# How does Thomas Minka arrive at this answer for the following Question?

In this paper by Thomas Minka the author gives the following example :

Suppose you, a Bostonian, have entered the New Hampshire lottery along with 999 people from New Hampshire. The prize will be awarded to exactly one of the 1000 people. By sheer luck, you obtain a computer printout listing 998 participants; each name is marked "no prize", and yours is not among them. Should your chances of winning increase from 1/1000 to 1/2? Under normal circumstances, yes. But suppose while poring anxiously over the list you discover the query that produced it: "Print the names of any 998 New Hampshire residents who did not win." Since you are from Boston, the list could not possibly have had you on it. Thus it is completely irrelevant to you; your probability of winning is still 1/1000.

I don't undestand why he arrives at 1/1000 ? If I reduce the number of people from Hampshire to be 2. I can represent the total set of events by $\Omega = \{ (1,0,0);(0,1,0);(0,0,1)\}$. With $(Boston,Hampshire1,Hampshire2)$ and a $1$ indicating a win for the particular person.

Now if i have a list saying $Hampshire1$ lost, clearly the set of all possible events is reduced to : $\Omega = \{ (1,0,0);(0,0,1)\}$. Thus the the probability of winning given that list does change...or no??

• I wonder what Minka's answer would be if "998" were changed to "999". If it remains 1/1000 (which obviously it should not be), how could he possibly justify it; and if it were to change, how he could possibly argue that 1/1000 really is a correct answer to the question as stated. I would love to be able to place bets with him on this with 999:1 odds! – whuber Dec 27 '16 at 22:02
• I interpret Minka's answer in terms of the "Odds Bayes Rule", i.e. if $W$ = "win" and $L$ = "list", then $\frac{\Pr[W|L]}{\Pr[\sim\!W|L]}=\frac{\Pr[L|W]}{\Pr[L|\sim\!W]}\frac{\Pr[W]}{\Pr[\sim\!W]}$. So for a list of 998 NH residents the Bayes factor would be $1$. And for 999 NH residents the Bayes factor would be $1/0$. – GeoMatt22 Dec 28 '16 at 16:59

The event of the particular $998$ New Hampshirites appearing on the list is* independent of the event of the Bostonian winning, and thus the conditional probability of the Bostonian winning equals the marginal probability of the Bostonian winning.
In your simplified (3 players) case, you condition on "Hampshire 1 lost" but the information we actually have is "A computer search for one losing player from Hampshire returned Hampshire 1". This is analogous to the Monty Hall problem - think winning = winning, the Bostonian = the door we originally pick, the list we discover = the doors opened by game host. Finding* out the list was produced by querying $998$ non-winning New Hampshirites = the standard assumptions of Monty Hall, i.e., the host always reveals a door not-initially-selected not-containing-the-prize.
Furthermore, we assume the event "We find these printed results of this query" can be handled just like the event "The results of this query are..." This necessitates that the "We find" part is independent of everything - it could be that someone run the query for $998$ non-winning New Hampshirites with the intention to get the list of all non-winning ones.