probability games: simulating a friend's chance to win a condo bid

A friend of mine big on a condo and lost, ending up 3rd/9. The friend was convinced the bid had been too low. I wasn't so convinced: after all, what if the two people who'd bid higher hadn't bid, and no other higher bidders had shown up? It's an inherently probabilistic process. If the goal was to win at all cost, you'd just bid the most you could possibly spend every single time.

So to estimate the chances that bid would have won the bid requires estimating two probability distributions: the probability distribution of the number of bids and the probability distribution of the values of those bids. Without any information available on these, it's necessary to guess.

The first approximation is that bids are uncorrelated: whether someone bids, and how much, is independent of how many others do so.

The next approximation is that the bids are statistically representative of the probability distributions. So not counting the friend, there were 8 other bids. If the process were repeated 1000 times, there'd be around 8000 total other bids. Within each iteration, the number of bids would have a probability distribution well approximated by Poisson statistics.

Next, I won't try to estimate a probability distribution of bid values. I don't need to. Bids are binary: either they're better or they're worse than the friend's bid. Based on this outcome, there is a 25% (2/8) a given bid would be better than the friend's bid.

So I need to calculate, for all possible number of bidders, the probability that the friend's bid would win given a 75% chance of being better than any competitor, then sum up these results weighted by the probability of that number of bidders.

Given N other bidders, the probability that the friend's bid would have won would be α^N, where α is the probability the friend's bid is better than a given other bid. In this case we estimate that at 75%.

Then there's the probability that there were N other bidders. Poisson statistics say that probability is approximated by (λ^N / N!) exp(-λ), where &lambda is the expectation value for other bidders. In this case I'm assuming that to be 8. N! is N factorial.

So the probability the bid would have won is represented by the infinite sum:

sum N from 0 to infinity [ (λ^N / N!) exp(-λ) α^N ] =

sum N from 0 to infinity [ ((λα)^N / N!) exp(-λ) ]

Plugging in the numbers yields 13.5%.

If I had been lazy and assumed the number of bids was fixed at 8, the estimate would have been 10%. It's not surprising this was a bit lower: there's two ways to get lucky. One is the competitors happen to bid lower. The other would be the number of competitors was less. The 10% would exclude the latter possibility.

So 13.5% may not be so bad a probability. If the friend bid on 8 places, there's a good chance with that record he gets one of them, and at a lower price than if he raised his bid high enough to get a higher probability of success. But then again there's a lot of unjustified assumptions here, and those assumptions have only a chance of being accurate.

An example of the limitation of this approach is it would lead to the conclusion the "winner" of the bidding had a 100% chance to do so, which is of course incorrect. A more sophisticated approach would need to be adopted to do a better job with his chances. However, for the 3rd place bid I suspect the approach here does considerably better.