What's wrong with this approach for finding expected number of throws before 2 identical throws?
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I am trying to find the expected number of throws before a die shows up the same twice in a row.
The definition of expected value is $$E(n) = sum_nnP(n)$$
I have seen other answers elsewhere on how to solve this problem, but my approach is thus:
Succeeding on the $n^{th}$ throw requires that you failed on the $(n-1)^{th}$ throw. So it makes sense that $$P(n) = frac{1}{6}(1-P(n-1))$$ since, after $n-1$ throws, you must roll what was rolled on the $(n-1)^{th}$ throw, and there is a $1/6$ chance of that. Having Wolfram do the heavy lifting with $P(2) = 1/6$ I get $$P(n) = frac{1}{7}(-frac{1}{6})^n((-6)^n+6)$$
So it seems to me that $$E(n) = sum_{n=2}^infty nP(n)$$ but, apparently, this is divergent. So what is wrong with this approach?
probability recurrence-relations dice
asked Dec 14 '18 at 5:26
john morrisonjohn morrison
605
$endgroup$
add a comment |
$begingroup$
I am trying to find the expected number of throws before a die shows up the same twice in a row.
The definition of expected value is $$E(n) = sum_nnP(n)$$
I have seen other answers elsewhere on how to solve this problem, but my approach is thus:
Succeeding on the $n^{th}$ throw requires that you failed on the $(n-1)^{th}$ throw. So it makes sense that $$P(n) = frac{1}{6}(1-P(n-1))$$ since, after $n-1$ throws, you must roll what was rolled on the $(n-1)^{th}$ throw, and there is a $1/6$ chance of that. Having Wolfram do the heavy lifting with $P(2) = 1/6$ I get $$P(n) = frac{1}{7}(-frac{1}{6})^n((-6)^n+6)$$
So it seems to me that $$E(n) = sum_{n=2}^infty nP(n)$$ but, apparently, this is divergent. So what is wrong with this approach?
probability recurrence-relations dice
asked Dec 14 '18 at 5:26
john morrisonjohn morrison
605
$endgroup$
2
$begingroup$
The answer will be $7$. The first throw cannot trigger the end, but from then on each throw has a $frac{1}{6}$ chance to match the throw previous regardless what it happened to be and so this becomes effectively the same as trying to calculate the expected number of throws until rolling a $1$ (after taking the off-by-one error into account).
$endgroup$
– JMoravitz
Dec 14 '18 at 5:32
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Okay, thank you for the comment. But I am not only interested in the answer; I am also interested in why my approach doesn't work. Can you shed light on that?
$endgroup$
– john morrison
Dec 14 '18 at 5:34
$begingroup$
The probability that your first success occurs on the $n$'th throw is $frac{1}{6}$ (the probability that the $n$'th throw matched the $n-1$'st) times the probability that the first $n-1$ attempts were all failures. Note that the probability that the first $n-1$ attempts are all failures is not the same thing as $1$ minus the probability that the $n-1$'st throw is the first success. Instead $1-P(n-1)$ also includes in it the chance that the first success occurred within the earlier attempts before the $n-1$'st throw.
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– JMoravitz
Dec 14 '18 at 5:44
1
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Rather, for $n>1$ you have $P(n) = frac{1}{6}times left(frac{5}{6}right)^{n-2}$
$endgroup$
– JMoravitz
Dec 14 '18 at 5:45
$begingroup$
Perfect sense. Perhaps you should make this an answer?
$endgroup$
– john morrison
Dec 14 '18 at 5:58
add a comment |
$begingroup$
I am trying to find the expected number of throws before a die shows up the same twice in a row.
The definition of expected value is $$E(n) = sum_nnP(n)$$
I have seen other answers elsewhere on how to solve this problem, but my approach is thus:
Succeeding on the $n^{th}$ throw requires that you failed on the $(n-1)^{th}$ throw. So it makes sense that $$P(n) = frac{1}{6}(1-P(n-1))$$ since, after $n-1$ throws, you must roll what was rolled on the $(n-1)^{th}$ throw, and there is a $1/6$ chance of that. Having Wolfram do the heavy lifting with $P(2) = 1/6$ I get $$P(n) = frac{1}{7}(-frac{1}{6})^n((-6)^n+6)$$
So it seems to me that $$E(n) = sum_{n=2}^infty nP(n)$$ but, apparently, this is divergent. So what is wrong with this approach?
probability recurrence-relations dice
asked Dec 14 '18 at 5:26
john morrisonjohn morrison
605
$endgroup$
I am trying to find the expected number of throws before a die shows up the same twice in a row.
