and with the buyer offering a start of 0 would get for £33.33.

A couple of observations:

(1) The values 50 and 100 are not important. Call them a and b. The limit will always be some fixed fraction of the way from a to b, namely, a + k (b-a), for some constant k. So, for example, we can simplify the problem by setting a = 0 and b = 1; then the limit is k.

(2) The positions of the buyer and seller are symmetric. Each picks a number in the center of the current interval.

Now, if we start with a = 0, b = 1, then the answer is k. The buyer bids 1/2, and the positions of the buyer & seller switch, so that a = 1/2 and b = 0. The limit is now (1/2) + k (0 – 1/2).

Solve:

k = (1/2) + k (0 – 1/2)

k = 1/3

I was doing some of the math, and I suddenly realized this is a sort of modified Fibbonaci sequence. Here is why.

Say the seller first names x, then the buyer counters with y. In the puzzle, the seller’s response is not the sum (like in the Fibonacci sequence) but the average of the last two terms 0.5 (x+y). The buyer’s price is again the average of the last two terms in the sequence. The final price is the limit of this sequence.

Discussion question 4 is an extension for a weighted average of w. The sequence is x, y, x (1-w) + wy, and so on. The general formula is escaping me for the moment but it seems like it will match with Tristan’s answer.

Ignoring the first move, the roles of first and second movers are reversed; hence, they meet at 1/3 of the difference between 75 and 50.

seller=100.0
buyer=50.0
while((seller – buyer) >= 0.01) do
seller = (buyer + (seller – buyer) / 2.0)
seller = (seller * 100).round / 100.0
break if (seller – buyer) <= 0.01
buyer = (seller + buyer) / 2
end

puts "Seller #{seller} Buyer #{buyer}"