Double or Halve Riddle
$begingroup$
You start off with a $100 pool. Every second, your pool doubles in value with probability $p$ or gets cut in half with probability $1-p.$ If you drop below 98.76543210¢ at any point, you lose and the host of the game show laughs at you.
Furthermore, you stubbornly refuse to stop playing. What's the probability that you eventually lose? Note: Assume that you and the host are both immortal, and that thermodynamic restrictions do not prevent gameplay in the longue duree.
sequence
asked 6 hours ago
Display nameDisplay name
1,010217
$endgroup$
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$begingroup$
You start off with a $100 pool. Every second, your pool doubles in value with probability $p$ or gets cut in half with probability $1-p.$ If you drop below 98.76543210¢ at any point, you lose and the host of the game show laughs at you.
Furthermore, you stubbornly refuse to stop playing. What's the probability that you eventually lose? Note: Assume that you and the host are both immortal, and that thermodynamic restrictions do not prevent gameplay in the longue duree.
sequence
asked 6 hours ago
Display nameDisplay name
1,010217
$endgroup$
add a comment |
$begingroup$
You start off with a $100 pool. Every second, your pool doubles in value with probability $p$ or gets cut in half with probability $1-p.$ If you drop below 98.76543210¢ at any point, you lose and the host of the game show laughs at you.
Furthermore, you stubbornly refuse to stop playing. What's the probability that you eventually lose? Note: Assume that you and the host are both immortal, and that thermodynamic restrictions do not prevent gameplay in the longue duree.
sequence
asked 6 hours ago
Display nameDisplay name
1,010217
$endgroup$
You start off with a $100 pool. Every second, your pool doubles in value with probability $p$ or gets cut in half with probability $1-p.$ If you drop below 98.76543210¢ at any point, you lose and the host of the game show laughs at you.
Furthermore, you stubbornly refuse to stop playing. What's the probability that you eventually lose? Note: Assume that you and the host are both immortal, and that thermodynamic restrictions do not prevent gameplay in the longue duree.
sequence
sequence
asked 6 hours ago
Display nameDisplay name
1,010217
asked 6 hours ago
Display nameDisplay name
1,010217
edited 5 hours ago
Display name
asked 6 hours ago
Display nameDisplay name
1,010217
asked 6 hours ago
Display nameDisplay name
1,010217
asked 6 hours ago
Display nameDisplay name
1,010217
1,010217
add a comment |
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1 Answer
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$begingroup$
Meta: This is in effect a standard question in the theory of random walks. Does that make it off-topic as a textbook problem? I don't think so. The relevant meta-question is this one whose excellent and highest-voted answer gives this closely related question as an example of something that's a "math puzzle" rather than a "math problem".
OK then. Assuming that there's no weird business with rounding of fractional cents, this question is equivalent to the following: You start with a number $x=7$. You repeatedly add 1 (with probability $p$) or subtract 1 (with probability $1-p$). What's the probability that you eventually reach $x=0$? (Why 7? Because 6 halvings of $100 take you to $1.5625 (above the threshold in the question) and 7 halvings take you to $0.78125 (below the threshold).)
And the answer, which you can find in any suitable textbook, is
The probability is 1 if $pleqfrac12$; otherwise it's $bigl(frac{1-p}{p}bigr)^7$.
If you don't want to go out and find a suitable textbook, you can also deduce it from the accepted answer to that question I linked above: see that formula for $c_1,c_2$ and the way in which the answer to the original question is derived from it, and then put $a=-7$ instead of $a=-1$.
Meta again: Does that mean that this should have been closed as a duplicate? I think it's just different enough not to be: you can figure out the answer to this one from the answer to that one, but you can't just see it there.
answered 4 hours ago
Gareth McCaughan♦Gareth McCaughan
61.7k3153238
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1 Answer
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$begingroup$
Meta: This is in effect a standard question in the theory of random walks. Does that make it off-topic as a textbook problem? I don't think so. The relevant meta-question is this one whose excellent and highest-voted answer gives this closely related question as an example of something that's a "math puzzle" rather than a "math problem".
OK then. Assuming that there's no weird business with rounding of fractional cents, this question is equivalent to the following: You start with a number $x=7$. You repeatedly add 1 (with probability $p$) or subtract 1 (with probability $1-p$). What's the probability that you eventually reach $x=0$? (Why 7? Because 6 halvings of $100 take you to $1.5625 (above the threshold in the question) and 7 halvings take you to $0.78125 (below the threshold).)
And the answer, which you can find in any suitable textbook, is
The probability is 1 if $pleqfrac12$; otherwise it's $bigl(frac{1-p}{p}bigr)^7$.
If you don't want to go out and find a suitable textbook, you can also deduce it from the accepted answer to that question I linked above: see that formula for $c_1,c_2$ and the way in which the answer to the original question is derived from it, and then put $a=-7$ instead of $a=-1$.