The definition of expected value is $$E(n) = sum_nnP(n)$$
I have seen other answers elsewhere on how to solve this problem, but my approach is thus:
Succeeding on the $n^{th}$ throw requires that you failed on the $(n-1)^{th}$ throw. So it makes sense that $$P(n) = frac{1}{6}(1-P(n-1))$$ since, after $n-1$ throws, you must roll what was rolled on the $(n-1)^{th}$ throw, and there is a $1/6$ chance of that. Having Wolfram do the heavy lifting with $P(2) = 1/6$ I get $$P(n) = frac{1}{7}(-frac{1}{6})^n((-6)^n+6)$$
So it seems to me that $$E(n) = sum_{n=2}^infty nP(n)$$ but, apparently, this is divergent. So what is wrong with this approach?
probability recurrence-relations dice
probability recurrence-relations dice
asked Dec 14 '18 at 5:26
john morrisonjohn morrison
605
asked Dec 14 '18 at 5:26
john morrisonjohn morrison
605
asked Dec 14 '18 at 5:26
john morrisonjohn morrison
605
asked Dec 14 '18 at 5:26
john morrisonjohn morrison
605
asked Dec 14 '18 at 5:26
john morrisonjohn morrison
605
605
2
$begingroup$
The answer will be $7$. The first throw cannot trigger the end, but from then on each throw has a $frac{1}{6}$ chance to match the throw previous regardless what it happened to be and so this becomes effectively the same as trying to calculate the expected number of throws until rolling a $1$ (after taking the off-by-one error into account).
$endgroup$
– JMoravitz
Dec 14 '18 at 5:32
$begingroup$
Okay, thank you for the comment. But I am not only interested in the answer; I am also interested in why my approach doesn't work. Can you shed light on that?
$endgroup$
– john morrison
Dec 14 '18 at 5:34
$begingroup$
The probability that your first success occurs on the $n$'th throw is $frac{1}{6}$ (the probability that the $n$'th throw matched the $n-1$'st) times the probability that the first $n-1$ attempts were all failures. Note that the probability that the first $n-1$ attempts are all failures is not the same thing as $1$ minus the probability that the $n-1$'st throw is the first success. Instead $1-P(n-1)$ also includes in it the chance that the first success occurred within the earlier attempts before the $n-1$'st throw.
$endgroup$
– JMoravitz
Dec 14 '18 at 5:44
1
$begingroup$
Rather, for $n>1$ you have $P(n) = frac{1}{6}times left(frac{5}{6}right)^{n-2}$
$endgroup$
– JMoravitz
Dec 14 '18 at 5:45
$begingroup$
Perfect sense. Perhaps you should make this an answer?
$endgroup$
– john morrison
Dec 14 '18 at 5:58
add a comment |
2
$begingroup$
The answer will be $7$. The first throw cannot trigger the end, but from then on each throw has a $frac{1}{6}$ chance to match the throw previous regardless what it happened to be and so this becomes effectively the same as trying to calculate the expected number of throws until rolling a $1$ (after taking the off-by-one error into account).
$endgroup$
– JMoravitz
Dec 14 '18 at 5:32
$begingroup$
Okay, thank you for the comment. But I am not only interested in the answer; I am also interested in why my approach doesn't work. Can you shed light on that?
$endgroup$
– john morrison
Dec 14 '18 at 5:34
$begingroup$
The probability that your first success occurs on the $n$'th throw is $frac{1}{6}$ (the probability that the $n$'th throw matched the $n-1$'st) times the probability that the first $n-1$ attempts were all failures. Note that the probability that the first $n-1$ attempts are all failures is not the same thing as $1$ minus the probability that the $n-1$'st throw is the first success. Instead $1-P(n-1)$ also includes in it the chance that the first success occurred within the earlier attempts before the $n-1$'st throw.
$endgroup$
– JMoravitz
Dec 14 '18 at 5:44
1
$begingroup$
Rather, for $n>1$ you have $P(n) = frac{1}{6}times left(frac{5}{6}right)^{n-2}$
$endgroup$
– JMoravitz
Dec 14 '18 at 5:45
$begingroup$
Perfect sense. Perhaps you should make this an answer?
