Sitting Together

If Bob chose any spot on the circular table, Alice would have 2 choices no matter what, which gives us 10 possibilities.

If Bob chose any 3 of the middle spots, Alice would have 2 choices, but if Bob chose the outer 2 spots, Alice would only have 2 choices, giving us 8 total possibilities.

Relevant wiki: Rule of Product - Intermediate

Let's look at the circular table first. Suppose Alice is seated first. There are four remaining seats, two of which are next to Alice. The probability of Bob ending up next to Alice is therefore $\frac{2}{4}=\frac{1}{2}$ .

Now look at the rectangular table. We will look at two mutually exclusive cases.

In the first case Alice is seated at the end of the table. The probability of this happening is $\frac{2}{5}$ . In this case there is only one of the four remaining seats next to Alice, so the probability of Bob sitting next to Alice is $\frac{1}{4}$

In the second case Alice is seated at one of the three middle seats. The probability of this happening is $\frac{3}{5}$ . In this case there are two of the four remaining seats next to Alice, so the probability of Bob sitting next to Alice is $\frac{2}{4}=\frac{1}{2}$

Combining the two cases we see that the probability of Bob being next to Alice at the rectangular table is

$\frac{2}{5} \times \frac{1}{4} + \frac{3}{5} \times \frac{1}{2}=\frac{2}{5}$

Since this is clearly less than a half, Alice and Bob are most likely to sit together at the round table.

table B is actually a transformation of Table A.

but for table A----every time you will get at least 2 persons beside you wherever you take seat.

but for table B---if you sit at the 2 corners you will get only 1 person beside you.

so, getting 2 persons every time have more probability to sit next to each other than getting 1 person.

There are 4! Arrangements in the circular table with 2x3! of them where Bob and Alice are together. There are 5! Arrangements in second case out of which 2x4! have Bob and Alice together.

The configuration of table B translates naturally to a configuration of table A (a one-to-one mapping; a bijection).

If the pair sits together at table B, they'll also sit together at table A. But there is a configuration of table A where they sit together, but in the corresponding configuration of table B they sit on opposite ends.

Therefore table B has fewer configurations where they sit together.

They are same arrangement. I have this discussion with my kids often who want to sit "next" to me or my wife. The table b seat at the end are "next" to each other along the outside. If you eliminate this option A has a more likely occurance.

One can look at table-A as simply being a graph where the chairs are connected (as next to each other) in 5 ways... In table-B there are only 4 such connections (for the same arrangement of nodes) This clearly implies that there are more ways to sit Alice and Bob together with Table-A arrangement since there are more ways to connect their nodes (i.e. chairs)

Ergo: Table-A is the better chance.

Round would have 2 by both your sides, whichever seat you choose. So friends could cuddle well in this set up. So table A

End seats have a lower chance of success.

Loops are endless.

Use loops.