A mathematical dinner!!

It's a pity that the probability of being served last is uniform, since this means that people can get to the answer without having to work hard ( $49$ outcomes are all equally likely, so their common probability is $\tfrac{1}{49}$ ). Let us see why the probability of being served last is the same for all guests.

Consider the symmetric random walk on the integers, and let $N\ge4$ be a positive integer. If we let $q_{N,j}$ be the probability of reaching the point $N$ before the point $1$ , given that the walk starts at $j$ (for $1 \le j \le N$ ). Then it is clear that $q_{N,1}=0$ and $q_{N,N}=1$ , while conditioning on the outcome of the first move gives $q_{N,j} \; =\; \tfrac12\big[q_{N,j-1} + q_{N,j+1}\big] \qquad 2 \le j \le N-1$ Solving this recurrence relation gives $q_{N,j} = \frac{j-1}{N-1}$ for $1 \le j \le N$ .

Now let $p_{N,j}$ be the probability of reaching both $2$ and $N$ before either $1$ or $N+1$ , given that the walk starts at $j$ (for $2 \le j \le N$ . Conditioning on the outcome of the first move gives $p_{N,j} \; = \;\tfrac12\big[p_{N,j-1} + p_{N,j+1}\big] \qquad 3 \le j \le N-1$ Now $p_{N,2} = q_{N,2} = \frac{1}{N-1}$ . By symmetry it is also true that $p_{N,N} = q_{N,2} = \frac{1}{N-1}$ . Thus we deduce that $p_{N,j} \; = \; \frac{1}{N-1} \qquad 2 \le j \le N$ But $p_{N,j}$ is the probability that the person sitting in seat $1$ in a circular seating arrangement with $N$ chairs gets served last, when the host is in seat $j$ . Since this does not depend on $j$ , we deduce that all guests have the same probability $\frac{1}{N-1}$ of being served last, irrespective of their position at the table. Thus the desired probability for this problem is $\tfrac{1}{49}$ , making the answer $\boxed{50}$ .

The head of the table is always served first. Therefore the head is not considered in the equation. The total number of people waiting for spaghetti (n) is therefore 49 and when considering any individual at the table, the expression is 1/49. M=1 so M+N=1+49=50

For the case of n = 3, it is obvious that the two people not at the head of the table have equal 1/2 chances of being the last served (BTLS). For the case of n = 4, label the diners as A,B,C,D (with A being the head), and consider D’s probability of BTLS. The various paths of spaghetti that allow D to be the last served are: ABC..., ABABC..., ABABABC..., etc. (1) The sum of the probabilities of these is 1/2^2 +1/2^4+1/2^6..... = 1/3 By symmetry, B also has a 1/3 chance of BLTS, and then that leaves a 1/3 chance for C. Hence, B, C, and D all have equal 1/3 chances of BLTS. The probabilities for n = 5 are a bit tedious to calculate in this same manner, so at this point we will (for lack of a better option) make the following guess:

Two things must happen to a given diner for BTLS: (1) The plate must approach the given diner from the right or left and reach the person next to that diner. (2) The plate must then reverse its direction and make its way (in whatever manner) all the way around the table until it reaches the person on the other side of the given diner.

For any of the (non-head) diners, the probability that the first of these conditions will be satisfied is 1. This condition will therefore not differentiate the probabilities of BTLS.

Given that (1) has happened, there is some definite probability of (2) happening, independent of where the diner is located. This is true because the probability of traveling all the way around the table does not depend on where this traveling starts. Hence, (2) also does not differentiate between the n − 1 (non-head) probabilities of BTLS.

Thus, all the n − 1 (non-head) probabilities of BTLS are equal, and are therefore equal to 1/(n − 1).

Thus for 50 diners probability of BTLS is 1/49.