How often will two random integers between one and three (boundary inclusive) be the same?
Clash Royale CLAN TAG#URR8PPP
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
add a comment |Â
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
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.
â lulu
Aug 6 at 0:06
add a comment |Â
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
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
edited Aug 5 at 23:56
asked Aug 5 at 23:39
user92272
11
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.
â lulu
Aug 6 at 0:06
add a comment |Â
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.
â 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.
â 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.
â lulu
Aug 6 at 0:06
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
If the numbers are integers then 1/3, but how does this relate to monty hall at all?
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.
â user92272
Aug 5 at 23:46
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If the numbers are integers then 1/3, but how does this relate to monty hall at all?
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.
â user92272
Aug 5 at 23:46
add a comment |Â
up vote
1
down vote
If the numbers are integers then 1/3, but how does this relate to monty hall at all?
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.
â user92272
Aug 5 at 23:46
add a comment |Â
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?
If the numbers are integers then 1/3, but how does this relate to monty hall at all?
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.
â user92272
Aug 5 at 23:46
add a comment |Â
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.
â 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.
â 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.
â user92272
Aug 5 at 23:46
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873436%2fhow-often-will-two-random-integers-between-one-and-three-boundary-inclusive-be%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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.
â lulu
Aug 6 at 0:06