Expected number of sequences
$begingroup$
Problem: What is the expected number of sequences of $3$ heads in $50$ tosses of a coin?
I am a bit confused about this problem in my course. So far we defined expectation value as:
$$E[X]= sum_{x=0 }^n x cdot P[X=x] $$
Which tells us the expected value, or mean of a certain experiment. However, now they go and just pick a certain value without explaining how this is done. Normally instead of 3, it would say "$X$". How is this done?
I know that there are $2^{50}$ possible sequences as every entry can be either heads or tails.
probability-theory expected-value
asked Dec 11 '18 at 15:53
Wesley StrikWesley Strik
2,007423
$endgroup$
|
show 2 more comments
$begingroup$
Problem: What is the expected number of sequences of $3$ heads in $50$ tosses of a coin?
I am a bit confused about this problem in my course. So far we defined expectation value as:
$$E[X]= sum_{x=0 }^n x cdot P[X=x] $$
Which tells us the expected value, or mean of a certain experiment. However, now they go and just pick a certain value without explaining how this is done. Normally instead of 3, it would say "$X$". How is this done?
I know that there are $2^{50}$ possible sequences as every entry can be either heads or tails.
probability-theory expected-value
asked Dec 11 '18 at 15:53
Wesley StrikWesley Strik
2,007423
$endgroup$
$begingroup$
Some clarification might be needed. Suppose your string was $HHHH$. Is that two sequences of three heads?
$endgroup$
– lulu
Dec 11 '18 at 15:54
$begingroup$
I think they count that as 1. So far we have only been counting $3$ and then we stop.
$endgroup$
– Wesley Strik
Dec 11 '18 at 15:56
$begingroup$
Well, you need to clarify that point. Either way, though, you can do it with indicator variables. For $iin {1,cdots, 48}$ let $X_i$ be the indicator variable which is $1$ if a "good" block (however defined) starts at the $i^{th}$ slot. Easy to compute the expected value of $X_i$ and then use Linearity.
$endgroup$
– lulu
Dec 11 '18 at 15:57
$begingroup$
This sounds way too advanced for a first lecture into introductory probability theory. We only know linearity and coin tosses.
$endgroup$
– Wesley Strik
Dec 11 '18 at 16:00
1
$begingroup$
I don't see an easier way of doing the problem. And indicator variables are pretty easy to use. I suggest looking that up.
$endgroup$
– lulu
Dec 11 '18 at 16:05
|
show 2 more comments
$begingroup$
Problem: What is the expected number of sequences of $3$ heads in $50$ tosses of a coin?
I am a bit confused about this problem in my course. So far we defined expectation value as:
$$E[X]= sum_{x=0 }^n x cdot P[X=x] $$
Which tells us the expected value, or mean of a certain experiment. However, now they go and just pick a certain value without explaining how this is done. Normally instead of 3, it would say "$X$". How is this done?
I know that there are $2^{50}$ possible sequences as every entry can be either heads or tails.
probability-theory expected-value
asked Dec 11 '18 at 15:53
Wesley StrikWesley Strik
2,007423
$endgroup$
Problem: What is the expected number of sequences of $3$ heads in $50$ tosses of a coin?
I am a bit confused about this problem in my course. So far we defined expectation value as:
$$E[X]= sum_{x=0 }^n x cdot P[X=x] $$
Which tells us the expected value, or mean of a certain experiment. However, now they go and just pick a certain value without explaining how this is done. Normally instead of 3, it would say "$X$". How is this done?
I know that there are $2^{50}$ possible sequences as every entry can be either heads or tails.
probability-theory expected-value
probability-theory expected-value
asked Dec 11 '18 at 15:53
Wesley StrikWesley Strik
2,007423
asked Dec 11 '18 at 15:53
Wesley StrikWesley Strik
2,007423
edited Dec 11 '18 at 16:50
Wesley Strik
asked Dec 11 '18 at 15:53
Wesley StrikWesley Strik
2,007423
asked Dec 11 '18 at 15:53
Wesley StrikWesley Strik
2,007423
asked Dec 11 '18 at 15:53
Wesley StrikWesley Strik
2,007423
2,007423
$begingroup$
Some clarification might be needed. Suppose your string was $HHHH$. Is that two sequences of three heads?
