number guessing game [duplicate]
$begingroup$
This question already has an answer here:
Game Theory: What are Best Strategies for High-Low game (game details are below)?
1 answer
Someone gets to choose a number between $1$ and $100$. I than have to try and guess by guessing a number and being told if the number is higher lower or equal. If I get the number on my first try I get $50$ on my second try $40$ and so on. Now if the other person chooses the number randomly the game is pretty clear to me. But will the other person choose the number randomly? The obvious guessing pattern is $50, 75$(or $25$) and so on. So it seems maybe the person wouldn't want to choose the numbers $50,75,25$ and maybe also not $12,13,87,88,62,63,37,38$. This logic could then be continued to deliver even more numbers that maybe shouldn't be chosen as often.
So my question is where does the equillibrium lie? Is a number chosen at random and if not what is the probability distribution and what is my guessing strategy?
probability game-theory
edited Dec 11 '18 at 2:03
Ross Millikan
297k23198371
asked Dec 11 '18 at 1:06
Jagol95Jagol95
1637
$endgroup$
marked as duplicate by Community♦ Dec 11 '18 at 9:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
|
show 2 more comments
$begingroup$
This question already has an answer here:
Game Theory: What are Best Strategies for High-Low game (game details are below)?
1 answer
Someone gets to choose a number between $1$ and $100$. I than have to try and guess by guessing a number and being told if the number is higher lower or equal. If I get the number on my first try I get $50$ on my second try $40$ and so on. Now if the other person chooses the number randomly the game is pretty clear to me. But will the other person choose the number randomly? The obvious guessing pattern is $50, 75$(or $25$) and so on. So it seems maybe the person wouldn't want to choose the numbers $50,75,25$ and maybe also not $12,13,87,88,62,63,37,38$. This logic could then be continued to deliver even more numbers that maybe shouldn't be chosen as often.
So my question is where does the equillibrium lie? Is a number chosen at random and if not what is the probability distribution and what is my guessing strategy?
probability game-theory
edited Dec 11 '18 at 2:03
Ross Millikan
297k23198371
asked Dec 11 '18 at 1:06
Jagol95Jagol95
1637
$endgroup$
marked as duplicate by Community♦ Dec 11 '18 at 9:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
This can be viewed a problem in game theory. I have no idea how to analyze it, but I added the game-theory tag in hope of attracting the attention of someone who does.
$endgroup$
– saulspatz
Dec 11 '18 at 1:23
$begingroup$
So you can only get paid if you guess within five rounds?
$endgroup$
– Ross Millikan
Dec 11 '18 at 2:01
$begingroup$
yes, and if it takes 7 rounds you lose 10
$endgroup$
– Jagol95
Dec 11 '18 at 2:06
$begingroup$
@Jagol95 I upvoted because I like the original problem and I like your analysis. I wonder if this problem should be attacked by a computer program. You could (for example) somehow identify 5 distinct strategies, and then identify the vulnerabilities of each strategy; in fact your analysis constitutes the beginning of such an approach. You could then assign probabilities to each of the 5 strategies (e.g. select each strategy with frequency 20%), and then evaluate the vulnerability of your combined strategy.
$endgroup$
– user2661923
Dec 11 '18 at 2:38
$begingroup$
interesting problem! Can someone (who has time :)) find the solution to the corresponding game for small $n$ starting from 3 and maybe we can see if there's a pattern.
$endgroup$
– zoidberg
Dec 11 '18 at 6:27
|
show 2 more comments
$begingroup$
This question already has an answer here:
Game Theory: What are Best Strategies for High-Low game (game details are below)?
1 answer
Someone gets to choose a number between $1$ and $100$. I than have to try and guess by guessing a number and being told if the number is higher lower or equal. If I get the number on my first try I get $50$ on my second try $40$ and so on. Now if the other person chooses the number randomly the game is pretty clear to me. But will the other person choose the number randomly? The obvious guessing pattern is $50, 75$(or $25$) and so on. So it seems maybe the person wouldn't want to choose the numbers $50,75,25$ and maybe also not $12,13,87,88,62,63,37,38$. This logic could then be continued to deliver even more numbers that maybe shouldn't be chosen as often.
So my question is where does the equillibrium lie? Is a number chosen at random and if not what is the probability distribution and what is my guessing strategy?
probability game-theory
edited Dec 11 '18 at 2:03
Ross Millikan
297k23198371
asked Dec 11 '18 at 1:06
Jagol95Jagol95
1637
$endgroup$
This question already has an answer here:
Game Theory: What are Best Strategies for High-Low game (game details are below)?
