Billy Pang
tuxedo_man at hotmail.com
Fri Aug 26 14:20:48 CDT 2005
this three doors problem reminds me of this year's NHL (National Hockey League) entry draft. Every year, an entry draft is held for the 30 teams to draft young players for their team. Every player is ranked and this year, the top ranked player (Sydney Crosby) is considered the next "great one". The team who gets first pick at the entry draft will pick Mr. Crosby for sure (in a sense, it would be like winning the lottery). Ok, long story short, they decided to use the "ping pong" method of determining which teams gets to go first. every team either gets one, two or three ping pong balls in the pool. there are 30 teams. Because the first ping pong ball selected gets first pick in the entry draft, the teams with three ping pong balls have a greater chance of getting the coveted first pick than the other teams with only two or one ping pong ball. I watched the draft live as it took place. To increase drama, they select the ping pong balls behind closed doors to determine the order of the thiry teams; then they come out and reveal the order in reverse (starting with 30, 29, 28, etc...); however, i realized that if they pick the ping pong balls in reverse order (ie. first ball they picked is 30th pick), the teams with three ping pong balls would no longer be favorites because they have a greater chance of being selected. In a sense, this situation is similiar but not exactly the same as the three doors problem. in the beginning, when you ave three doors to choose from, you have 33.3% of winning or 66.7% of losing. you either win or don't win, there is no in-between. therefore if you pick door A, you have 33% of winning. another of way saying it is, if you don't pick door A, you have 66.7% of winning (note the difference between this sentence and first sentence of this paragraph). yes, if you picture the other two doors as a group of doors and the host reveals to you that one of those doors does not have the winning prize, the group of doors still has 66.7% of winning but now the host has taken out a door in question for you and therefore the other door has 66.7% chance of winning (since the door just opened by the host has 0% chance of winning; 0% + 66.7% = 66.7%) I'm not sure if i am explaining it correctly. I can see reasons for both why you should swap and why you should not swap. the difference lies in on the level where you calculate the probabililty of winning. Billy >From: "Scott Marcus" <marcus at tsstech.com> >Reply-To: Discussion of Hardware and Software >issues<dba-tech at databaseadvisors.com> >To: "Discussion of Hardware and Software >issues"<dba-tech at databaseadvisors.com> >Subject: RE: [dba-Tech] The Three Doors Problem >Date: Fri, 26 Aug 2005 11:02:59 -0400 > >John, > >The host's intentions have nothing to do with the math. The fact is that >the host revealed a non-winning door. Therefore your odds of winning are >2/3 if you switch. > >Scott Marcus >IT Programmer >TSS Technologies Inc. >www.tss.com > > >-----Original Message----- >From: dba-tech-bounces at databaseadvisors.com >[mailto:dba-tech-bounces at databaseadvisors.com] On Behalf Of John W. >Colby >Sent: Friday, August 26, 2005 10:52 AM >To: 'Discussion of Hardware and Software issues' >Subject: RE: [dba-Tech] The Three Doors Problem > >It does indeed state that, what it does NOT state is that the host >INTENTIONALLY PICKED a non-winning door. > > >I am the host of a TV program and you are the guest. This is the deal: > > >there are 3 doors. Behind one of them is $100 million. Behind the other >two >are a dead catfish and a dead pickerel respectively. > > >I invite you to select a door. You choose any one of the three: call it >x. > > >I open another door, and say, Had you selected door y, you would have >won a >dead catfish. > >That is the original message, word for word, as stated in Arthur's >email. > >The last sentence is the crux of the matter... > > >I open another door, and say, Had you selected door y, you would have >won a >dead catfish. > >Had he said: > >"I intentionally pick a door that is a loser and say..." > >THEN the solution being discussed is indeed valid. It his use of the >knowledge he has that is the issue. > >If the host flips a coin, and uses the result of the coin toss to pick >one >of the other doors, then you are back to 50/50. He is not telling you >anything, he is just eliminating a door, that just happens to be a >loser. > >John W. Colby >www.ColbyConsulting.com > >Contribute your unused CPU cycles to a good cause: >http://folding.stanford.edu/ > >-----Original Message----- >From: dba-tech-bounces at databaseadvisors.com >[mailto:dba-tech-bounces at databaseadvisors.com] On Behalf Of Scott Marcus >Sent: Friday, August 26, 2005 10:36 AM >To: Discussion of Hardware and Software issues >Subject: RE: [dba-Tech] The Three Doors Problem > > >John, > >The problem was stated originally that the host revealed a non-winning >door. > >Scott Marcus >IT Programmer >TSS Technologies Inc. >www.tss.com >_______________________________________________ >dba-Tech mailing list >dba-Tech at databaseadvisors.com >http://databaseadvisors.com/mailman/listinfo/dba-tech >Website: http://www.databaseadvisors.com > > > >_______________________________________________ >dba-Tech mailing list >dba-Tech at databaseadvisors.com >http://databaseadvisors.com/mailman/listinfo/dba-tech >Website: http://www.databaseadvisors.com >_______________________________________________ >dba-Tech mailing list >dba-Tech at databaseadvisors.com >http://databaseadvisors.com/mailman/listinfo/dba-tech >Website: http://www.databaseadvisors.com