How would one calculate the probabilities or odds of an event as the event takes place. flipping a coin for example, the first player to reach 5 successful flips.

The probablities of either is 50% at the beginning but as the game goes on the real probilities change and therefore the odds.

Heads tails probability heads/tails
0 0 50%/50%
1 0 ?/?
2 0 ?/?
2 1 ?/?
etc

Thanks
Sacha

When I try to apply the model in your paper to a situation where one observer has a probability of 30%, and another gives 99%, I end up using a weighting term that says that the 99% must correspond to stronger confidence or more observations. That seems to have a similar effect to using geometric means; the result is closer to the more extreme belief than the arithmetic mean would suggest.

The closing line on page 17 is: “the relative stock that an agent places on its own prior probability [...] serves to attenuate demand by a multiplicative factor.” This doesn’t seem a pure linear model to me, but maybe I’m not reading it correctly.

.00001 .0043 .065, .25 .37 .25 .065 .0043 .00001

Which doesn’t match either the geometric or the arithmetic mean the way I’ve calculated them. I’ll have to appeal to a higher authority to see how this should be calculated. The good news is that the calculated probability of drawing blue is 50%, so not everything is lost.

If anyone wants to look over my spreadsheet to help figure out what’s going on, I’ll be glad to provide copies. The email address is my last name, and the domain is mydruthers dot com.

p = P/(1-P), q = Q/(1-Q), r = sqrt( p * q ).

To convert r back from odds to probability you compute R = r/(1+r).

With P=.3, and Q=.8, I get p=.429, q=4, r=1.3, and R=.57.

I guess I’m dumb, but why it the geometric mean 57%? The geometric mean of 30 and 80 is 49, and the Bayesian update of 30% with 80% evidence should be 63% by my calculation. I can’t get 57%.

p(h|e) = p(e|h) * p(h) / p(e)

Alice has to start with an initial hypothesis that assigns a probability to every possibility from 0 to 10 red balls. Let’s assume each possible distribution was equally likely. That means the prior has 1/11 as the probability of each number from 0 (all blue) to 10 (all red).

After drawing 6 blues and 4 reds, Alice will rule out 0 and 10, and will update the likelihood of each of the other outcomes. Her estimate of an even split will be

p(drawing 6 blues|even split) * p(even split) / p(drawing 6 blues)

The prior probability for an even split is 1/11.

The probability of drawing 6 blues if the urn is evenly split is
(10 choose 6) * .5^10 = 10!/6!(10-6)!/1024 = 210/1024 =~ .205

The prior probility of drawing 6 balls (over all the possible distributions of balls) was .083 (if I’ve calculated correctly.)

So Alice’s posterior for 5 Blue/5 Red is .205*.083*11 =~ .226

The full probability distribution is (sorry, I can’t make a table here.)

# blue 1 2 3 4 5 6 7 8 9
prob 0.00015 0.006 0.04 0.12 0.23 0.28 0.22 0.097 0.012

(all blue and all red are impossible)

With that distribution of probabilities, Alice’s estimate of the probability of drawing a blue ball is .58. By symmetry, Bob’s estimate is .42. But more interesting, his probability distribution is the inverse of Alice’s.

0.00015 0.006 0.04 0.12 0.23 0.28 0.22 0.097 0.012

When we want to combine the estimates, we have to combine the individual probabilities.

Combined using arithmetic mean:
0 0.0062 0.0514 0.13 0.199 0.226 0.199 0.13 0.051 0.00621 0

Combined using Geometric mean:
0 0.0016 0.029 0.113 0.22 0.27 0.22 0.113 0.029 0.0016 0

Both of these combinations pass my first cross check, which is that they’re both symmetric. I’m not sure how to figure out which is better, except to compute out the same probabilities using 20 draws. It’ll take me a little while, even though I have the spreadsheet started. I thought sharing these results now would give people a chance to show where I’m already wrong, or help me work out how to compare these estimates.

One other idea on the distortion of extreme odds has been widely discussed by players of the FX game. That game uses play money and gives players initial fixed sums (which they can then grow or shrink according to how well they do). If a claim has a price of $FX 0.10, and 10 players have $FX 100 to bet on this, one player betting YES can match the bets of 9 players betting NO. In these extreme price ranges, players betting on the long shots have more “market power” than those betting the other way.

1) Alice & Bob are trying to estimate the percentage of red balls in
an urn. Alice draws 10 balls (without replacing), 4 are red, 6 are
blue — she estimates 40% red. Bob draws 10 more balls (without
replacing), 6 are red, 4 are blue — he estimates 60% red. If they
combine their evidence, they in effect have 20 balls, 10 of which are
red, and 10 of which are blue, so to combine their estimates, clearly
the arithmetic mean is right way to go, yielding an estimate of 50%
red. (Geometric mean gives slightly less than 50%.)

2) Alice & Bob are trying to estimate the probability that all the
balls in the urn are red. Alice draws a bunch of red balls and guesses
that the probability is fairly high. Bob draws one ball — a blue one,
and gives a probability of 0. The right way to aggregate their
probabilities here is clearly not the arithmetic mean, but rather the
geometric.

3) This time let’s just place 2 balls in the urn. Alice and Bob are
trying to predict the probability that they are both red (prior is 25%).
Alice draws a red one and replaces it. She estimates the probability at 50%.
Bob draws one ball, red, and replaces it, estimating the probability of both
red at 50% also. If they combine their info from the two independent trials,
they ought to come up with 2/3 which neither the geometric nor arithmetic
mean will give.