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- From: rar@csl.sri.com (Bob Riemenschneider)
- Newsgroups: sci.math
- Subject: Re: Equation needed
- Date: 9 Nov 92 19:29:30
- Organization: Computer Science Lab, SRI International, Menlo Park, CA.
- Lines: 160
- Distribution: USA
- Message-ID: <1dna5vINN3ac@roche.csl.sri.com>
- References: <1dbud7INNsli@hpsdde.sdd.hp.com>
- NNTP-Posting-Host: birch.csl.sri.com
- In-reply-to: regard@hpsdde.sdd.hp.com's message of 5 Nov 1992 11:58:31 -0800
-
- In article <1dbud7INNsli@hpsdde.sdd.hp.com> regard@hpsdde.sdd.hp.com (Adrienne Regard) writes:
-
- > From: regard@hpsdde.sdd.hp.com (Adrienne Regard)
- > Newsgroups: sci.math
- > Subject: Equation needed
- >
- > ... The person has a better chance of winning the
- > grand prize if he switches, but I don't know how to write the equation to
- > express the probabilities, and I've got a very excited friend who wants
- > proof on this subject. ...
-
- If your only objective is to convince someone who doesn't believe it,
- you're out of luck. There's been a lot of discussion of this in the
- past, and no one ever seems to convince anyone (s)he's wrong. Let me argue for
- switching using only one "equation", the definition of conditional probability
-
- P(A | B) = P(A & B) / P(B)
-
- Here's my model. Three independent choices are made, at random. Carol
- flips a three-sided fair coin to decide where to put the grand prize;
- call the outcomes GP1, GP2, and GP3. Monty flips a two-sided coin to
- decide which goat to reveal if you pick the door with the grand prize
- behind it; call the outcomes F1 (for "reveal the lower numbered goat")
- and F2 (for "reveal the higher numbered goat"). You flip a three-sided
- fair coin to decide which door to pick; call the outcomes C1, C2, and
- C3. All eighteen outcomes
-
- GP1 & C1 & F1
- GP1 & C1 & F2
- GP1 & C2 & F1
- GP1 & C2 & F2
- GP1 & C3 & F1
- GP1 & C3 & F2
- GP2 & C1 & F1
- GP2 & C1 & F2
- GP2 & C2 & F1
- GP2 & C2 & F2
- GP2 & C3 & F1
- GP2 & C3 & F2
- GP3 & C1 & F1
- GP3 & C1 & F2
- GP3 & C2 & F1
- GP3 & C2 & F2
- GP3 & C3 & F1
- GP3 & C3 & F2
-
- are equally likely before the flips. Hence
-
- P(GP1 & C1 & F1) = 1/18
- P(GP1 & C1 & F2) = 1/18
- P(GP1 & C2 & F1) = 1/18
- P(GP1 & C2 & F2) = 1/18
- P(GP1 & C3 & F1) = 1/18
- P(GP1 & C3 & F2) = 1/18
- P(GP2 & C1 & F1) = 1/18
- P(GP2 & C1 & F2) = 1/18
- P(GP2 & C2 & F1) = 1/18
- P(GP2 & C2 & F2) = 1/18
- P(GP2 & C3 & F1) = 1/18
- P(GP2 & C3 & F2) = 1/18
- P(GP3 & C1 & F1) = 1/18
- P(GP3 & C1 & F2) = 1/18
- P(GP3 & C2 & F1) = 1/18
- P(GP3 & C2 & F2) = 1/18
- P(GP3 & C3 & F1) = 1/18
- P(GP3 & C3 & F2) = 1/18
-
- Now everyone flips, and the outcome is determined once you switch or
- decline to switch. First, you look at your coin -- suppose it's C1
- -- and that allows you to update the possible outcomes' probabilities.