$endgroup$
– lulu
Dec 11 '18 at 15:54
$begingroup$
I think they count that as 1. So far we have only been counting $3$ and then we stop.
$endgroup$
– Wesley Strik
Dec 11 '18 at 15:56
$begingroup$
Well, you need to clarify that point. Either way, though, you can do it with indicator variables. For $iin {1,cdots, 48}$ let $X_i$ be the indicator variable which is $1$ if a "good" block (however defined) starts at the $i^{th}$ slot. Easy to compute the expected value of $X_i$ and then use Linearity.
$endgroup$
– lulu
Dec 11 '18 at 15:57
$begingroup$
This sounds way too advanced for a first lecture into introductory probability theory. We only know linearity and coin tosses.
$endgroup$
– Wesley Strik
Dec 11 '18 at 16:00
1
$begingroup$
I don't see an easier way of doing the problem. And indicator variables are pretty easy to use. I suggest looking that up.
$endgroup$
– lulu
Dec 11 '18 at 16:05
|
show 2 more comments
$begingroup$
Some clarification might be needed. Suppose your string was $HHHH$. Is that two sequences of three heads?
$endgroup$
– lulu
Dec 11 '18 at 15:54
$begingroup$
I think they count that as 1. So far we have only been counting $3$ and then we stop.
$endgroup$
– Wesley Strik
Dec 11 '18 at 15:56
$begingroup$
Well, you need to clarify that point. Either way, though, you can do it with indicator variables. For $iin {1,cdots, 48}$ let $X_i$ be the indicator variable which is $1$ if a "good" block (however defined) starts at the $i^{th}$ slot. Easy to compute the expected value of $X_i$ and then use Linearity.
$endgroup$
– lulu
Dec 11 '18 at 15:57
$begingroup$
This sounds way too advanced for a first lecture into introductory probability theory. We only know linearity and coin tosses.
$endgroup$
– Wesley Strik
Dec 11 '18 at 16:00
1
$begingroup$
I don't see an easier way of doing the problem. And indicator variables are pretty easy to use. I suggest looking that up.
$endgroup$
– lulu
Dec 11 '18 at 16:05
$begingroup$
Some clarification might be needed. Suppose your string was $HHHH$. Is that two sequences of three heads?
$endgroup$
– lulu
Dec 11 '18 at 15:54
$begingroup$
Some clarification might be needed. Suppose your string was $HHHH$. Is that two sequences of three heads?
$endgroup$
– lulu
Dec 11 '18 at 15:54
$begingroup$
I think they count that as 1. So far we have only been counting $3$ and then we stop.
$endgroup$
– Wesley Strik
Dec 11 '18 at 15:56
$begingroup$
I think they count that as 1. So far we have only been counting $3$ and then we stop.
$endgroup$
– Wesley Strik
Dec 11 '18 at 15:56
$begingroup$
Well, you need to clarify that point. Either way, though, you can do it with indicator variables. For $iin {1,cdots, 48}$ let $X_i$ be the indicator variable which is $1$ if a "good" block (however defined) starts at the $i^{th}$ slot. Easy to compute the expected value of $X_i$ and then use Linearity.
$endgroup$
– lulu
Dec 11 '18 at 15:57
$begingroup$
Well, you need to clarify that point. Either way, though, you can do it with indicator variables. For $iin {1,cdots, 48}$ let $X_i$ be the indicator variable which is $1$ if a "good" block (however defined) starts at the $i^{th}$ slot. Easy to compute the expected value of $X_i$ and then use Linearity.
$endgroup$
– lulu
Dec 11 '18 at 15:57
$begingroup$
This sounds way too advanced for a first lecture into introductory probability theory. We only know linearity and coin tosses.
