Mixing
Guessing the last answer in the test
Clash Royale CLAN TAG#URR8PPP
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Let's imagine someone passing an exam like shown and filling out the correct answers in tasks 1-11. Then, this someone stops and tries to guess an answer in the last one task (he do not know the right answer). Which answer should he choose?
Since the answers to the tasks were randomly distributed, does it mean that the probability of A12-3 is maximal? Or, since the answers are not dependent on each other, the probability is still 1/4 for any and no matter which line should be chosen?
probability
edited Jul 9 '17 at 21:20
Henry
92.6k468146
asked Jul 9 '17 at 21:16
Jarro
236
add a comment |Â
up vote
1
down vote
favorite
Let's imagine someone passing an exam like shown and filling out the correct answers in tasks 1-11. Then, this someone stops and tries to guess an answer in the last one task (he do not know the right answer). Which answer should he choose?
Since the answers to the tasks were randomly distributed, does it mean that the probability of A12-3 is maximal? Or, since the answers are not dependent on each other, the probability is still 1/4 for any and no matter which line should be chosen?
probability
edited Jul 9 '17 at 21:20
Henry
92.6k468146
asked Jul 9 '17 at 21:16
Jarro
236
1
Let's say I flip a fair coin 99 times and get heads every time. Should I guess tails next? On the other hand, you could of course argue that no well set test will have all answers $A$, for example, even if the options are randomly selected - the test setters will regenerate the options (to prevent candidates which guess 'all $A$' from succeeding) and therefore they isn't a 'fair' chance of all combinations appearing. So perhaps going for the option which is represented the least is the best idea. It all depends on how the test setters are randomly generating the options.
– Shuri2060
Jul 9 '17 at 21:24
Baye's Theorem works pretty well if we don't know the probability that the answer choices are in random order.
– Nairit Sarkar
Jul 9 '17 at 21:48
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let's imagine someone passing an exam like shown and filling out the correct answers in tasks 1-11. Then, this someone stops and tries to guess an answer in the last one task (he do not know the right answer). Which answer should he choose?
Since the answers to the tasks were randomly distributed, does it mean that the probability of A12-3 is maximal? Or, since the answers are not dependent on each other, the probability is still 1/4 for any and no matter which line should be chosen?
probability
edited Jul 9 '17 at 21:20
Henry
92.6k468146
asked Jul 9 '17 at 21:16
Jarro
236
Let's imagine someone passing an exam like shown and filling out the correct answers in tasks 1-11. Then, this someone stops and tries to guess an answer in the last one task (he do not know the right answer). Which answer should he choose?
Since the answers to the tasks were randomly distributed, does it mean that the probability of A12-3 is maximal? Or, since the answers are not dependent on each other, the probability is still 1/4 for any and no matter which line should be chosen?
probability
edited Jul 9 '17 at 21:20
Henry
92.6k468146
asked Jul 9 '17 at 21:16
Jarro
236
edited Jul 9 '17 at 21:20
Henry
92.6k468146
edited Jul 9 '17 at 21:20
Henry
92.6k468146
edited Jul 9 '17 at 21:20
Henry
92.6k468146
92.6k468146
asked Jul 9 '17 at 21:16
Jarro
236
asked Jul 9 '17 at 21:16
Jarro
236
asked Jul 9 '17 at 21:16
Jarro
236
236
1
Let's say I flip a fair coin 99 times and get heads every time. Should I guess tails next? On the other hand, you could of course argue that no well set test will have all answers $A$, for example, even if the options are randomly selected - the test setters will regenerate the options (to prevent candidates which guess 'all $A$' from succeeding) and therefore they isn't a 'fair' chance of all combinations appearing. So perhaps going for the option which is represented the least is the best idea. It all depends on how the test setters are randomly generating the options.
– Shuri2060
Jul 9 '17 at 21:24
Baye's Theorem works pretty well if we don't know the probability that the answer choices are in random order.