1 answer
Someone gets to choose a number between $1$ and $100$. I than have to try and guess by guessing a number and being told if the number is higher lower or equal. If I get the number on my first try I get $50$ on my second try $40$ and so on. Now if the other person chooses the number randomly the game is pretty clear to me. But will the other person choose the number randomly? The obvious guessing pattern is $50, 75$(or $25$) and so on. So it seems maybe the person wouldn't want to choose the numbers $50,75,25$ and maybe also not $12,13,87,88,62,63,37,38$. This logic could then be continued to deliver even more numbers that maybe shouldn't be chosen as often.
So my question is where does the equillibrium lie? Is a number chosen at random and if not what is the probability distribution and what is my guessing strategy?
This question already has an answer here:
Game Theory: What are Best Strategies for High-Low game (game details are below)?
1 answer
probability game-theory
probability game-theory
edited Dec 11 '18 at 2:03
Ross Millikan
297k23198371
asked Dec 11 '18 at 1:06
Jagol95Jagol95
1637
edited Dec 11 '18 at 2:03
Ross Millikan
297k23198371
asked Dec 11 '18 at 1:06
Jagol95Jagol95
1637
edited Dec 11 '18 at 2:03
Ross Millikan
297k23198371
edited Dec 11 '18 at 2:03
Ross Millikan
297k23198371
edited Dec 11 '18 at 2:03
Ross Millikan
297k23198371
297k23198371
asked Dec 11 '18 at 1:06
Jagol95Jagol95
1637
asked Dec 11 '18 at 1:06
Jagol95Jagol95
1637
asked Dec 11 '18 at 1:06
Jagol95Jagol95
1637
1637
marked as duplicate by Community♦ Dec 11 '18 at 9:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Community♦ Dec 11 '18 at 9:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
This can be viewed a problem in game theory. I have no idea how to analyze it, but I added the game-theory tag in hope of attracting the attention of someone who does.
$endgroup$
– saulspatz
Dec 11 '18 at 1:23
$begingroup$
So you can only get paid if you guess within five rounds?
$endgroup$
– Ross Millikan
Dec 11 '18 at 2:01
$begingroup$
yes, and if it takes 7 rounds you lose 10
$endgroup$
– Jagol95
Dec 11 '18 at 2:06
$begingroup$
@Jagol95 I upvoted because I like the original problem and I like your analysis. I wonder if this problem should be attacked by a computer program. You could (for example) somehow identify 5 distinct strategies, and then identify the vulnerabilities of each strategy; in fact your analysis constitutes the beginning of such an approach. You could then assign probabilities to each of the 5 strategies (e.g. select each strategy with frequency 20%), and then evaluate the vulnerability of your combined strategy.
$endgroup$
– user2661923
Dec 11 '18 at 2:38
$begingroup$
interesting problem! Can someone (who has time :)) find the solution to the corresponding game for small $n$ starting from 3 and maybe we can see if there's a pattern.
$endgroup$
– zoidberg
Dec 11 '18 at 6:27
|
show 2 more comments
$begingroup$
This can be viewed a problem in game theory. I have no idea how to analyze it, but I added the game-theory tag in hope of attracting the attention of someone who does.
$endgroup$
– saulspatz
Dec 11 '18 at 1:23
$begingroup$
So you can only get paid if you guess within five rounds?
$endgroup$
– Ross Millikan
Dec 11 '18 at 2:01
$begingroup$
yes, and if it takes 7 rounds you lose 10
$endgroup$
– Jagol95
Dec 11 '18 at 2:06
$begingroup$
@Jagol95 I upvoted because I like the original problem and I like your analysis. I wonder if this problem should be attacked by a computer program. You could (for example) somehow identify 5 distinct strategies, and then identify the vulnerabilities of each strategy; in fact your analysis constitutes the beginning of such an approach. You could then assign probabilities to each of the 5 strategies (e.g. select each strategy with frequency 20%), and then evaluate the vulnerability of your combined strategy.
$endgroup$
– user2661923
Dec 11 '18 at 2:38
$begingroup$
interesting problem! Can someone (who has time :)) find the solution to the corresponding game for small $n$ starting from 3 and maybe we can see if there's a pattern.
$endgroup$
– zoidberg
Dec 11 '18 at 6:27
$begingroup$
This can be viewed a problem in game theory. I have no idea how to analyze it, but I added the game-theory tag in hope of attracting the attention of someone who does.