$endgroup$
– Wesley Strik
Dec 11 '18 at 16:00
$begingroup$
This sounds way too advanced for a first lecture into introductory probability theory. We only know linearity and coin tosses.
$endgroup$
– Wesley Strik
Dec 11 '18 at 16:00
1
1
$begingroup$
I don't see an easier way of doing the problem. And indicator variables are pretty easy to use. I suggest looking that up.
$endgroup$
– lulu
Dec 11 '18 at 16:05
$begingroup$
I don't see an easier way of doing the problem. And indicator variables are pretty easy to use. I suggest looking that up.
$endgroup$
– lulu
Dec 11 '18 at 16:05
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
This is a straight forward exercise in the use of indicator variables.
Note that a string of the form $HHH$ can start anywhere from the first slot to the $48^{th}$. For $iin {1,cdots, 48}$ let $X_i$ be the indicator variable for the $i^{th}$ slot. Thus, $X_i=1$ if a good sequence begins on the $i^{th}$ slot and $X_i=0$ otherwise.
By Linearity $$E=Ebig [sum_{i=1}^{48}X_ibig ] =sum_{i=1}^{48}E[X_i]$$
Now, the $X_i$ don't all have the same expectation, $X_1$ and $X_{48}$ are different than all the others (which all equal each other).
To handle $X_1$ note that the only good sequence that starts in the first slot is $HHHT$. Thus the probability of starting with a good sequence is $frac 1{16}$, so $E[X_1]=frac 1{16}$. A similar computation shows that $E[X_{48}]=frac 1{16}$ as well.
For $1<i<48$ we get a good sequence starting in slot $i$ by $THHHT$, where the first $H$ is in the $i^{th}$ slot. Thus the probability that a good string starts in slot $i$ is $frac 1{32}$
Combining all this we see that $$boxed {E=2times frac 1{16}+46times frac 1{32}=frac {25}{16}=1.5625}$$
answered Dec 11 '18 at 16:16
lulululu
41.2k24979
$endgroup$
$begingroup$
Why is $TTT$ a good string? As I read the problem we need $THHHT$ if we are in the middle. We can delete one $T$ at the ends.
$endgroup$
– Ross Millikan
Dec 11 '18 at 16:23
$begingroup$
@RossMillikan My computation takes all that into account, or at least it intends to. My point was that a $3-$ sequence, either $HHH$ or $TTT$ can start anywhere from slot $1$ to slot $48$.
$endgroup$
– lulu
Dec 11 '18 at 17:30
$begingroup$
@RossMillikan Oh, sorry. for some reason I read the problem as looking for three of a kind in a row. Now I see that only Heads is accepted. I'll edit accordingly.
$endgroup$
– lulu
Dec 11 '18 at 17:32
$begingroup$
@RossMillikan reviewing the edit history, the problem originally did accept $HHH$ or $TTT$, the OP changed it at some point after I had posted my solution. Anyway, I have edited my solution to correspond to the question in its present form.
$endgroup$
– lulu
Dec 11 '18 at 17:35
add a comment |
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1 Answer
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$begingroup$
This is a straight forward exercise in the use of indicator variables.
Note that a string of the form $HHH$ can start anywhere from the first slot to the $48^{th}$. For $iin {1,cdots, 48}$ let $X_i$ be the indicator variable for the $i^{th}$ slot. Thus, $X_i=1$ if a good sequence begins on the $i^{th}$ slot and $X_i=0$ otherwise.
By Linearity $$E=Ebig [sum_{i=1}^{48}X_ibig ] =sum_{i=1}^{48}E[X_i]$$
Now, the $X_i$ don't all have the same expectation, $X_1$ and $X_{48}$ are different than all the others (which all equal each other).
To handle $X_1$ note that the only good sequence that starts in the first slot is $HHHT$. Thus the probability of starting with a good sequence is $frac 1{16}$, so $E[X_1]=frac 1{16}$. A similar computation shows that $E[X_{48}]=frac 1{16}$ as well.
