How often will two random integers between one and three (boundary inclusive) be the same?

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Assuming the answer is one out of three? And, also, would this be a good distillation of how to understand the "Monty Hall" problem?




probability monty-hall




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  • The connection with the Monty Hall problem seems slender. In that problem, everything depends on the superior knowledge of the host. You get information from the fact that Monty does not open a certain door. Hard to connect that up to a truly random draw.
    – lulu
    Aug 6 at 0:06














up vote
-2
down vote

favorite












Assuming the answer is one out of three? And, also, would this be a good distillation of how to understand the "Monty Hall" problem?




probability monty-hall




share|cite|improve this question





















  • The connection with the Monty Hall problem seems slender. In that problem, everything depends on the superior knowledge of the host. You get information from the fact that Monty does not open a certain door. Hard to connect that up to a truly random draw.
    – lulu
    Aug 6 at 0:06












up vote
-2
down vote

favorite









up vote
-2
down vote

favorite











Assuming the answer is one out of three? And, also, would this be a good distillation of how to understand the "Monty Hall" problem?




probability monty-hall




share|cite|improve this question













Assuming the answer is one out of three? And, also, would this be a good distillation of how to understand the "Monty Hall" problem?





probability monty-hall





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 5 at 23:56
























asked Aug 5 at 23:39









user92272

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11











  • The connection with the Monty Hall problem seems slender. In that problem, everything depends on the superior knowledge of the host. You get information from the fact that Monty does not open a certain door. Hard to connect that up to a truly random draw.
    – lulu
    Aug 6 at 0:06
















  • The connection with the Monty Hall problem seems slender. In that problem, everything depends on the superior knowledge of the host. You get information from the fact that Monty does not open a certain door. Hard to connect that up to a truly random draw.
    – lulu
    Aug 6 at 0:06















The connection with the Monty Hall problem seems slender. In that problem, everything depends on the superior knowledge of the host. You get information from the fact that Monty does not open a certain door. Hard to connect that up to a truly random draw.
– lulu
Aug 6 at 0:06




The connection with the Monty Hall problem seems slender. In that problem, everything depends on the superior knowledge of the host. You get information from the fact that Monty does not open a certain door. Hard to connect that up to a truly random draw.
– lulu
Aug 6 at 0:06










1 Answer
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If the numbers are integers then 1/3, but how does this relate to monty hall at all?






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  • Thanks Dean. So I've been making a concerted effort to simplify Monty Hall into the easiest way to understand (for myself at least). The lightbulb for me was in the fact that it only makes sense to NOT shift to the remaining door, if you actually happened to have guessed correctly initially, but that only happens 33% of the time, so you are always advantaged by switching. So in attempting to distill that down even further, success rate in shifting to the remaining door would be the same as the probability of two random numbers between one and three NOT being the same, which would be 2/3.
    – user92272
    Aug 5 at 23:46











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up vote
1
down vote













If the numbers are integers then 1/3, but how does this relate to monty hall at all?






share|cite|improve this answer





















  • Thanks Dean. So I've been making a concerted effort to simplify Monty Hall into the easiest way to understand (for myself at least). The lightbulb for me was in the fact that it only makes sense to NOT shift to the remaining door, if you actually happened to have guessed correctly initially, but that only happens 33% of the time, so you are always advantaged by switching. So in attempting to distill that down even further, success rate in shifting to the remaining door would be the same as the probability of two random numbers between one and three NOT being the same, which would be 2/3.
    – user92272
    Aug 5 at 23:46















up vote
1
down vote













If the numbers are integers then 1/3, but how does this relate to monty hall at all?






share|cite|improve this answer





















  • Thanks Dean. So I've been making a concerted effort to simplify Monty Hall into the easiest way to understand (for myself at least). The lightbulb for me was in the fact that it only makes sense to NOT shift to the remaining door, if you actually happened to have guessed correctly initially, but that only happens 33% of the time, so you are always advantaged by switching. So in attempting to distill that down even further, success rate in shifting to the remaining door would be the same as the probability of two random numbers between one and three NOT being the same, which would be 2/3.
    – user92272
    Aug 5 at 23:46













up vote
1
down vote










up vote
1
down vote









If the numbers are integers then 1/3, but how does this relate to monty hall at all?






share|cite|improve this answer













If the numbers are integers then 1/3, but how does this relate to monty hall at all?







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Aug 5 at 23:41









Dean Yang

1388




1388











  • Thanks Dean. So I've been making a concerted effort to simplify Monty Hall into the easiest way to understand (for myself at least). The lightbulb for me was in the fact that it only makes sense to NOT shift to the remaining door, if you actually happened to have guessed correctly initially, but that only happens 33% of the time, so you are always advantaged by switching. So in attempting to distill that down even further, success rate in shifting to the remaining door would be the same as the probability of two random numbers between one and three NOT being the same, which would be 2/3.
    – user92272
    Aug 5 at 23:46

















  • Thanks Dean. So I've been making a concerted effort to simplify Monty Hall into the easiest way to understand (for myself at least). The lightbulb for me was in the fact that it only makes sense to NOT shift to the remaining door, if you actually happened to have guessed correctly initially, but that only happens 33% of the time, so you are always advantaged by switching. So in attempting to distill that down even further, success rate in shifting to the remaining door would be the same as the probability of two random numbers between one and three NOT being the same, which would be 2/3.
    – user92272
    Aug 5 at 23:46
















Thanks Dean. So I've been making a concerted effort to simplify Monty Hall into the easiest way to understand (for myself at least). The lightbulb for me was in the fact that it only makes sense to NOT shift to the remaining door, if you actually happened to have guessed correctly initially, but that only happens 33% of the time, so you are always advantaged by switching. So in attempting to distill that down even further, success rate in shifting to the remaining door would be the same as the probability of two random numbers between one and three NOT being the same, which would be 2/3.
– user92272
Aug 5 at 23:46





Thanks Dean. So I've been making a concerted effort to simplify Monty Hall into the easiest way to understand (for myself at least). The lightbulb for me was in the fact that it only makes sense to NOT shift to the remaining door, if you actually happened to have guessed correctly initially, but that only happens 33% of the time, so you are always advantaged by switching. So in attempting to distill that down even further, success rate in shifting to the remaining door would be the same as the probability of two random numbers between one and three NOT being the same, which would be 2/3.
– user92272
Aug 5 at 23:46













 

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