$endgroup$
– saulspatz
Dec 11 '18 at 1:23
$begingroup$
This can be viewed a problem in game theory. I have no idea how to analyze it, but I added the game-theory tag in hope of attracting the attention of someone who does.
$endgroup$
– saulspatz
Dec 11 '18 at 1:23
$begingroup$
So you can only get paid if you guess within five rounds?
$endgroup$
– Ross Millikan
Dec 11 '18 at 2:01
$begingroup$
So you can only get paid if you guess within five rounds?
$endgroup$
– Ross Millikan
Dec 11 '18 at 2:01
$begingroup$
yes, and if it takes 7 rounds you lose 10
$endgroup$
– Jagol95
Dec 11 '18 at 2:06
$begingroup$
yes, and if it takes 7 rounds you lose 10
$endgroup$
– Jagol95
Dec 11 '18 at 2:06
$begingroup$
@Jagol95 I upvoted because I like the original problem and I like your analysis. I wonder if this problem should be attacked by a computer program. You could (for example) somehow identify 5 distinct strategies, and then identify the vulnerabilities of each strategy; in fact your analysis constitutes the beginning of such an approach. You could then assign probabilities to each of the 5 strategies (e.g. select each strategy with frequency 20%), and then evaluate the vulnerability of your combined strategy.
$endgroup$
– user2661923
Dec 11 '18 at 2:38
$begingroup$
@Jagol95 I upvoted because I like the original problem and I like your analysis. I wonder if this problem should be attacked by a computer program. You could (for example) somehow identify 5 distinct strategies, and then identify the vulnerabilities of each strategy; in fact your analysis constitutes the beginning of such an approach. You could then assign probabilities to each of the 5 strategies (e.g. select each strategy with frequency 20%), and then evaluate the vulnerability of your combined strategy.
$endgroup$
– user2661923
Dec 11 '18 at 2:38
$begingroup$
interesting problem! Can someone (who has time :)) find the solution to the corresponding game for small $n$ starting from 3 and maybe we can see if there's a pattern.
$endgroup$
– zoidberg
Dec 11 '18 at 6:27
$begingroup$
interesting problem! Can someone (who has time :)) find the solution to the corresponding game for small $n$ starting from 3 and maybe we can see if there's a pattern.
$endgroup$
– zoidberg
Dec 11 '18 at 6:27
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
We can compute the expected value if the number is chosen randomly. You have $0.01$ chance to win $50$, $0.02$ chance to win $40$, $0.04$ chance to win $30$, $0.08$ chance to win $20$, $0.16$ chance to win $10$, $0.32$ chance to win $0$, and $0.37$ to lose $10$, giving an expected value of $2$, so you should play.
I don't see an easy way to compute the payoff if your opponent chooses the number to hurt you. You can start with any number in the range $[36,64]$ and guarantee success in $7$ rounds. If you follow bisection, both $1$ and $100$ are guaranteed to take the maximum number of choices, so your opponent should choose one of those. Of course, if you recognize it, you should guess those first, despite the chance you have to go through $9$ guesses.
answered Dec 11 '18 at 6:30
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
Say your opponent is really trying to hurt you. They assume you bisect and choose 1 or 100. Hence you choose 1 or 100. But they think a level up, they know you're picking 1 or 100, and they can just pick one of the numbers 6 bisections in, like 2, 3, 98, 99. But then maybe you will pick one of 2, 3, 98 ,99. Both players can keep analyzing but if you try doing this "backwards-bisection", the risk seems higher than the reward. You get very little information per guess, and by symmetry you have to guess twice (if 1 didn't work, you have to guess 100, if 2 didn't work, etc.).
$endgroup$
– palmpo
Dec 11 '18 at 20:51
$begingroup$
And say they both keep analyzing the game one level higher, I'm curious, is there a point where both players have strategies that cannot be taken advantage of?
$endgroup$
– palmpo
Dec 11 '18 at 20:53
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We can compute the expected value if the number is chosen randomly. You have $0.01$ chance to win $50$, $0.02$ chance to win $40$, $0.04$ chance to win $30$, $0.08$ chance to win $20$, $0.16$ chance to win $10$, $0.32$ chance to win $0$, and $0.37$ to lose $10$, giving an expected value of $2$, so you should play.
