein. (I found similar problems to this one but want to make sure I understand this concept.)
You have a well-shuffed 52-card deck. On average, how many pairs of adjacent cards are there such that both cards are red?
Create indicator random variable I, where I=1 if both red, else 0
Cards could be RR, RB, BB, BR, so probability of indicator variable success is:
- P(I=0)=$frac{3}{4}$
- P(I=1)=$frac{1}{4}$
You don't have to check the 1st card alone, so there are 52-1 cards to check. Expectation is therefore:
$E(I) = sum_{i=1}^{51} P(I_i=1) = 51(0*frac{3}{4}+1frac{1}{4}) = frac{51}{4}$
Questions:
1. Is the logic correct?
2. Can someone verify and further explain why sum from 1 to 51 is correct (or incorrect if I'm wrong)
3. Is the nomenclature for what to sum correct?
(Edit: edited P(I=1) and P(I=0) per comment below)
random-variables card-games
random-variables card-games
asked Dec 1 '18 at 19:58
user603569user603569
728
asked Dec 1 '18 at 19:58
user603569user603569
728
edited Dec 1 '18 at 20:27
user603569
asked Dec 1 '18 at 19:58
user603569user603569
728
asked Dec 1 '18 at 19:58
user603569user603569
728
asked Dec 1 '18 at 19:58
user603569user603569
728
728
add a comment |
add a comment |
1 Answer
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Apart from what looks likes a typo flipping $P(I_i = 1)$ and $P(I_i=0)$, this looks okay. You should also be careful with being explicit about your random variables; as you define $I$, you should have $E(I) = frac{1}{4}$ (what you really want is to define a separate $I_i$ for each $1 leq i leq 51$).
The justification for the sum comes from linearity of expectation. If you define the $I_i$'s as described, and let $I$ be the random variable denoting the total number of positions where this occurs, we have:
$$Eleft(Iright) = Eleft(sum_{i=1}^{51} I_i right) = sum_{i=1}^{51} Eleft(I_i right) = sum_{i=1}^{51} Pr(I_i = 1) = frac{51}{4}$$
answered Dec 1 '18 at 20:20
plattyplatty
3,370320
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$begingroup$
Good catch on P(I=1) and P(I=0); I'll edit in the question in case anyone gets confused by that. On semantics: is there any issue with writing $sum_{i=2}^{52}$ instead?
$endgroup$
– user603569
Dec 1 '18 at 20:26
1
$begingroup$
Not really; this depends on how you set up your indicators (does $I_i$ represent $i$ and $i+1$ both being red, or $i$ and $i-1$ both being red?) You should get the same answer either way.
$endgroup$
– platty
Dec 1 '18 at 20:27
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
Apart from what looks likes a typo flipping $P(I_i = 1)$ and $P(I_i=0)$, this looks okay. You should also be careful with being explicit about your random variables; as you define $I$, you should have $E(I) = frac{1}{4}$ (what you really want is to define a separate $I_i$ for each $1 leq i leq 51$).
The justification for the sum comes from linearity of expectation. If you define the $I_i$'s as described, and let $I$ be the random variable denoting the total number of positions where this occurs, we have:
$$Eleft(Iright) = Eleft(sum_{i=1}^{51} I_i right) = sum_{i=1}^{51} Eleft(I_i right) = sum_{i=1}^{51} Pr(I_i = 1) = frac{51}{4}$$
answered Dec 1 '18 at 20:20
plattyplatty
3,370320
$endgroup$
$begingroup$
Good catch on P(I=1) and P(I=0); I'll edit in the question in case anyone gets confused by that. On semantics: is there any issue with writing $sum_{i=2}^{52}$ instead?
$endgroup$
– user603569
Dec 1 '18 at 20:26
1
$begingroup$
Not really; this depends on how you set up your indicators (does $I_i$ represent $i$ and $i+1$ both being red, or $i$ and $i-1$ both being red?) You should get the same answer either way.
$endgroup$
– platty
Dec 1 '18 at 20:27
add a comment |
$begingroup$
Apart from what looks likes a typo flipping $P(I_i = 1)$ and $P(I_i=0)$, this looks okay. You should also be careful with being explicit about your random variables; as you define $I$, you should have $E(I) = frac{1}{4}$ (what you really want is to define a separate $I_i$ for each $1 leq i leq 51$).