Meta again: Does that mean that this should have been closed as a duplicate? I think it's just different enough not to be: you can figure out the answer to this one from the answer to that one, but you can't just see it there.
answered 4 hours ago
Gareth McCaughan♦Gareth McCaughan
61.7k3153238
$endgroup$
add a comment |
$begingroup$
Meta: This is in effect a standard question in the theory of random walks. Does that make it off-topic as a textbook problem? I don't think so. The relevant meta-question is this one whose excellent and highest-voted answer gives this closely related question as an example of something that's a "math puzzle" rather than a "math problem".
OK then. Assuming that there's no weird business with rounding of fractional cents, this question is equivalent to the following: You start with a number $x=7$. You repeatedly add 1 (with probability $p$) or subtract 1 (with probability $1-p$). What's the probability that you eventually reach $x=0$? (Why 7? Because 6 halvings of $100 take you to $1.5625 (above the threshold in the question) and 7 halvings take you to $0.78125 (below the threshold).)
And the answer, which you can find in any suitable textbook, is
The probability is 1 if $pleqfrac12$; otherwise it's $bigl(frac{1-p}{p}bigr)^7$.
If you don't want to go out and find a suitable textbook, you can also deduce it from the accepted answer to that question I linked above: see that formula for $c_1,c_2$ and the way in which the answer to the original question is derived from it, and then put $a=-7$ instead of $a=-1$.
Meta again: Does that mean that this should have been closed as a duplicate? I think it's just different enough not to be: you can figure out the answer to this one from the answer to that one, but you can't just see it there.
answered 4 hours ago
Gareth McCaughan♦Gareth McCaughan
61.7k3153238
$endgroup$
add a comment |
$begingroup$
Meta: This is in effect a standard question in the theory of random walks. Does that make it off-topic as a textbook problem? I don't think so. The relevant meta-question is this one whose excellent and highest-voted answer gives this closely related question as an example of something that's a "math puzzle" rather than a "math problem".
OK then. Assuming that there's no weird business with rounding of fractional cents, this question is equivalent to the following: You start with a number $x=7$. You repeatedly add 1 (with probability $p$) or subtract 1 (with probability $1-p$). What's the probability that you eventually reach $x=0$? (Why 7? Because 6 halvings of $100 take you to $1.5625 (above the threshold in the question) and 7 halvings take you to $0.78125 (below the threshold).)
And the answer, which you can find in any suitable textbook, is
The probability is 1 if $pleqfrac12$; otherwise it's $bigl(frac{1-p}{p}bigr)^7$.
If you don't want to go out and find a suitable textbook, you can also deduce it from the accepted answer to that question I linked above: see that formula for $c_1,c_2$ and the way in which the answer to the original question is derived from it, and then put $a=-7$ instead of $a=-1$.
Meta again: Does that mean that this should have been closed as a duplicate? I think it's just different enough not to be: you can figure out the answer to this one from the answer to that one, but you can't just see it there.
answered 4 hours ago
Gareth McCaughan♦Gareth McCaughan
61.7k3153238
$endgroup$
Meta: This is in effect a standard question in the theory of random walks. Does that make it off-topic as a textbook problem? I don't think so. The relevant meta-question is this one whose excellent and highest-voted answer gives this closely related question as an example of something that's a "math puzzle" rather than a "math problem".
OK then. Assuming that there's no weird business with rounding of fractional cents, this question is equivalent to the following: You start with a number $x=7$. You repeatedly add 1 (with probability $p$) or subtract 1 (with probability $1-p$). What's the probability that you eventually reach $x=0$? (Why 7? Because 6 halvings of $100 take you to $1.5625 (above the threshold in the question) and 7 halvings take you to $0.78125 (below the threshold).)
And the answer, which you can find in any suitable textbook, is
The probability is 1 if $pleqfrac12$; otherwise it's $bigl(frac{1-p}{p}bigr)^7$.
If you don't want to go out and find a suitable textbook, you can also deduce it from the accepted answer to that question I linked above: see that formula for $c_1,c_2$ and the way in which the answer to the original question is derived from it, and then put $a=-7$ instead of $a=-1$.
Meta again: Does that mean that this should have been closed as a duplicate? I think it's just different enough not to be: you can figure out the answer to this one from the answer to that one, but you can't just see it there.
answered 4 hours ago
Gareth McCaughan♦Gareth McCaughan
61.7k3153238
answered 4 hours ago
Gareth McCaughan♦Gareth McCaughan
61.7k3153238
answered 4 hours ago
Gareth McCaughan♦Gareth McCaughan
61.7k3153238
answered 4 hours ago
Gareth McCaughan♦Gareth McCaughan
61.7k3153238
61.7k3153238
add a comment |
add a comment |
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