– Nairit Sarkar
Jul 9 '17 at 21:48
add a comment |Â
1
Let's say I flip a fair coin 99 times and get heads every time. Should I guess tails next? On the other hand, you could of course argue that no well set test will have all answers $A$, for example, even if the options are randomly selected - the test setters will regenerate the options (to prevent candidates which guess 'all $A$' from succeeding) and therefore they isn't a 'fair' chance of all combinations appearing. So perhaps going for the option which is represented the least is the best idea. It all depends on how the test setters are randomly generating the options.
– Shuri2060
Jul 9 '17 at 21:24
Baye's Theorem works pretty well if we don't know the probability that the answer choices are in random order.
– Nairit Sarkar
Jul 9 '17 at 21:48
1
1
Let's say I flip a fair coin 99 times and get heads every time. Should I guess tails next? On the other hand, you could of course argue that no well set test will have all answers $A$, for example, even if the options are randomly selected - the test setters will regenerate the options (to prevent candidates which guess 'all $A$' from succeeding) and therefore they isn't a 'fair' chance of all combinations appearing. So perhaps going for the option which is represented the least is the best idea. It all depends on how the test setters are randomly generating the options.
– Shuri2060
Jul 9 '17 at 21:24
Let's say I flip a fair coin 99 times and get heads every time. Should I guess tails next? On the other hand, you could of course argue that no well set test will have all answers $A$, for example, even if the options are randomly selected - the test setters will regenerate the options (to prevent candidates which guess 'all $A$' from succeeding) and therefore they isn't a 'fair' chance of all combinations appearing. So perhaps going for the option which is represented the least is the best idea. It all depends on how the test setters are randomly generating the options.
– Shuri2060
Jul 9 '17 at 21:24
Baye's Theorem works pretty well if we don't know the probability that the answer choices are in random order.
– Nairit Sarkar
Jul 9 '17 at 21:48
Baye's Theorem works pretty well if we don't know the probability that the answer choices are in random order.
– Nairit Sarkar
Jul 9 '17 at 21:48
add a comment |Â
2 Answers
2
active
oldest
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up vote
1
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accepted
"Probability" depends highly on the probability distribution. In this case it's unknown. We can make assumptions about this unknown distribution, but these are no more than guesses. Here are three possibilities:
Perhaps (3) is not a valid choice; 1 means "true", 2 means "false", and 4 means "other". In this case the probability of 3 is zero.
Perhaps the author of the exam noticed that there haven't been any (3) answers so far, and rearranged the answers to force an answer of 3. In this case the probability of 3 is one.
Perhaps the probability is truly equal for all four answers, on every question. The previous data bears no influence on the answer to the last question in that case. The probability of 3 is one fourth.
Generally on multiple-choice questions, a student can eliminate one or more of the options as definitely wrong, and choose randomly from the remainder. If all four answers appear equally plausible, then my advice is to study more instead of trying to look for patterns among the previous answers. :-)
answered Jul 9 '17 at 21:21
vadim123
73.5k895183
add a comment |Â
up vote
0
down vote
Oh, so that's a classical mistake I've almost made, Gambler's fallacy.
answered 4 hours ago
Jarro
236
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
"Probability" depends highly on the probability distribution. In this case it's unknown. We can make assumptions about this unknown distribution, but these are no more than guesses. Here are three possibilities:
Perhaps (3) is not a valid choice; 1 means "true", 2 means "false", and 4 means "other". In this case the probability of 3 is zero.
Perhaps the author of the exam noticed that there haven't been any (3) answers so far, and rearranged the answers to force an answer of 3. In this case the probability of 3 is one.
Perhaps the probability is truly equal for all four answers, on every question. The previous data bears no influence on the answer to the last question in that case. The probability of 3 is one fourth.
Generally on multiple-choice questions, a student can eliminate one or more of the options as definitely wrong, and choose randomly from the remainder. If all four answers appear equally plausible, then my advice is to study more instead of trying to look for patterns among the previous answers. :-)
answered Jul 9 '17 at 21:21
vadim123
73.5k895183
add a comment |Â
up vote
1
down vote
accepted
"Probability" depends highly on the probability distribution. In this case it's unknown. We can make assumptions about this unknown distribution, but these are no more than guesses. Here are three possibilities:
Perhaps (3) is not a valid choice; 1 means "true", 2 means "false", and 4 means "other". In this case the probability of 3 is zero.
