An intelligent team, ‘Smarty Pants’, applying to a competition completes an evaluation form early. Since they are feeling generous they randomly spread their assignment to all 10 teams applying from their school by successively having one team forward it to some other, which then forwards to some other team, and so on. A team can pass the form on to anyone except the team they got it from.

a. By how many different paths can an assignment travel through the group of 10 in n transfers?

b. What is the probability that team ‘Dunce’ receives it in the third transfer?

c. What is the probability that team ‘Smarty Pants’ receives it in the third transfer?

  • $\begingroup$ When you ask about probabilities - should we assume that in each transfer, all 8 possible teams have the same chance of being selected as the next one to get the assignment? $\endgroup$
    – G0BLiN
    Oct 24, 2014 at 17:32

3 Answers 3


a. The first transfer can go to any team except the one that already has it, so for n = 1, there are 9 possible paths. After that, each team has 8 options, so for n = 2, 9x8=72 different paths. In general there are:

$Paths = 9\times 8^{(n-1)}$

b. The probability that team Dunce gets the form on the third pass is controlled by which team has it after the first and second. The only way that Dunce can get it is if they are not the one who had it after transfer 1, ($P_1=8/9$), nor after transfer 2 ($P_2=7/8$, since the form cannot go back to either Smarty Pants or Team B), and they ARE passed it in transfer 3, ($P_3 = 1/8$). The odds of all of these happening are:

$P_4 = P_1 \times P_2\times P_3 = 56/576 = 7/72$

c. On the third transfer (i.e. Smarty Pants gives it to Team B (1), Team B gives it to Team C (2), then Team C gives it back to Smarty Pants), there is no possibility that Team Smarty Pants was Team B or Team C. Using the same designations as above, $P_1 = 1$ and $P_2=1$. The probability that they get it back after transfer three is:

$P_3 = 1/8$.

  • 1
    $\begingroup$ Good, quick answer and a nice explanation $\endgroup$
    – skv
    Oct 24, 2014 at 17:10
  • 1
    $\begingroup$ for b. they can't have it after the first transfer or after the second transfer. $\endgroup$
    – Bozman
    Oct 24, 2014 at 17:13
  • $\begingroup$ @bozman Good point, reworking b. $\endgroup$ Oct 24, 2014 at 17:17


Assuming 'n' does include the first transfer to 'Smarty Pants' there will be an initial 9 choices for team 'Smarty Pants' to forward to and then 8 choices for every receiving team thereafter. So the answer is:



An easy way to think about this question is: Any team has a 1/8 chance of receiving the forward as long as they are NOT the current or previous team. So using conditional probability we can phrase it as:

P(Dunce is 3rd) = P(Dunce is 3rd|Dunce is not 2nd)* P(Dunce is not 2nd) = (1/8)*(8/9) = 1/9

We do not need to go further since we know there is 100% probability of Dunce NOT being the first to receive.


We know that 'Smarty Pants' is the first team and teams cannot forward back. So regardless of who is 2nd, 'Smarty Pants' cannot be third.

P(Smary is 3rd) = 0

  • $\begingroup$ I was uncertain about the numbering of the transfers as well. I assumed a transfer to mean a move of the paper from one team to the next. $\endgroup$ Oct 24, 2014 at 17:26
  • $\begingroup$ Agreed, the question is ambiguous about that which is why I started my answer with an assumption. $\endgroup$
    – Leo
    Oct 24, 2014 at 17:28
  • $\begingroup$ I believe it wasn't transferred to them. All teams already have the blank assignment, they just answered theirs. $\endgroup$
    – Bozman
    Oct 24, 2014 at 17:30

At any time n the assignment is in one of 3 states

  1. Smarty Pants owns the assignment.
  2. Someone else owns the assignment and got it from Smarty Pants.
  3. Someone else owns the assignment and got it from someone else.

Let's define the probability of each state after n transfers:

$p_{n,s} = P(assignment\ is\ in\ state\ s\ after\ n\ tranfers)$

$p_{0,1} = 1,\ p_{0,2} = p_{0,3} = 0$

From state $1$ the assignment always goes to state $2$. From state $2$ the assignment always goes to state $3$. From state $3$, it goes to state $1$ with probability $1\over 8$ and to state $3$ with probability $7\over 8$. This translates to the formulas:

$p_{n+1,1} = {1\over 8} p_{n,3}$
$p_{n+1,2} = p_{n,1}$
$p_{n+1,3} = p_{n,2} + {7\over 8} p_{n,3}$

We can compute the first 3 transfers:

$p_{0,1} = 1,\ p_{0,2} = p_{0,3} = 0$

$p_{1,1} = 0,\ p_{1,2} = 1,\ p_{1,3} = 0$

$p_{2,1} = 0,\ p_{2,2} = 0,\ p_{2,3} = 1$

$p_{3,1} = {1\over 8},\ p_{3,2} = 0,\ p_{3,3} = {7\over 8}$

And we have the answers:


Smarty Pants gets it back after 3 transfers with probability $p_{3,1} = {1\over 8}$.


Other teams have the assignment after 3 transfers with probability $p_{3,3} = {7\over 8}$ collectively. Each of the 9 teams has it with probability ${7\over 8}\times{1\over 9} = {7\over72}$


As everybody else already mentioned, the first step has 9 possibilities, the others 8, so the number of paths for the assignment is $8\times 9^{n-1}$

  • $\begingroup$ And I should have read the other answers more carefully. The correct answer was given already... $\endgroup$
    – Florian F
    Oct 25, 2014 at 9:56

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