I don't see an easy way to compute the payoff if your opponent chooses the number to hurt you. You can start with any number in the range $[36,64]$ and guarantee success in $7$ rounds. If you follow bisection, both $1$ and $100$ are guaranteed to take the maximum number of choices, so your opponent should choose one of those. Of course, if you recognize it, you should guess those first, despite the chance you have to go through $9$ guesses.
answered Dec 11 '18 at 6:30
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
Say your opponent is really trying to hurt you. They assume you bisect and choose 1 or 100. Hence you choose 1 or 100. But they think a level up, they know you're picking 1 or 100, and they can just pick one of the numbers 6 bisections in, like 2, 3, 98, 99. But then maybe you will pick one of 2, 3, 98 ,99. Both players can keep analyzing but if you try doing this "backwards-bisection", the risk seems higher than the reward. You get very little information per guess, and by symmetry you have to guess twice (if 1 didn't work, you have to guess 100, if 2 didn't work, etc.).
$endgroup$
– palmpo
Dec 11 '18 at 20:51
$begingroup$
And say they both keep analyzing the game one level higher, I'm curious, is there a point where both players have strategies that cannot be taken advantage of?
$endgroup$
– palmpo
Dec 11 '18 at 20:53
add a comment |
$begingroup$
We can compute the expected value if the number is chosen randomly. You have $0.01$ chance to win $50$, $0.02$ chance to win $40$, $0.04$ chance to win $30$, $0.08$ chance to win $20$, $0.16$ chance to win $10$, $0.32$ chance to win $0$, and $0.37$ to lose $10$, giving an expected value of $2$, so you should play.
I don't see an easy way to compute the payoff if your opponent chooses the number to hurt you. You can start with any number in the range $[36,64]$ and guarantee success in $7$ rounds. If you follow bisection, both $1$ and $100$ are guaranteed to take the maximum number of choices, so your opponent should choose one of those. Of course, if you recognize it, you should guess those first, despite the chance you have to go through $9$ guesses.
answered Dec 11 '18 at 6:30
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
Say your opponent is really trying to hurt you. They assume you bisect and choose 1 or 100. Hence you choose 1 or 100. But they think a level up, they know you're picking 1 or 100, and they can just pick one of the numbers 6 bisections in, like 2, 3, 98, 99. But then maybe you will pick one of 2, 3, 98 ,99. Both players can keep analyzing but if you try doing this "backwards-bisection", the risk seems higher than the reward. You get very little information per guess, and by symmetry you have to guess twice (if 1 didn't work, you have to guess 100, if 2 didn't work, etc.).
$endgroup$
– palmpo
Dec 11 '18 at 20:51
$begingroup$
And say they both keep analyzing the game one level higher, I'm curious, is there a point where both players have strategies that cannot be taken advantage of?
$endgroup$
– palmpo
Dec 11 '18 at 20:53
add a comment |
$begingroup$
We can compute the expected value if the number is chosen randomly. You have $0.01$ chance to win $50$, $0.02$ chance to win $40$, $0.04$ chance to win $30$, $0.08$ chance to win $20$, $0.16$ chance to win $10$, $0.32$ chance to win $0$, and $0.37$ to lose $10$, giving an expected value of $2$, so you should play.
I don't see an easy way to compute the payoff if your opponent chooses the number to hurt you. You can start with any number in the range $[36,64]$ and guarantee success in $7$ rounds. If you follow bisection, both $1$ and $100$ are guaranteed to take the maximum number of choices, so your opponent should choose one of those. Of course, if you recognize it, you should guess those first, despite the chance you have to go through $9$ guesses.
answered Dec 11 '18 at 6:30
Ross MillikanRoss Millikan
297k23198371
$endgroup$
We can compute the expected value if the number is chosen randomly. You have $0.01$ chance to win $50$, $0.02$ chance to win $40$, $0.04$ chance to win $30$, $0.08$ chance to win $20$, $0.16$ chance to win $10$, $0.32$ chance to win $0$, and $0.37$ to lose $10$, giving an expected value of $2$, so you should play.
I don't see an easy way to compute the payoff if your opponent chooses the number to hurt you. You can start with any number in the range $[36,64]$ and guarantee success in $7$ rounds. If you follow bisection, both $1$ and $100$ are guaranteed to take the maximum number of choices, so your opponent should choose one of those. Of course, if you recognize it, you should guess those first, despite the chance you have to go through $9$ guesses.
answered Dec 11 '18 at 6:30
Ross MillikanRoss Millikan
297k23198371
answered Dec 11 '18 at 6:30
Ross MillikanRoss Millikan
297k23198371
answered Dec 11 '18 at 6:30
Ross MillikanRoss Millikan
297k23198371
answered Dec 11 '18 at 6:30
Ross MillikanRoss Millikan
297k23198371
297k23198371
$begingroup$
Say your opponent is really trying to hurt you. They assume you bisect and choose 1 or 100. Hence you choose 1 or 100. But they think a level up, they know you're picking 1 or 100, and they can just pick one of the numbers 6 bisections in, like 2, 3, 98, 99. But then maybe you will pick one of 2, 3, 98 ,99. Both players can keep analyzing but if you try doing this "backwards-bisection", the risk seems higher than the reward. You get very little information per guess, and by symmetry you have to guess twice (if 1 didn't work, you have to guess 100, if 2 didn't work, etc.).