Perhaps the author of the exam noticed that there haven't been any (3) answers so far, and rearranged the answers to force an answer of 3. In this case the probability of 3 is one.
Perhaps the probability is truly equal for all four answers, on every question. The previous data bears no influence on the answer to the last question in that case. The probability of 3 is one fourth.
Generally on multiple-choice questions, a student can eliminate one or more of the options as definitely wrong, and choose randomly from the remainder. If all four answers appear equally plausible, then my advice is to study more instead of trying to look for patterns among the previous answers. :-)
answered Jul 9 '17 at 21:21
vadim123
73.5k895183
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
"Probability" depends highly on the probability distribution. In this case it's unknown. We can make assumptions about this unknown distribution, but these are no more than guesses. Here are three possibilities:
Perhaps (3) is not a valid choice; 1 means "true", 2 means "false", and 4 means "other". In this case the probability of 3 is zero.
Perhaps the author of the exam noticed that there haven't been any (3) answers so far, and rearranged the answers to force an answer of 3. In this case the probability of 3 is one.
Perhaps the probability is truly equal for all four answers, on every question. The previous data bears no influence on the answer to the last question in that case. The probability of 3 is one fourth.
Generally on multiple-choice questions, a student can eliminate one or more of the options as definitely wrong, and choose randomly from the remainder. If all four answers appear equally plausible, then my advice is to study more instead of trying to look for patterns among the previous answers. :-)
answered Jul 9 '17 at 21:21
vadim123
73.5k895183
"Probability" depends highly on the probability distribution. In this case it's unknown. We can make assumptions about this unknown distribution, but these are no more than guesses. Here are three possibilities:
Perhaps (3) is not a valid choice; 1 means "true", 2 means "false", and 4 means "other". In this case the probability of 3 is zero.
Perhaps the author of the exam noticed that there haven't been any (3) answers so far, and rearranged the answers to force an answer of 3. In this case the probability of 3 is one.
Perhaps the probability is truly equal for all four answers, on every question. The previous data bears no influence on the answer to the last question in that case. The probability of 3 is one fourth.
Generally on multiple-choice questions, a student can eliminate one or more of the options as definitely wrong, and choose randomly from the remainder. If all four answers appear equally plausible, then my advice is to study more instead of trying to look for patterns among the previous answers. :-)
answered Jul 9 '17 at 21:21
vadim123
73.5k895183
answered Jul 9 '17 at 21:21
vadim123
73.5k895183
answered Jul 9 '17 at 21:21
vadim123
73.5k895183
answered Jul 9 '17 at 21:21
vadim123
73.5k895183
73.5k895183
add a comment |Â
add a comment |Â
up vote
0
down vote
Oh, so that's a classical mistake I've almost made, Gambler's fallacy.
answered 4 hours ago
Jarro
236
add a comment |Â
up vote
0
down vote
Oh, so that's a classical mistake I've almost made, Gambler's fallacy.
answered 4 hours ago
Jarro
236
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Oh, so that's a classical mistake I've almost made, Gambler's fallacy.
answered 4 hours ago
Jarro
236
Oh, so that's a classical mistake I've almost made, Gambler's fallacy.
answered 4 hours ago
Jarro
236
answered 4 hours ago
Jarro
236
answered 4 hours ago
Jarro
236
answered 4 hours ago
Jarro
236
236
add a comment |Â
add a comment |Â
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1
Let's say I flip a fair coin 99 times and get heads every time. Should I guess tails next? On the other hand, you could of course argue that no well set test will have all answers $A$, for example, even if the options are randomly selected - the test setters will regenerate the options (to prevent candidates which guess 'all $A$' from succeeding) and therefore they isn't a 'fair' chance of all combinations appearing. So perhaps going for the option which is represented the least is the best idea. It all depends on how the test setters are randomly generating the options.
– Shuri2060
Jul 9 '17 at 21:24
Baye's Theorem works pretty well if we don't know the probability that the answer choices are in random order.
– Nairit Sarkar
Jul 9 '17 at 21:48