judy benjamin is a sleeping beauty, modulo monty hall
DESCRIPTION
Judy Benjamin is a Sleeping Beauty, modulo Monty Hall. Luc Bovens Progic Conference September 2009. Written work. Bovens, L. “Judy Benjamin is a Sleeping Beauty” Analysis , 2010. CV version Published version 1 and 2 - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/1.jpg)
Judy Benjamin is a Sleeping Beauty, modulo Monty Hall
Luc Bovens
Progic Conference
September 2009
![Page 2: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/2.jpg)
Written work
• Bovens, L. “Judy Benjamin is a Sleeping Beauty” Analysis, 2010. – CV version– Published version 1 and 2
• Bovens, L. and Ferreira, J.L. “Monty Hall drives a wedge between Judy Benjamin and the Sleeping Beauty: a reply to Bovens” Analysis (forthcoming)
![Page 3: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/3.jpg)
JB and SB
• Judy Benjamin Problem– Van Fraassen, B. 1981, A Problem for
Relative Information Minimizers in Probability Kinematics. British Journal for the Philosophy of Science, 32, 4, 375-9.
• Sleeping Beauty Problem– Elga, A. 2000. Self-locating belief and the
Sleeping Beauty problem. Analysis 60: 143–47.
![Page 4: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/4.jpg)
‘If you are in R, then P(S)=p’(for p = 0)
B R
Q ¼ ¼
S ¼ ¼
B R
Q ? ?
S ? 0
![Page 5: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/5.jpg)
SB’: ‘If He, then P(Tu)=0’
Ta He
Mo ¼ ¼
Tu ¼ ¼
Ta He
Mo ? ?
Tu ? 0
![Page 6: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/6.jpg)
SB ½’ers
• Beauty did not learn anything new upon awakening, so why should she change her beliefs?– Response: She did learn something new in
SB’!
![Page 7: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/7.jpg)
½’ers
B R
Q ¼ ½
S ¼ 0
Ta He
Mo ¼ ½
Tu ¼ 0
Justification: Bradley’s Adams conditioning: Learning a conditional should not affect your credence for its antecedent.
![Page 8: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/8.jpg)
1/3’ers
Ta He
Mo 1/3 1/3
Tu 1/3 0
B R
Q 1/3 1/3
S 1/3 0
3/14/3
2/12/1
))&((
)()/)&(())&(/(
SRP
RPRSRPSRRP
![Page 9: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/9.jpg)
RAA of Adams Conditioning
• What did SB’ learn?– If Heads, then ¬Tu– If Tu, then ¬Heads
• So by Adams Conditioning: – ? => 0– So if Beauty is told in
addition that Tails came up, then she should infer that Tu. Absurd
Ta He
Mo ? 1/2
Tu 1/2 0
![Page 10: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/10.jpg)
Symmetry
• P(B) = P(Q)• P(Q|R) = 1 • P(Q|B) = 1/2
• P(Q) = P(Q|B)P(B) + P(Q|R)(1-P(B))
=>P(B) = 2/3
B R
Q ? ?
S ? 0
![Page 11: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/11.jpg)
‘Judy Benjamin is a Monty Hall.’(Jose Luis Ferreira)
=> joint work with J.L. Ferreira
![Page 12: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/12.jpg)
Monty Hall• Three doors. One door has a car, two doors have donkeys behind them.
You pick door x. Monty Hall opens up door y and a donkey walks out. What is the probability that the door is behind door x?
• P(CX | DY) =
• “The change represented by [the formula for conditional probability] is defensible and justifiable only on the basis of the protocol that tells circumstances under which the [new information] will be acquired.” (Shafer, G. “Conditional Probability”, 1985)
• Two protocols:
1. Monty Hall opens up one of the two remaining doors and this door is bound to have a donkey behind it.
2. Monty Hall opens up one of the two remaining doors – it may or may not have a donkey behind it.
2/13/2
3/11
)(
)()/(
DYP
CXPCXDYP
![Page 13: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/13.jpg)
Monty Hall-1
P(Info/@) @ =
C in X C in Y C in Z
Info = “D in Y” ½ 0 1
“D in Z” ½ 1 0
![Page 14: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/14.jpg)
Monty Hall - 1
P(Info/@) @ =
C in X C in Y C in Z
Info = “D in Y”
½ 0 1
“D in Z”
½ 1 0
P(@=CX | Info = DY) =
3/1)2/101(3/1
3/12/1
)(
)(@)/@(
DYInfoP
CXPCXDYInfoP
![Page 15: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/15.jpg)
Monty Hall - 2
P(Info/@) @ =
C in X C in Y C in Z
Info = “D in Y” ½ 0 ½
“C in Y” 0 ½ 0
“D in Z” ½ ½ 0
“C in Z” 0 0 ½
![Page 16: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/16.jpg)
Monty Hall - 2
P(Info/@) @ =
C in X C in Y C in Z
Info = “D in Y”
½ 0 ½
“C in Y”
0 ½ 0
“D in Z”
½ ½ 0
“C in Z”
0 0 ½
P(@=CX | Info =DY) =
2/1)2/102/1(3/1
3/12/1
)(
)(@)/@DY(
DYInfoP
CXPCXInfoP
![Page 17: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/17.jpg)
Judy Benjamin
• The informer checks SE and reports to JB whether she is there or not.– {“SE”, “¬SE”}
• The informer checks all quadrants and informs JB of one quadrant where she is not—suppose that the machine can only provide one true negative.– {“¬NE”, “¬SE”, “¬NW”, “¬SW”}
• The informer checks East and informs JB of one quadrant where she is not—suppose that the machine can only provide one true negative.– {“E→¬N”, “E →¬S”}
![Page 18: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/18.jpg)
Judy Benjamin
P(Info | @) @ = JB is in
NW SW NE SE
I
n
f
o=
“E→¬N” ½ ½ 0 1
“E→¬S” ½ ½ 1 0
![Page 19: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/19.jpg)
Judy Benjamin
P(Info | @) @ = JB is in
NW
SW
NE
SE
I
n
f
o=
“E→¬N” ½ ½ 0 1
“E→¬S” ½ ½ 1 0
P(@=W | Info=“E→¬S”) =
2 × P(@=NW | Info=“E→¬S”) =
2/1)102/12/1(4/1
4/12/12
)(
)(@)/@(2
SEInfoP
NWPNWSEInfoP
![Page 20: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/20.jpg)
Sleeping Beauty’’
• SB is put to sleep on Su knowing– A fair coin has been tossed– There will be awakenings on Mo or Tu– One combination is ruled out, but we do not
know which– Amnesia is induced after each awakening of
all the info gained over and above the Su info
• After the awakening, info is provided which combination is ruled out.
![Page 21: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/21.jpg)
Sleeping Beauty’’
P(Info | @) @ = SB is in
NW SW NE SE
I
n
f
o=
“¬NW” 0 1/3 1/3 1/3
“¬SW” 1/3 0 1/3 1/3
“¬NE” 1/3 1/3 0 1/3
“¬SE” 1/3 1/3 1/3 0
![Page 22: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/22.jpg)
Sleeping Beauty”
P(Info | @) @ = SB is in
NW SW NE SE
I
n
f
o
=
“¬NW” 0 1/3 1/3 1/3
“¬SW” 1/3 0 1/3 1/3
“¬NE” 1/3 1/3 0 1/3
“¬SE” 1/3 1/3 1/3 0
P(@=W / INFO=¬SE) =
2 × P(@=NW / INFO=¬SE) =
3/2)3/103/13/1(4/1
4/13/12
)(
)(@)/@(2
SEInfoP
NWPNWSEInfoP
![Page 23: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/23.jpg)
Judy Benjamin• The informer checks SE and reports to JB whether she is there or
not.– {“SE”, “¬SE”} – P(@=W | Info = “¬SE”) = 2/3
• The informer checks all quadrants and informs JB of one quadrant where she is not.– {“¬NE”, “¬SE”, “¬NW”, “¬SW”} – P(@=W | Info = “¬SE”) = 2/3
• The informer checks East and informs JB of one quadrant where she is not—suppose that the machine can only provide one true negative– {“E→¬N”, “E →¬S”}– P(@=W | Info = “E→¬S”) = 1/2
![Page 24: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/24.jpg)