-
- P(GPi & Cj & Fk | C1) = P(GPi & Cj & Fk & C1) / P(C1)
-
- So
-
- P(GP1 & C1 & F1 | C1) = P(GP1 & C1 & F1 & C1) / P(C1) = (1/18)/(1/3) = 1/6
- P(GP1 & C1 & F2 | C1) = P(GP1 & C1 & F2 & C1) / P(C1) = (1/18)/(1/3) = 1/6
- P(GP1 & C2 & F1 | C1) = P(GP1 & C2 & F1 & C1) / P(C1) = 0/(1/3) = 0
- P(GP1 & C2 & F2 | C1) = P(GP1 & C2 & F2 & C1) / P(C1) = 0/(1/3) = 0
- P(GP1 & C3 & F1 | C1) = P(GP1 & C3 & F1 & C1) / P(C1) = 0/(1/3) = 0
- P(GP1 & C3 & F2 | C1) = P(GP1 & C3 & F2 & C1) / P(C1) = 0/(1/3) = 0
- P(GP2 & C1 & F1 | C1) = P(GP2 & C1 & F1 & C1) / P(C1) = (1/18)/(1/3) = 1/6
- P(GP2 & C1 & F2 | C1) = P(GP2 & C1 & F2 & C1) / P(C1) = (1/18)/(1/3) = 1/6
- P(GP2 & C2 & F1 | C1) = P(GP2 & C2 & F1 & C1) / P(C1) = 0/(1/3) = 0
- P(GP2 & C2 & F2 | C1) = P(GP2 & C2 & F2 & C1) / P(C1) = 0/(1/3) = 0
- P(GP2 & C3 & F1 | C1) = P(GP2 & C3 & F1 & C1) / P(C1) = 0/(1/3) = 0
- P(GP2 & C3 & F2 | C1) = P(GP2 & C3 & F2 & C1) / P(C1) = 0/(1/3) = 0
- P(GP3 & C1 & F1 | C1) = P(GP3 & C1 & F1 & C1) / P(C1) = (1/18)/(1/3) = 1/6
- P(GP3 & C1 & F2 | C1) = P(GP3 & C1 & F2 & C1) / P(C1) = (1/18)/(1/3) = 1/6
- P(GP3 & C2 & F1 | C1) = P(GP3 & C2 & F1 & C1) / P(C1) = 0/(1/3) = 0
- P(GP3 & C2 & F2 | C1) = P(GP3 & C2 & F2 & C1) / P(C1) = 0/(1/3) = 0
- P(GP3 & C3 & F1 | C1) = P(GP3 & C3 & F1 & C1) / P(C1) = 0/(1/3) = 0
- P(GP3 & C3 & F2 | C1) = P(GP3 & C3 & F2 & C1) / P(C1) = 0/(1/3) = 0
-
- Next, Monty reveals not the outcome of the coin flip, but a door with a
- goat behind it. Suppose he opens Door Number 2.
-
- P(GPi & Cj & Fk | C1 & D2)
- = P(GPi & Cj & Fk & C1 & D2) / P(C1 & D2)
-
- so we just need to compute P(C1 & D2). Now
-
- GP1 & C1 & F1 => D2
- GP1 & C1 & F2 => D3
- GP1 & C2 & F1 => D3
- GP1 & C2 & F2 => D3
- GP1 & C3 & F1 => D2
- GP1 & C3 & F2 => D2
- GP2 & C1 & F1 => D3
- GP2 & C1 & F2 => D3
- GP2 & C2 & F1 => D1
- GP2 & C2 & F2 => D3
- GP2 & C3 & F1 => D1
- GP2 & C3 & F2 => D1
- GP3 & C1 & F1 => D2
- GP3 & C1 & F2 => D2
- GP3 & C2 & F1 => D1
- GP3 & C2 & F2 => D1
- GP3 & C3 & F1 => D1
- GP3 & C3 & F2 => D2
-
- C1 & D2 are true in cases 1, 13, and 14 only, so P(C1 & D2) = 3/18.
- Therefore,
-
- P(GP1 & C1 & F1 | C1 & D2) = (1/18)/(3/18) = 1/3
- P(GP1 & C1 & F2 | C1 & D2) = 0/(3/18)
- P(GP1 & C2 & F1 | C1 & D2) = 0/(3/18)
- P(GP1 & C2 & F2 | C1 & D2) = 0/(3/18)
- P(GP1 & C3 & F1 | C1 & D2) = 0/(3/18)
- P(GP1 & C3 & F2 | C1 & D2) = 0/(3/18)
- P(GP2 & C1 & F1 | C1 & D2) = 0/(3/18)
- P(GP2 & C1 & F2 | C1 & D2) = 0/(3/18)
- P(GP2 & C2 & F1 | C1 & D2) = 0/(3/18)
- P(GP2 & C2 & F2 | C1 & D2) = 0/(3/18)
- P(GP2 & C3 & F1 | C1 & D2) = 0/(3/18)
- P(GP2 & C3 & F2 | C1 & D2) = 0/(3/18)
- P(GP3 & C1 & F1 | C1 & D2) = (1/18)/(3/18) = 1/3
- P(GP3 & C1 & F2 | C1 & D2) = (1/18)/(3/18) = 1/3
- P(GP3 & C2 & F1 | C1 & D2) = 0/(3/18)
- P(GP3 & C2 & F2 | C1 & D2) = 0/(3/18)
- P(GP3 & C3 & F1 | C1 & D2) = 0/(3/18)
- P(GP3 & C3 & F2 | C1 & D2) = 0/(3/18)
-
- So, should you switch? There is one chance in three that you have chosen
- the grand prize and should not switch; there are two chances in three
- that that you have not chosen the grand prize and should switch.
- Therefore, you should switch. The bottom line comes out the same for
- the other possibilities -- C1 and D3, C2 and D1, C2 and D3, C3 and D1,
- and C3 and D2 -- so, no matter what happens, you should switch. Q.E.D.
-
- The argument that there's no point in switching -- after choosing, you
- know the next thing that happens is that Monty will open a door to reveal
- a goat, so his doing so provides no reason to switch -- is compelling.
- (But I'll be happy to play, say, 10000 rounds of this game against your
- friend, with me using the "always switch" strategy, him/her using the
- "always stand" strategy, and me making her/his pots 10% larger than
- (s)he makes mine, any time, for any stakes (s)he'd care to name!!)
-
- -- rar
-
-