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Is there a known formula to calculate perceived value? If I am auctioning some popular product that is worth $100 (shelf price), how much would people bid if they:

  1. Have a 1 in C chance of winning the product.
  2. Will forfeit P of their bid amount if they lose.

B = Average bid.

V = Shelf value of the product.

C = # of bidders.

P = Percentage of bid amount forfeited for losing.

I'm looking for a formula in which I can plug in values for the variables V, C, and P.

I'm not too "mathy" so I don't really know what tags to apply to this question, sorry about that.

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2 Answers 2

Another alternative approach to this (if you don't want to confuse yourself with long equations or if you need to do it quickly on your mobile calculator ....etc):

1) you get the ratio between probability of losing to winning
2) then multiply that ratio by the Percentage of bid amount forfeited for losing.
3) Then You can simply multiply the result by "V" and you will get the same result of what you'd expect people to bid at most ..

- Check this out - Using the first example mentioned above:

Step 1) Get the ratio between probability of losing to winning = 4 (since there are 5 bidders)

Step 2) Multiply 4x0.05 = 0.2 (the Percentage of bid amount forfeited for losing gets magnified directly proportional to the number of bidders against you (the ratio)

Step 3) 100 - (100 *0.2) = 80 $ so you'd expect people to bid at most that much (more or less same answer as mentioned above)

  • Try it with the second example:

Step 1) 0.667/0.333= 2

Step 2) so the percentage of bid amount forfeited for losing is doubled = 2 * 0.02 = 0.04

Step 3) 150-(150*0.04) = 144 $

More or less same result as mentioned above

Cheers Guys

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This is a pretty simplified model, but with the assumptions you've given, you could think about the inequality $$(1/c)(v - b) + (1 - 1/c)(-p b) > 0$$

This is something like the expected change in your assets if you pretended that was v dollars you were bidding on. With probability $1/c$ you win \$v minus your bid. With probability $1 - 1/c$ you lose $p$ times your bid.

So for example, if there were $c=5$ people bidding, $v=100$ dollars value, and you lose 5% of the value of your bid when you lose, so $p=0.05$. Then we'd solve for $b$ in

$$(1/5)(100 - b) + (1 - 1/5)(-0.05 b) > 0$$

which is satisfied when $b<83.3333$, so you'd expect people to bid at most that much.

For another example, if there were 3 people bidding on \$150 value, and they'd lose 2% of their bid when they lost, then we'd solve $$(1/3)(150 - b) + (1 - 1/3)(-0.02 b) > 0$$

and find that it is satisfied for $b<144.231$.

I would guess that in reality it would be more complicated because there's probably some game theory stuff going on as well as people having different risk tolerances. I'm by no means an expert on auctions.

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Thanks for all the detail in your answer - I'll think it over! –  Josh M. May 27 '11 at 2:57
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