Mixing
Probabilty of a specific 4 digit number
Clash Royale CLAN TAG#URR8PPP
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0;
up vote
1
down vote
favorite
I'm having an argument with a colleague about the probability of a 4 digit number.
The assumption is that the number is randomly and independently created.
Assume the probability of generating the number 4109.
Now, my colleague reckons that it's a simple P(4109) = 1/10 * 1/10 * 1/10 * 1/10. (a bit like simply tossing a coin).
But I'm not interested in the probability of generating the numbers 0149, 9014, 4910 etc etc.
Obviously the probability of that specific sequence, 4109 is a probability found by calculating the permutation, not simple combination....
Please confirm my believe or point me to some proper material to support this?
Kind regards!
probability permutation-test
asked 1 hour ago
amateurjustin
334
add a comment |Â
up vote
1
down vote
favorite
I'm having an argument with a colleague about the probability of a 4 digit number.
The assumption is that the number is randomly and independently created.
Assume the probability of generating the number 4109.
Now, my colleague reckons that it's a simple P(4109) = 1/10 * 1/10 * 1/10 * 1/10. (a bit like simply tossing a coin).
But I'm not interested in the probability of generating the numbers 0149, 9014, 4910 etc etc.
Obviously the probability of that specific sequence, 4109 is a probability found by calculating the permutation, not simple combination....
Please confirm my believe or point me to some proper material to support this?
Kind regards!
probability permutation-test
asked 1 hour ago
amateurjustin
334
The probabiliy of generating the number 4109 if we are generating all possible combinations of 4 digit numbers is $1/10^4$, since there is exactly 1 such number out of all the combinations. Was that the qeustion?
– user2974951
1 hour ago
@user2974951: It looks like you're using the term "combinations" where you actually mean "permutations". This is a little confusing, especially in the context of this particular question.
– Ruben van Bergen
1 hour ago
So technically, 1/10^4 is the permutation, and not a combination?
– amateurjustin
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm having an argument with a colleague about the probability of a 4 digit number.
The assumption is that the number is randomly and independently created.
Assume the probability of generating the number 4109.
Now, my colleague reckons that it's a simple P(4109) = 1/10 * 1/10 * 1/10 * 1/10. (a bit like simply tossing a coin).
But I'm not interested in the probability of generating the numbers 0149, 9014, 4910 etc etc.
Obviously the probability of that specific sequence, 4109 is a probability found by calculating the permutation, not simple combination....
Please confirm my believe or point me to some proper material to support this?
Kind regards!
probability permutation-test
asked 1 hour ago
amateurjustin
334
I'm having an argument with a colleague about the probability of a 4 digit number.
The assumption is that the number is randomly and independently created.
Assume the probability of generating the number 4109.
Now, my colleague reckons that it's a simple P(4109) = 1/10 * 1/10 * 1/10 * 1/10. (a bit like simply tossing a coin).
But I'm not interested in the probability of generating the numbers 0149, 9014, 4910 etc etc.
Obviously the probability of that specific sequence, 4109 is a probability found by calculating the permutation, not simple combination....
Please confirm my believe or point me to some proper material to support this?
Kind regards!
probability permutation-test
probability permutation-test
asked 1 hour ago
amateurjustin
334
asked 1 hour ago
amateurjustin
334
asked 1 hour ago
amateurjustin
334
asked 1 hour ago
amateurjustin
334
asked 1 hour ago
amateurjustin
334
334
The probabiliy of generating the number 4109 if we are generating all possible combinations of 4 digit numbers is $1/10^4$, since there is exactly 1 such number out of all the combinations. Was that the qeustion?
– user2974951
1 hour ago
@user2974951: It looks like you're using the term "combinations" where you actually mean "permutations". This is a little confusing, especially in the context of this particular question.
– Ruben van Bergen
1 hour ago
So technically, 1/10^4 is the permutation, and not a combination?
– amateurjustin
1 hour ago
add a comment |Â
The probabiliy of generating the number 4109 if we are generating all possible combinations of 4 digit numbers is $1/10^4$, since there is exactly 1 such number out of all the combinations. Was that the qeustion?
– user2974951
1 hour ago
@user2974951: It looks like you're using the term "combinations" where you actually mean "permutations". This is a little confusing, especially in the context of this particular question.
– Ruben van Bergen
1 hour ago
So technically, 1/10^4 is the permutation, and not a combination?
– amateurjustin
1 hour ago
The probabiliy of generating the number 4109 if we are generating all possible combinations of 4 digit numbers is $1/10^4$, since there is exactly 1 such number out of all the combinations. Was that the qeustion?
– user2974951
1 hour ago
The probabiliy of generating the number 4109 if we are generating all possible combinations of 4 digit numbers is $1/10^4$, since there is exactly 1 such number out of all the combinations. Was that the qeustion?
– user2974951
1 hour ago
@user2974951: It looks like you're using the term "combinations" where you actually mean "permutations". This is a little confusing, especially in the context of this particular question.
– Ruben van Bergen
1 hour ago
@user2974951: It looks like you're using the term "combinations" where you actually mean "permutations". This is a little confusing, especially in the context of this particular question.