$endgroup$
– john morrison
Dec 14 '18 at 5:58
2
2
$begingroup$
The answer will be $7$. The first throw cannot trigger the end, but from then on each throw has a $frac{1}{6}$ chance to match the throw previous regardless what it happened to be and so this becomes effectively the same as trying to calculate the expected number of throws until rolling a $1$ (after taking the off-by-one error into account).
$endgroup$
– JMoravitz
Dec 14 '18 at 5:32
$begingroup$
The answer will be $7$. The first throw cannot trigger the end, but from then on each throw has a $frac{1}{6}$ chance to match the throw previous regardless what it happened to be and so this becomes effectively the same as trying to calculate the expected number of throws until rolling a $1$ (after taking the off-by-one error into account).
$endgroup$
– JMoravitz
Dec 14 '18 at 5:32
$begingroup$
Okay, thank you for the comment. But I am not only interested in the answer; I am also interested in why my approach doesn't work. Can you shed light on that?
$endgroup$
– john morrison
Dec 14 '18 at 5:34
$begingroup$
Okay, thank you for the comment. But I am not only interested in the answer; I am also interested in why my approach doesn't work. Can you shed light on that?
$endgroup$
– john morrison
Dec 14 '18 at 5:34
$begingroup$
The probability that your first success occurs on the $n$'th throw is $frac{1}{6}$ (the probability that the $n$'th throw matched the $n-1$'st) times the probability that the first $n-1$ attempts were all failures. Note that the probability that the first $n-1$ attempts are all failures is not the same thing as $1$ minus the probability that the $n-1$'st throw is the first success. Instead $1-P(n-1)$ also includes in it the chance that the first success occurred within the earlier attempts before the $n-1$'st throw.
$endgroup$
– JMoravitz
Dec 14 '18 at 5:44
$begingroup$
The probability that your first success occurs on the $n$'th throw is $frac{1}{6}$ (the probability that the $n$'th throw matched the $n-1$'st) times the probability that the first $n-1$ attempts were all failures. Note that the probability that the first $n-1$ attempts are all failures is not the same thing as $1$ minus the probability that the $n-1$'st throw is the first success. Instead $1-P(n-1)$ also includes in it the chance that the first success occurred within the earlier attempts before the $n-1$'st throw.
$endgroup$
– JMoravitz
Dec 14 '18 at 5:44
1
1
$begingroup$
Rather, for $n>1$ you have $P(n) = frac{1}{6}times left(frac{5}{6}right)^{n-2}$
$endgroup$
– JMoravitz
Dec 14 '18 at 5:45
$begingroup$
Rather, for $n>1$ you have $P(n) = frac{1}{6}times left(frac{5}{6}right)^{n-2}$
$endgroup$
– JMoravitz
Dec 14 '18 at 5:45
$begingroup$
Perfect sense. Perhaps you should make this an answer?
$endgroup$
– john morrison
Dec 14 '18 at 5:58
$begingroup$
Perfect sense. Perhaps you should make this an answer?
$endgroup$
– john morrison
Dec 14 '18 at 5:58
add a comment |
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$begingroup$
The answer will be $7$. The first throw cannot trigger the end, but from then on each throw has a $frac{1}{6}$ chance to match the throw previous regardless what it happened to be and so this becomes effectively the same as trying to calculate the expected number of throws until rolling a $1$ (after taking the off-by-one error into account).
$endgroup$
– JMoravitz
Dec 14 '18 at 5:32
$begingroup$
Okay, thank you for the comment. But I am not only interested in the answer; I am also interested in why my approach doesn't work. Can you shed light on that?
$endgroup$
– john morrison
Dec 14 '18 at 5:34
$begingroup$
The probability that your first success occurs on the $n$'th throw is $frac{1}{6}$ (the probability that the $n$'th throw matched the $n-1$'st) times the probability that the first $n-1$ attempts were all failures. Note that the probability that the first $n-1$ attempts are all failures is not the same thing as $1$ minus the probability that the $n-1$'st throw is the first success. Instead $1-P(n-1)$ also includes in it the chance that the first success occurred within the earlier attempts before the $n-1$'st throw.
$endgroup$
– JMoravitz
Dec 14 '18 at 5:44
1
$begingroup$
Rather, for $n>1$ you have $P(n) = frac{1}{6}times left(frac{5}{6}right)^{n-2}$
$endgroup$
– JMoravitz
Dec 14 '18 at 5:45
$begingroup$
Perfect sense. Perhaps you should make this an answer?
$endgroup$
– john morrison
Dec 14 '18 at 5:58