Selling an extra ticket – a math bargaining puzzle

I came across a fun puzzle at a brain teaser forum BrainDen.com.

Here is the problem:

Sean wants to attend a basketball game. Unfortunately, he could not get tickets before it was sold out, but he has located someone that has an extra ticket which he is willing to sell to Sean. The seller begins by asking $100 for the ticket. Sean counters by offering $50. The seller always counters by asking the average of the two offers. Sean always counters by the average of his previous offer and the seller’s new offer.

At what price will Sean and the seller finally agree upon?

The bidding and selling prices are interesting because each round’s price is dependent on the previous ones.

You may wish to solve this on your own. Below are a couple solution methods.

(For a game theory model on this, where buyer and seller have different preferences, see this article about patience in negotiations)

Method 1: spreadsheet

I like using numerical methods for problems like this. It’s easy and fun to find an answer quickly.

I whipped up a quick table in Open Office SCalc and here is what I got:

graph ticket prices

The answer is $66.67 and the convergence happens in about 7 rounds.

(A small technical note is the answer depends on when the rounding is done. When I rounded each price to the nearest penny step-by-step, I found the price to be $66.69. I’m going to ignore this small detail and assume the rounding takes place only at the last round.)

Method 2: a series formula

The user psychic_mind on BrainDen.com provides a clever way to solve the puzzle.

The trick is seeing a pattern in the bids.

In the first round, the seller names the starting price of $100 and the buyer counters with $50, or half of the seller’s price. We can also write this the following way:

Price after one round = $100 (1 – 1/2)

In the second round, the seller names a price equal the average of the last two. This is $75 = $100 (1 – 1/2 + 1/4). In other words, the seller is adding back half the difference between his last price and the buyer’s last bid. The buyer then counters using the same technique to get $62.5 = (1 – 1/2 + 1/4 – 1/8). The buyer is thus subtracting half of the difference between the current seller’s price and his last price.

Price after two rounds = $100 (1 – 1/2 + 1/4 – 1/8)

The pattern is becoming clear that we have an infinite alternating series with a factor of -1/2:

I love the compactness of this method, and the answer is again shown to be $66.67.

Discussion questions

1. Consider the same problem with the seller starting at $100 and the buyer $0. At what price will they agree now?

2. Consider the same problem but with the buyer going first. That is, the buyer first bids $50, the seller counters $100 and so on as before but with the buyer going first in each round. Will they agree to the same price or different price?

3. Assume the seller goes first again. Now suppose that instead of choosing the average of the last prices, the seller and buyer choose another constant. Say each chooses a price that is 1/3 of the distance between his last price and the other’s last bid. For instance, the seller’s first counter is $100 (1 – 1/3) + $50 * 1/3 = $83.33 and the buyer’s counter to that is $50 (1 – 1/3) + $83.33 * 1/3 = $61.11. What will the final price in this case be? (I have solved this numerically but not analytically, so please share if you figure out a formula)

4. Replace 1/3 in the last problem with a general multiplier of 0 < x < 1.



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  • Tristan

    Hi Presh,
    This is a great series of questions.
    Here is a closed form for the general case:
    Price=Y_1-((Y_1-Y_2))/(2-X)
    Where
    Y1 is the initial bid
    Y2 is the second bid
    X is the weighting ratio between bids

    For example in the original problem Y1=$100, Y2=$50 and X=1/2 (equal weighting between the previous bids, i.e. the arithmetic average). Then the Price= $100-($50/(3/2))=$66.67.
    Discussion problem 3, Y1=$100, Y2=$50 and X=1/3. Then Price=$100-($50/(5/3))=$70.

  • http://ceilingfish.net Tom

    Presh,
    Thanks for the fun problem. Solved it in ruby:

    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}"

  • Someone

    Intuitively, the ‘second mover’ always takes half the step of the first mover. So, the two bidders move towards each other, but the second one moves twice as fast as the first one. Result: in the limit, they will meet at 1/3 of the difference between 50 and 100.

    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.

  • http://www.mindyourdecisions.com/blog/ Presh Talwalkar

    Thanks all for comments, especially Tristan’s explicit formula!

    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.

  • Glenn G. Chappell

    Another method.

    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

  • http://www.mynext.co.uk steve

    Very nice, do like the convergence.

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

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