– Ruben van Bergen
1 hour ago
So technically, 1/10^4 is the permutation, and not a combination?
– amateurjustin
1 hour ago
So technically, 1/10^4 is the permutation, and not a combination?
– amateurjustin
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
Your colleague is right. If your digits are randomly and independently selected (assuming equal likelihood for each digit), the chance that your number is 4109 is nothing more complicated than:
the chance that the first digit is 4
the chance that the second digit is 1
the chance that the third digit is 0
the chance that the fourth digit is 9
all multiplied together. That's $(1/10)^4$, as your colleague says. Where would permutations come into it?
You said in the question that you're "not interested in the probability of generating the numbers 0149, 9014, 4910 etc etc", i.e. the set of all 4-digit numbers where the digits are 0,1,4,9 in any order, but that isn't what $(1/10) ^ 4$ tells you - to calculate that you'd need:
the probability that the first digit is 0,1,4, or 9 = 4/10
the probability that the second digit is one of the three not yet selected = 3/10
the probability that the third digit is one of the two not yet selected = 2/10
the probability that the final digit is the remaining one from the set of 0,1,4,9 = 1/10
i.e. $prod_i=1^4 fraci10$
You could then worry about permutations to pick 4109 specifically, but this is needless complication in comparison the way your colleague said.
answered 1 hour ago
Chris H
99928
add a comment |Â
up vote
3
down vote
Your colleague is correct. Think of it this way: there are 10,000 ($10^4$) 4-digit numbers: 0000-9999. Your target number is one of those 10,000. Thus, if you picked a 4-digit number randomly, you'd have a one in 10,000 chance of picking that number (or any other specific 4-digit number).
You're right that what you want to calculate here is a permutation, not a "simple" (as you called it) combination. But permutations are actually simpler to calculate than combinations. In a permutation, you're essentially just counting all possible outcomes. In a combination, you additionally have to calculate how many of those outcomes are the same combination of elements in a different order, and then adjust your calculation for that.
answered 1 hour ago
Ruben van Bergen
3,4291817
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Your colleague is right. If your digits are randomly and independently selected (assuming equal likelihood for each digit), the chance that your number is 4109 is nothing more complicated than:
the chance that the first digit is 4
the chance that the second digit is 1
the chance that the third digit is 0
the chance that the fourth digit is 9
all multiplied together. That's $(1/10)^4$, as your colleague says. Where would permutations come into it?
You said in the question that you're "not interested in the probability of generating the numbers 0149, 9014, 4910 etc etc", i.e. the set of all 4-digit numbers where the digits are 0,1,4,9 in any order, but that isn't what $(1/10) ^ 4$ tells you - to calculate that you'd need:
the probability that the first digit is 0,1,4, or 9 = 4/10
the probability that the second digit is one of the three not yet selected = 3/10
the probability that the third digit is one of the two not yet selected = 2/10
the probability that the final digit is the remaining one from the set of 0,1,4,9 = 1/10
i.e. $prod_i=1^4 fraci10$
You could then worry about permutations to pick 4109 specifically, but this is needless complication in comparison the way your colleague said.
answered 1 hour ago
Chris H
99928
add a comment |Â
up vote
4
down vote
Your colleague is right. If your digits are randomly and independently selected (assuming equal likelihood for each digit), the chance that your number is 4109 is nothing more complicated than:
the chance that the first digit is 4
the chance that the second digit is 1
the chance that the third digit is 0
the chance that the fourth digit is 9
all multiplied together. That's $(1/10)^4$, as your colleague says. Where would permutations come into it?
You said in the question that you're "not interested in the probability of generating the numbers 0149, 9014, 4910 etc etc", i.e. the set of all 4-digit numbers where the digits are 0,1,4,9 in any order, but that isn't what $(1/10) ^ 4$ tells you - to calculate that you'd need:
the probability that the first digit is 0,1,4, or 9 = 4/10
the probability that the second digit is one of the three not yet selected = 3/10
the probability that the third digit is one of the two not yet selected = 2/10
the probability that the final digit is the remaining one from the set of 0,1,4,9 = 1/10
i.e. $prod_i=1^4 fraci10$
You could then worry about permutations to pick 4109 specifically, but this is needless complication in comparison the way your colleague said.
answered 1 hour ago
Chris H
99928
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Your colleague is right. If your digits are randomly and independently selected (assuming equal likelihood for each digit), the chance that your number is 4109 is nothing more complicated than:
the chance that the first digit is 4
the chance that the second digit is 1
the chance that the third digit is 0
the chance that the fourth digit is 9
all multiplied together. That's $(1/10)^4$, as your colleague says. Where would permutations come into it?