The justification for the sum comes from linearity of expectation. If you define the $I_i$'s as described, and let $I$ be the random variable denoting the total number of positions where this occurs, we have:
$$Eleft(Iright) = Eleft(sum_{i=1}^{51} I_i right) = sum_{i=1}^{51} Eleft(I_i right) = sum_{i=1}^{51} Pr(I_i = 1) = frac{51}{4}$$
answered Dec 1 '18 at 20:20
plattyplatty
3,370320
$endgroup$
$begingroup$
Good catch on P(I=1) and P(I=0); I'll edit in the question in case anyone gets confused by that. On semantics: is there any issue with writing $sum_{i=2}^{52}$ instead?
$endgroup$
– user603569
Dec 1 '18 at 20:26
1
$begingroup$
Not really; this depends on how you set up your indicators (does $I_i$ represent $i$ and $i+1$ both being red, or $i$ and $i-1$ both being red?) You should get the same answer either way.
$endgroup$
– platty
Dec 1 '18 at 20:27
add a comment |
$begingroup$
Apart from what looks likes a typo flipping $P(I_i = 1)$ and $P(I_i=0)$, this looks okay. You should also be careful with being explicit about your random variables; as you define $I$, you should have $E(I) = frac{1}{4}$ (what you really want is to define a separate $I_i$ for each $1 leq i leq 51$).
The justification for the sum comes from linearity of expectation. If you define the $I_i$'s as described, and let $I$ be the random variable denoting the total number of positions where this occurs, we have:
$$Eleft(Iright) = Eleft(sum_{i=1}^{51} I_i right) = sum_{i=1}^{51} Eleft(I_i right) = sum_{i=1}^{51} Pr(I_i = 1) = frac{51}{4}$$
answered Dec 1 '18 at 20:20
plattyplatty
3,370320
$endgroup$
Apart from what looks likes a typo flipping $P(I_i = 1)$ and $P(I_i=0)$, this looks okay. You should also be careful with being explicit about your random variables; as you define $I$, you should have $E(I) = frac{1}{4}$ (what you really want is to define a separate $I_i$ for each $1 leq i leq 51$).
The justification for the sum comes from linearity of expectation. If you define the $I_i$'s as described, and let $I$ be the random variable denoting the total number of positions where this occurs, we have:
$$Eleft(Iright) = Eleft(sum_{i=1}^{51} I_i right) = sum_{i=1}^{51} Eleft(I_i right) = sum_{i=1}^{51} Pr(I_i = 1) = frac{51}{4}$$
answered Dec 1 '18 at 20:20
plattyplatty
3,370320
answered Dec 1 '18 at 20:20
plattyplatty
3,370320
answered Dec 1 '18 at 20:20
plattyplatty
3,370320
answered Dec 1 '18 at 20:20
plattyplatty
3,370320
3,370320
$begingroup$
Good catch on P(I=1) and P(I=0); I'll edit in the question in case anyone gets confused by that. On semantics: is there any issue with writing $sum_{i=2}^{52}$ instead?
$endgroup$
– user603569
Dec 1 '18 at 20:26
1
$begingroup$
Not really; this depends on how you set up your indicators (does $I_i$ represent $i$ and $i+1$ both being red, or $i$ and $i-1$ both being red?) You should get the same answer either way.
$endgroup$
– platty
Dec 1 '18 at 20:27
add a comment |
$begingroup$
Good catch on P(I=1) and P(I=0); I'll edit in the question in case anyone gets confused by that. On semantics: is there any issue with writing $sum_{i=2}^{52}$ instead?
$endgroup$
– user603569
Dec 1 '18 at 20:26
1
$begingroup$
Not really; this depends on how you set up your indicators (does $I_i$ represent $i$ and $i+1$ both being red, or $i$ and $i-1$ both being red?) You should get the same answer either way.
$endgroup$
– platty
Dec 1 '18 at 20:27
$begingroup$
Good catch on P(I=1) and P(I=0); I'll edit in the question in case anyone gets confused by that. On semantics: is there any issue with writing $sum_{i=2}^{52}$ instead?
$endgroup$
– user603569
Dec 1 '18 at 20:26
$begingroup$
Good catch on P(I=1) and P(I=0); I'll edit in the question in case anyone gets confused by that. On semantics: is there any issue with writing $sum_{i=2}^{52}$ instead?
$endgroup$
– user603569
Dec 1 '18 at 20:26
1
1
$begingroup$
Not really; this depends on how you set up your indicators (does $I_i$ represent $i$ and $i+1$ both being red, or $i$ and $i-1$ both being red?) You should get the same answer either way.
$endgroup$
– platty
Dec 1 '18 at 20:27
$begingroup$
Not really; this depends on how you set up your indicators (does $I_i$ represent $i$ and $i+1$ both being red, or $i$ and $i-1$ both being red?) You should get the same answer either way.
$endgroup$
– platty
Dec 1 '18 at 20:27
add a comment |
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