For $1<i<48$ we get a good sequence starting in slot $i$ by $THHHT$, where the first $H$ is in the $i^{th}$ slot. Thus the probability that a good string starts in slot $i$ is $frac 1{32}$
Combining all this we see that $$boxed {E=2times frac 1{16}+46times frac 1{32}=frac {25}{16}=1.5625}$$
answered Dec 11 '18 at 16:16
lulululu
41.2k24979
$endgroup$
$begingroup$
Why is $TTT$ a good string? As I read the problem we need $THHHT$ if we are in the middle. We can delete one $T$ at the ends.
$endgroup$
– Ross Millikan
Dec 11 '18 at 16:23
$begingroup$
@RossMillikan My computation takes all that into account, or at least it intends to. My point was that a $3-$ sequence, either $HHH$ or $TTT$ can start anywhere from slot $1$ to slot $48$.
$endgroup$
– lulu
Dec 11 '18 at 17:30
$begingroup$
@RossMillikan Oh, sorry. for some reason I read the problem as looking for three of a kind in a row. Now I see that only Heads is accepted. I'll edit accordingly.
$endgroup$
– lulu
Dec 11 '18 at 17:32
$begingroup$
@RossMillikan reviewing the edit history, the problem originally did accept $HHH$ or $TTT$, the OP changed it at some point after I had posted my solution. Anyway, I have edited my solution to correspond to the question in its present form.
$endgroup$
– lulu
Dec 11 '18 at 17:35
add a comment |
$begingroup$
This is a straight forward exercise in the use of indicator variables.
Note that a string of the form $HHH$ can start anywhere from the first slot to the $48^{th}$. For $iin {1,cdots, 48}$ let $X_i$ be the indicator variable for the $i^{th}$ slot. Thus, $X_i=1$ if a good sequence begins on the $i^{th}$ slot and $X_i=0$ otherwise.
By Linearity $$E=Ebig [sum_{i=1}^{48}X_ibig ] =sum_{i=1}^{48}E[X_i]$$
Now, the $X_i$ don't all have the same expectation, $X_1$ and $X_{48}$ are different than all the others (which all equal each other).
To handle $X_1$ note that the only good sequence that starts in the first slot is $HHHT$. Thus the probability of starting with a good sequence is $frac 1{16}$, so $E[X_1]=frac 1{16}$. A similar computation shows that $E[X_{48}]=frac 1{16}$ as well.
For $1<i<48$ we get a good sequence starting in slot $i$ by $THHHT$, where the first $H$ is in the $i^{th}$ slot. Thus the probability that a good string starts in slot $i$ is $frac 1{32}$
Combining all this we see that $$boxed {E=2times frac 1{16}+46times frac 1{32}=frac {25}{16}=1.5625}$$
answered Dec 11 '18 at 16:16
lulululu
41.2k24979
$endgroup$
$begingroup$
Why is $TTT$ a good string? As I read the problem we need $THHHT$ if we are in the middle. We can delete one $T$ at the ends.
$endgroup$
– Ross Millikan
Dec 11 '18 at 16:23
$begingroup$
@RossMillikan My computation takes all that into account, or at least it intends to. My point was that a $3-$ sequence, either $HHH$ or $TTT$ can start anywhere from slot $1$ to slot $48$.
$endgroup$
– lulu
Dec 11 '18 at 17:30
$begingroup$
@RossMillikan Oh, sorry. for some reason I read the problem as looking for three of a kind in a row. Now I see that only Heads is accepted. I'll edit accordingly.
$endgroup$
– lulu
Dec 11 '18 at 17:32
$begingroup$
@RossMillikan reviewing the edit history, the problem originally did accept $HHH$ or $TTT$, the OP changed it at some point after I had posted my solution. Anyway, I have edited my solution to correspond to the question in its present form.
$endgroup$
– lulu
Dec 11 '18 at 17:35
add a comment |
$begingroup$
This is a straight forward exercise in the use of indicator variables.