Judy Benjamin
P(Info | @) @ = SB is in
NW SW NE SE
I
n
f
o
=
“¬NW” 0 1/3 1/3 1/3
“¬SW” 1/3 0 1/3 1/3
“¬NE” 1/3 1/3 0 1/3
“¬SE” 1/3 1/3 1/3 0
P(@=W / INFO=¬SE) =
2 × P(@=NW / INFO=¬SE) =
3/2)3/103/13/1(4/1
4/13/12
)(
)(@)/@(2
SEInfoP
NWPNWSEInfoP
![Page 25: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/25.jpg)
Judy Benjamin
• The informer intends to check East, is unable to check NE due to the cloud cover, does check SE and reports to JB whether she is there or not.– {“SE”, “¬SE”} – {“SE”, “E →¬S”}– P(@=W | Info = “E →¬S”) = 2/3
• The informer checks East and informs JB of one quadrant where she is not—suppose that the machine can only provide one true negative– {“E→¬N”, “E →¬S”}– P(@=W | Info = “E→¬S”) = 1/2
![Page 26: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/26.jpg)
Judy Benjamin’
P(Info | @) @ = JB is in
NW
SW
NE
SE
I
n
f
o=
“SE” 0 0 0 1
“E→¬S” 1 1 1 0
P(@=W | Info=“E→¬S”) =
2 × P(@=NW | Info=“E→¬S”) =
3/2)0111(4/1
4/112
)(
)(@)/@(2
SEInfoP
NWPNWSEInfoP
![Page 27: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/27.jpg)
Sleeping Beauty
• Beauty is told on Su that a fair coin is flipped and that there are awakenings on Mo and Tu. Upon each awakening, she will be told some Heads quadrant in which she is not: – P(@ = Ta | Info = “¬He-Tu) = 1/2
• Upon awakening, the informer reveals to Beauty the structure of the game, viz. that the awakening may be taking place in any world-time quadrant, except for the quadrant Heads-Tu.– P(@ = Ta | Info = “¬He-Tu”) = 2/3
![Page 28: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/28.jpg)
Sleeping Beauty
P(Info | @) @ = JB is in
NW
SW
NE
SE
I
n
f
o=
“E→¬N” ½ ½ 0 1
“E→¬S” ½ ½ 1 0
P(@=W | Info=“E→¬S”) =
2 × P(@=NW | Info=“E→¬S”) =
2/1)102/12/1(4/1
4/12/12
)(
)(@)/@(2
SEInfoP
NWPNWSEInfoP
![Page 29: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/29.jpg)
Igor Douven and Jan-Willem Romeijn
‘A New Resolution to the Judy Benjamin Problem’
![Page 30: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/30.jpg)
‘If it Rains, then no Sd’
¬Rain Rain
¬Sd ¼ ¼
Sd ¼ ¼
¬Rain Rain
¬Sd ¼ ½
Sd ¼ 0
![Page 31: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/31.jpg)
‘If he robbed him, then he shot him’
¬Ro Ro
Sh .1 .1
¬Sh .1 .7
¬Ro Ro
Sh .1 .1
¬Sh .8 0
![Page 32: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/32.jpg)
‘If he robbed him, then he shot him’
¬Ro Ro
Sh ¼ ¼
¬Sh ¼ ¼
¬Ro Ro
Sh ¼ ¼
¬Sh ½ 0
![Page 33: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/33.jpg)
Modus Ponens
¬Rain Rain
¬Sd ¼ ½
Sd ¼ 0
•‘If it rains, then no sundowners’
•Keep Probability of rain fixed at ½
•Adjust probability of no Sundowners from ½ to ¾
•Probability of rain is more deeply epistemically entrenched than probability of Sundowners
![Page 34: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/34.jpg)
Modus Tollens
¬Ro Ro
Sh ¼ ¼
¬Sh ½ 0
•‘If he robbed him, then he shot him’
•Keep Probability of not-Shot fixed at ½
•Adjust probability of Robbed from ½ to 1/4
•Probability of not-Shot is more deeply epistemically entrenched than probability of Robbed
![Page 35: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/35.jpg)
Equal Epistemic Entrenchment?
B/He R/Ta
Q/
Mo
¼ ¼
+1/8
S/
Tu
¼ +1/8
0
B/He R/Ta
Q/
Mo
1/3 1/3
S/
Tu
1/3 0
![Page 36: Judy Benjamin is a Sleeping Beauty, modulo Monty Hall](https://reader035.vdocuments.mx/reader035/viewer/2022062518/56813f8f550346895daa7ede/html5/thumbnails/36.jpg)
Questions
• Epistemic entrenchment and probabilism?– Beta-functions?
• Epistemic entrenchment in the JB?
• Epistemic entrenchment in the SB?
• If not, then what?