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

• Bovens, L. and Ferreira, J.L. “Monty Hall drives a wedge between Judy Benjamin and the Sleeping Beauty: a reply to Bovens” Analysis (forthcoming)

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.

‘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

SB’: ‘If He, then P(Tu)=0’

Ta He

Mo ¼ ¼

Tu ¼ ¼

Ta He

Mo ? ?

Tu ? 0

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’!

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.

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

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

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

‘Judy Benjamin is a Monty Hall.’(Jose Luis Ferreira)

=> joint work with J.L. Ferreira

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

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

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

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

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

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”}

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

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

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.

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

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

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

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

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

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

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

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

Igor Douven and Jan-Willem Romeijn

‘A New Resolution to the Judy Benjamin Problem’

‘If it Rains, then no Sd’

¬Rain Rain

¬Sd ¼ ¼

Sd ¼ ¼

¬Rain Rain

¬Sd ¼ ½

Sd ¼ 0

‘If he robbed him, then he shot him’

¬Ro Ro

Sh .1 .1

¬Sh .1 .7

¬Ro Ro

Sh .1 .1

¬Sh .8 0

‘If he robbed him, then he shot him’

¬Ro Ro

Sh ¼ ¼

¬Sh ¼ ¼

¬Ro Ro

Sh ¼ ¼

¬Sh ½ 0

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

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

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

Questions

• Epistemic entrenchment and probabilism?– Beta-functions?

• Epistemic entrenchment in the JB?

• Epistemic entrenchment in the SB?

• If not, then what?