Note that a string of the form $HHH$ can start anywhere from the first slot to the $48^{th}$. For $iin {1,cdots, 48}$ let $X_i$ be the indicator variable for the $i^{th}$ slot. Thus, $X_i=1$ if a good sequence begins on the $i^{th}$ slot and $X_i=0$ otherwise.
By Linearity $$E=Ebig [sum_{i=1}^{48}X_ibig ] =sum_{i=1}^{48}E[X_i]$$
Now, the $X_i$ don't all have the same expectation, $X_1$ and $X_{48}$ are different than all the others (which all equal each other).
To handle $X_1$ note that the only good sequence that starts in the first slot is $HHHT$. Thus the probability of starting with a good sequence is $frac 1{16}$, so $E[X_1]=frac 1{16}$. A similar computation shows that $E[X_{48}]=frac 1{16}$ as well.
For $1<i<48$ we get a good sequence starting in slot $i$ by $THHHT$, where the first $H$ is in the $i^{th}$ slot. Thus the probability that a good string starts in slot $i$ is $frac 1{32}$
Combining all this we see that $$boxed {E=2times frac 1{16}+46times frac 1{32}=frac {25}{16}=1.5625}$$
answered Dec 11 '18 at 16:16
lulululu
41.2k24979
$endgroup$
This is a straight forward exercise in the use of indicator variables.
Note that a string of the form $HHH$ can start anywhere from the first slot to the $48^{th}$. For $iin {1,cdots, 48}$ let $X_i$ be the indicator variable for the $i^{th}$ slot. Thus, $X_i=1$ if a good sequence begins on the $i^{th}$ slot and $X_i=0$ otherwise.
By Linearity $$E=Ebig [sum_{i=1}^{48}X_ibig ] =sum_{i=1}^{48}E[X_i]$$
Now, the $X_i$ don't all have the same expectation, $X_1$ and $X_{48}$ are different than all the others (which all equal each other).
To handle $X_1$ note that the only good sequence that starts in the first slot is $HHHT$. Thus the probability of starting with a good sequence is $frac 1{16}$, so $E[X_1]=frac 1{16}$. A similar computation shows that $E[X_{48}]=frac 1{16}$ as well.
For $1<i<48$ we get a good sequence starting in slot $i$ by $THHHT$, where the first $H$ is in the $i^{th}$ slot. Thus the probability that a good string starts in slot $i$ is $frac 1{32}$
Combining all this we see that $$boxed {E=2times frac 1{16}+46times frac 1{32}=frac {25}{16}=1.5625}$$
answered Dec 11 '18 at 16:16
lulululu
41.2k24979
edited Dec 11 '18 at 17:33
answered Dec 11 '18 at 16:16
lulululu
41.2k24979
answered Dec 11 '18 at 16:16
lulululu
41.2k24979
answered Dec 11 '18 at 16:16
lulululu
41.2k24979
41.2k24979
$begingroup$
Why is $TTT$ a good string? As I read the problem we need $THHHT$ if we are in the middle. We can delete one $T$ at the ends.
$endgroup$
– Ross Millikan
Dec 11 '18 at 16:23
$begingroup$
@RossMillikan My computation takes all that into account, or at least it intends to. My point was that a $3-$ sequence, either $HHH$ or $TTT$ can start anywhere from slot $1$ to slot $48$.
$endgroup$
– lulu
Dec 11 '18 at 17:30
$begingroup$
@RossMillikan Oh, sorry. for some reason I read the problem as looking for three of a kind in a row. Now I see that only Heads is accepted. I'll edit accordingly.
$endgroup$
– lulu
Dec 11 '18 at 17:32
$begingroup$
@RossMillikan reviewing the edit history, the problem originally did accept $HHH$ or $TTT$, the OP changed it at some point after I had posted my solution. Anyway, I have edited my solution to correspond to the question in its present form.
$endgroup$
– lulu
Dec 11 '18 at 17:35
add a comment |
$begingroup$
Why is $TTT$ a good string? As I read the problem we need $THHHT$ if we are in the middle. We can delete one $T$ at the ends.