$endgroup$
– palmpo
Dec 11 '18 at 20:51
$begingroup$
And say they both keep analyzing the game one level higher, I'm curious, is there a point where both players have strategies that cannot be taken advantage of?
$endgroup$
– palmpo
Dec 11 '18 at 20:53
add a comment |
$begingroup$
Say your opponent is really trying to hurt you. They assume you bisect and choose 1 or 100. Hence you choose 1 or 100. But they think a level up, they know you're picking 1 or 100, and they can just pick one of the numbers 6 bisections in, like 2, 3, 98, 99. But then maybe you will pick one of 2, 3, 98 ,99. Both players can keep analyzing but if you try doing this "backwards-bisection", the risk seems higher than the reward. You get very little information per guess, and by symmetry you have to guess twice (if 1 didn't work, you have to guess 100, if 2 didn't work, etc.).
$endgroup$
– palmpo
Dec 11 '18 at 20:51
$begingroup$
And say they both keep analyzing the game one level higher, I'm curious, is there a point where both players have strategies that cannot be taken advantage of?
$endgroup$
– palmpo
Dec 11 '18 at 20:53
$begingroup$
Say your opponent is really trying to hurt you. They assume you bisect and choose 1 or 100. Hence you choose 1 or 100. But they think a level up, they know you're picking 1 or 100, and they can just pick one of the numbers 6 bisections in, like 2, 3, 98, 99. But then maybe you will pick one of 2, 3, 98 ,99. Both players can keep analyzing but if you try doing this "backwards-bisection", the risk seems higher than the reward. You get very little information per guess, and by symmetry you have to guess twice (if 1 didn't work, you have to guess 100, if 2 didn't work, etc.).
$endgroup$
– palmpo
Dec 11 '18 at 20:51
$begingroup$
Say your opponent is really trying to hurt you. They assume you bisect and choose 1 or 100. Hence you choose 1 or 100. But they think a level up, they know you're picking 1 or 100, and they can just pick one of the numbers 6 bisections in, like 2, 3, 98, 99. But then maybe you will pick one of 2, 3, 98 ,99. Both players can keep analyzing but if you try doing this "backwards-bisection", the risk seems higher than the reward. You get very little information per guess, and by symmetry you have to guess twice (if 1 didn't work, you have to guess 100, if 2 didn't work, etc.).
$endgroup$
– palmpo
Dec 11 '18 at 20:51
$begingroup$
And say they both keep analyzing the game one level higher, I'm curious, is there a point where both players have strategies that cannot be taken advantage of?
$endgroup$
– palmpo
Dec 11 '18 at 20:53
$begingroup$
And say they both keep analyzing the game one level higher, I'm curious, is there a point where both players have strategies that cannot be taken advantage of?
$endgroup$
– palmpo
Dec 11 '18 at 20:53
add a comment |
$begingroup$
This can be viewed a problem in game theory. I have no idea how to analyze it, but I added the game-theory tag in hope of attracting the attention of someone who does.
$endgroup$
– saulspatz
Dec 11 '18 at 1:23
$begingroup$
So you can only get paid if you guess within five rounds?
$endgroup$
– Ross Millikan
Dec 11 '18 at 2:01
$begingroup$
yes, and if it takes 7 rounds you lose 10
$endgroup$
– Jagol95
Dec 11 '18 at 2:06
$begingroup$
@Jagol95 I upvoted because I like the original problem and I like your analysis. I wonder if this problem should be attacked by a computer program. You could (for example) somehow identify 5 distinct strategies, and then identify the vulnerabilities of each strategy; in fact your analysis constitutes the beginning of such an approach. You could then assign probabilities to each of the 5 strategies (e.g. select each strategy with frequency 20%), and then evaluate the vulnerability of your combined strategy.
$endgroup$
– user2661923
Dec 11 '18 at 2:38
$begingroup$
interesting problem! Can someone (who has time :)) find the solution to the corresponding game for small $n$ starting from 3 and maybe we can see if there's a pattern.
$endgroup$
– zoidberg
Dec 11 '18 at 6:27