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Three people get into an empty elevator at the first floor of a building that has 10 floors. Each presses the button for their desired floor (unless one of the others has already pressed the button). Assume that they are equally likely to want to go to floors 2 through 10 (independently of each other). What is the probability that the buttons for 3 consecutive floors are pressed? (question from 'introduction to probability' by Bliztstein)

I thought the answer might be $$\frac{7 \cdot3!}{9^3}$$... but then I thought maybe the total number of ways is the bose-einstein value $n+k-1 \choose k$ (how many ways there are to choose k times from a set of n objects with replacement, if order doesn't matter).. so then the answer could also be $$\frac{7}{165}$$ Which is the correct answer, and why is the logic for the other answer wrong?

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    $\begingroup$ I think the $\frac{7\cdot 3!}{9^3}$ is the only defensible answer, given the assumptions. If we are told that exactly $3$ buttons are pressed, the answer changes. $\endgroup$ – André Nicolas Jun 28 '15 at 23:02
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    $\begingroup$ The problem with the unordered 3-tuples is that the 165 of them are not equally likely (under the assumptions of three independent uniformly distributed choices). For example, {3,4.7} is six times more likely to be chosen than {5,5,5}. $\endgroup$ – Ned Jun 28 '15 at 23:06
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I think there is only one reasonable interpretation of the problem. It leads to a probability model that produces your first answer.

Let our three passengers be A, B, and C. If A, B, C wish respectively to go to floors $p$, $q$, and $r$, record that fact as an ordered triple $(p,q,r)$. By the equally likely and independent part of the problem statement, all ordered triples $(p,q,r)$, where $p$, $q$, and $r$ range from $2$ to $10$, are equally likely. There are $9^3$ such triples. Whether a passenger, say C, happens not to press the elevator button to Floor $6$ because it has already been pressed is irrelevant.

We now count the favourables. There are $7$ ways to choose a collection of $3$ consecutive numbers in the interval from $2$ to $10$. For each of these collections, there are $3!$ ways in which A, B, and C might wish to get off at the floors in this collection, for a total of $7\cdot 3!$. For the probability, divide.

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  • $\begingroup$ whys is "Whether a passenger, say C, happens not to press the elevator button to Floor 6 because it has already been pressed is irrelevant.", isnt that relevant to the line " (unless one of the others has already pressed the button)" in question ? $\endgroup$ – bicepjai Dec 19 '18 at 8:48
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Call the passengers A,B,C.
Consider A presses 2, then for there to be three consecutive ( i.e. 2,3,4 ) buttons pressed B and C must press 3 and 4, this can occur in only two ways, with B pressing 3 and C pressing 4 or visa versa.

Repeating this method on floors 3,4,5,6,7 and 8 we have a total of 14 different ways of getting 3 consecutive buttons pressed where A gets out on a lower floor than the other two.
Therefore there is a total of 3 times 14 = 42 ways of getting 3 consecutive buttons pressed, the probability of which is $\Large\frac{42}{9^3}$
Much the same as André Nicolas' answer.
BTW if one of the passengers was an unaccompanied 4 year old, chances are almost certain that every damn button would be pressed, at least the ones he/she could reach.

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The first one $$ \dfrac{7*3!}{9^3} $$ is the correct answer.

We can not use Bose-Einstein here because it can not be used in naive probability expression according to 'introduction to probability' by Bliztstein, page 18. "The Bose-Einstein result should not be used in the naive definition of probability except in very special circumstances. For example, consider a survey where a sample of sizekis collected by choosing people from a population of size n one at a time, with replacement and with equal probabilities. Then then k ordered samples are equally likely, making the naive definition applicable, but the unordered samples (where all that matters is how many times each person was sampled) are not equally likely." Here, since each person presses which floor is an equal likely event, we cannot use Bose-Einstein.

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let's say we have A, B, and C people in the lift. A - is the event that passengers will press three consecutive numbers. P(A)= # of favourable outcomes/# of total outcomes Since there is replacement and order matters the # of total outcomes is 9^3 Now we should find the # of fav outcomes: there are 7 ways to get three consecutive numbers out of 2-10 => (234), (345), (456), (567), (678), (789), (8910) and 3! ways to shoose from three numbers Therefore, the number of fav outcomes is 7*3! P(A)=7*3!/9^3

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The number of possible outcomes for who is going to which floor is $9^3$. There are $7$ possibilities for which $3$ buttons are pressed such that there are consecutive floors: $(2,3,4),(3,4,5)...(8,9,10)$. For each of these $7$ possibilities, there are $3!$ ways to choose who is going to which floor. So by the naive definition, the probability is $${3!\times7\over9^3}= {42\over729}= {14\over243} = 0.05761316872 $$

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In the sequence $2,3,4,\cdots,10$ there are $7$ sequences of three consecutives numbers: $$(2,3,4), (3,4,5), (4,5,6), (5,6,7), (6,7,8), (7,8,9), (8,9,10)$$ and by outher side there are ${9 \choose 3}=84$ distinc modes od choose $3$ numbers in set ${2,3,\cdots,10}$, Then, in my opinion, the probability is $P=\frac{7}{84}$.

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  • $\begingroup$ This seems to exclude the possibility of any of the people being on the same floor $\endgroup$ – Dr Xorile Jun 29 '15 at 0:51
  • $\begingroup$ Really Dr. Xorile, If consider that people can come down on the same floor, then the total of modes is equal the number of non negative interger of equation $x_1+x_2+\cdots+x_9=3$, that is, ${9+3-1 \choose 3}=165$. In this case the correct probability is $P=\frac{7}{165}$. Thank you for letting me know! $\endgroup$ – Cgomes Jun 29 '15 at 1:26
  • $\begingroup$ But the 165 different multi-sets are not equally likely, so you can't just count 7 successes and divide by 165 to get the probability. The only obvious way to get equally likely outcomes is to take the outcomes as ORDERED triples, so the answer is the first version in the OP. $\endgroup$ – Ned Jun 29 '15 at 1:39

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