$endgroup$
– Ross Millikan
Dec 11 '18 at 16:23
$begingroup$
@RossMillikan My computation takes all that into account, or at least it intends to. My point was that a $3-$ sequence, either $HHH$ or $TTT$ can start anywhere from slot $1$ to slot $48$.
$endgroup$
– lulu
Dec 11 '18 at 17:30
$begingroup$
@RossMillikan Oh, sorry. for some reason I read the problem as looking for three of a kind in a row. Now I see that only Heads is accepted. I'll edit accordingly.
$endgroup$
– lulu
Dec 11 '18 at 17:32
$begingroup$
@RossMillikan reviewing the edit history, the problem originally did accept $HHH$ or $TTT$, the OP changed it at some point after I had posted my solution. Anyway, I have edited my solution to correspond to the question in its present form.
$endgroup$
– lulu
Dec 11 '18 at 17:35
$begingroup$
Why is $TTT$ a good string? As I read the problem we need $THHHT$ if we are in the middle. We can delete one $T$ at the ends.
$endgroup$
– Ross Millikan
Dec 11 '18 at 16:23
$begingroup$
Why is $TTT$ a good string? As I read the problem we need $THHHT$ if we are in the middle. We can delete one $T$ at the ends.
$endgroup$
– Ross Millikan
Dec 11 '18 at 16:23
$begingroup$
@RossMillikan My computation takes all that into account, or at least it intends to. My point was that a $3-$ sequence, either $HHH$ or $TTT$ can start anywhere from slot $1$ to slot $48$.
$endgroup$
– lulu
Dec 11 '18 at 17:30
$begingroup$
@RossMillikan My computation takes all that into account, or at least it intends to. My point was that a $3-$ sequence, either $HHH$ or $TTT$ can start anywhere from slot $1$ to slot $48$.
$endgroup$
– lulu
Dec 11 '18 at 17:30
$begingroup$
@RossMillikan Oh, sorry. for some reason I read the problem as looking for three of a kind in a row. Now I see that only Heads is accepted. I'll edit accordingly.
$endgroup$
– lulu
Dec 11 '18 at 17:32
$begingroup$
@RossMillikan Oh, sorry. for some reason I read the problem as looking for three of a kind in a row. Now I see that only Heads is accepted. I'll edit accordingly.
$endgroup$
– lulu
Dec 11 '18 at 17:32
$begingroup$
@RossMillikan reviewing the edit history, the problem originally did accept $HHH$ or $TTT$, the OP changed it at some point after I had posted my solution. Anyway, I have edited my solution to correspond to the question in its present form.
$endgroup$
– lulu
Dec 11 '18 at 17:35
$begingroup$
@RossMillikan reviewing the edit history, the problem originally did accept $HHH$ or $TTT$, the OP changed it at some point after I had posted my solution. Anyway, I have edited my solution to correspond to the question in its present form.
$endgroup$
– lulu
Dec 11 '18 at 17:35
add a comment |
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Some clarification might be needed. Suppose your string was $HHHH$. Is that two sequences of three heads?
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– lulu
Dec 11 '18 at 15:54
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I think they count that as 1. So far we have only been counting $3$ and then we stop.
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– Wesley Strik
Dec 11 '18 at 15:56
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Well, you need to clarify that point. Either way, though, you can do it with indicator variables. For $iin {1,cdots, 48}$ let $X_i$ be the indicator variable which is $1$ if a "good" block (however defined) starts at the $i^{th}$ slot. Easy to compute the expected value of $X_i$ and then use Linearity.
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– lulu
Dec 11 '18 at 15:57
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This sounds way too advanced for a first lecture into introductory probability theory. We only know linearity and coin tosses.
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– Wesley Strik
Dec 11 '18 at 16:00
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I don't see an easier way of doing the problem. And indicator variables are pretty easy to use. I suggest looking that up.
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– lulu
Dec 11 '18 at 16:05