You said in the question that you're "not interested in the probability of generating the numbers 0149, 9014, 4910 etc etc", i.e. the set of all 4-digit numbers where the digits are 0,1,4,9 in any order, but that isn't what $(1/10) ^ 4$ tells you - to calculate that you'd need:
the probability that the first digit is 0,1,4, or 9 = 4/10
the probability that the second digit is one of the three not yet selected = 3/10
the probability that the third digit is one of the two not yet selected = 2/10
the probability that the final digit is the remaining one from the set of 0,1,4,9 = 1/10
i.e. $prod_i=1^4 fraci10$
You could then worry about permutations to pick 4109 specifically, but this is needless complication in comparison the way your colleague said.
answered 1 hour ago
Chris H
99928
Your colleague is right. If your digits are randomly and independently selected (assuming equal likelihood for each digit), the chance that your number is 4109 is nothing more complicated than:
the chance that the first digit is 4
the chance that the second digit is 1
the chance that the third digit is 0
the chance that the fourth digit is 9
all multiplied together. That's $(1/10)^4$, as your colleague says. Where would permutations come into it?
You said in the question that you're "not interested in the probability of generating the numbers 0149, 9014, 4910 etc etc", i.e. the set of all 4-digit numbers where the digits are 0,1,4,9 in any order, but that isn't what $(1/10) ^ 4$ tells you - to calculate that you'd need:
the probability that the first digit is 0,1,4, or 9 = 4/10
the probability that the second digit is one of the three not yet selected = 3/10
the probability that the third digit is one of the two not yet selected = 2/10
the probability that the final digit is the remaining one from the set of 0,1,4,9 = 1/10
i.e. $prod_i=1^4 fraci10$
You could then worry about permutations to pick 4109 specifically, but this is needless complication in comparison the way your colleague said.
answered 1 hour ago
Chris H
99928
edited 59 mins ago
answered 1 hour ago
Chris H
99928
answered 1 hour ago
Chris H
99928
answered 1 hour ago
Chris H
99928
99928
add a comment |Â
add a comment |Â
up vote
3
down vote
Your colleague is correct. Think of it this way: there are 10,000 ($10^4$) 4-digit numbers: 0000-9999. Your target number is one of those 10,000. Thus, if you picked a 4-digit number randomly, you'd have a one in 10,000 chance of picking that number (or any other specific 4-digit number).
You're right that what you want to calculate here is a permutation, not a "simple" (as you called it) combination. But permutations are actually simpler to calculate than combinations. In a permutation, you're essentially just counting all possible outcomes. In a combination, you additionally have to calculate how many of those outcomes are the same combination of elements in a different order, and then adjust your calculation for that.
answered 1 hour ago
Ruben van Bergen
3,4291817
add a comment |Â
up vote
3
down vote
Your colleague is correct. Think of it this way: there are 10,000 ($10^4$) 4-digit numbers: 0000-9999. Your target number is one of those 10,000. Thus, if you picked a 4-digit number randomly, you'd have a one in 10,000 chance of picking that number (or any other specific 4-digit number).
You're right that what you want to calculate here is a permutation, not a "simple" (as you called it) combination. But permutations are actually simpler to calculate than combinations. In a permutation, you're essentially just counting all possible outcomes. In a combination, you additionally have to calculate how many of those outcomes are the same combination of elements in a different order, and then adjust your calculation for that.
answered 1 hour ago
Ruben van Bergen
3,4291817
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Your colleague is correct. Think of it this way: there are 10,000 ($10^4$) 4-digit numbers: 0000-9999. Your target number is one of those 10,000. Thus, if you picked a 4-digit number randomly, you'd have a one in 10,000 chance of picking that number (or any other specific 4-digit number).
You're right that what you want to calculate here is a permutation, not a "simple" (as you called it) combination. But permutations are actually simpler to calculate than combinations. In a permutation, you're essentially just counting all possible outcomes. In a combination, you additionally have to calculate how many of those outcomes are the same combination of elements in a different order, and then adjust your calculation for that.
answered 1 hour ago
Ruben van Bergen
3,4291817
Your colleague is correct. Think of it this way: there are 10,000 ($10^4$) 4-digit numbers: 0000-9999. Your target number is one of those 10,000. Thus, if you picked a 4-digit number randomly, you'd have a one in 10,000 chance of picking that number (or any other specific 4-digit number).
You're right that what you want to calculate here is a permutation, not a "simple" (as you called it) combination. But permutations are actually simpler to calculate than combinations. In a permutation, you're essentially just counting all possible outcomes. In a combination, you additionally have to calculate how many of those outcomes are the same combination of elements in a different order, and then adjust your calculation for that.
answered 1 hour ago
Ruben van Bergen
3,4291817
answered 1 hour ago
Ruben van Bergen
3,4291817
answered 1 hour ago
Ruben van Bergen
3,4291817
answered 1 hour ago
Ruben van Bergen
3,4291817
3,4291817
add a comment |Â
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%2fstats.stackexchange.com%2fquestions%2f372280%2fprobabilty-of-a-specific-4-digit-number%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 probabiliy of generating the number 4109 if we are generating all possible combinations of 4 digit numbers is $1/10^4$, since there is exactly 1 such number out of all the combinations. Was that the qeustion?
– user2974951
1 hour ago
@user2974951: It looks like you're using the term "combinations" where you actually mean "permutations". This is a little confusing, especially in the context of this particular question.
– Ruben van Bergen
1 hour ago
So technically, 1/10^4 is the permutation, and not a combination?
– amateurjustin
1 hour ago