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, –CV versionCV version –Published version 1 and 212 Bovens, L. and Ferreira, J.L. “Monty Hall drives a wedge between Judy Benjamin and the Sleeping Beauty: a reply to Bovens” Analysis (forthcoming)Monty Hall drives a wedge between Judy Benjamin and the Sleeping Beauty: a reply to Bovens

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, Sleeping Beauty Problem –Elga, A Self-locating belief and the Sleeping Beauty problem. Analysis 60: 143– 47.

‘If you are in R, then P(S)=p’ (for p = 0) BR Q¼¼ S¼¼ BR Q?? S?0

SB’: ‘If He, then P(Tu)=0’ TaHe Mo¼¼ Tu¼¼ TaHe 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 BR Q¼½ S¼0 TaHe Mo¼½ Tu¼0 Justification: Bradley’s Adams conditioning: Learning a conditional should not affect your credence for its antecedent.

1/3’ers TaHe Mo1/3 Tu1/30 BR Q S 0

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 TaHe Mo?1/2 Tu1/20

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 BR 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.

Monty Hall-1 = C in XC in YC in Z Info =“D in Y”½01 “D in Z”½10

Monty Hall - = C in XC in YC in Z Info =“D in Y” ½01 “D in Z” ½10 | Info = DY) =

Monty Hall - 2 = C in XC in YC in Z Info =“D in Y”½0½ “C in Y”0½0 “D in Z”½½0 “C in Z”00½

Monty Hall - = C in XC in YC in Z Info =“D in Y” ½0½ “C in Y” 0½0 “D in Z” ½½0 “C in Z” 00½ | Info =DY) =

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 NWSWNESE Info=Info= “E→¬N”½½01 “E→¬S”½½10

Judy Benjamin = JB is in NWNW SWSW NENE SESE Info=Info= “E→¬N” ½½01 “E→¬S” ½½10 | Info=“E→¬S”) = 2 × | Info=“E→¬S”) =

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 NWSWNESE Info=Info= “¬NW”01/3 “¬SW”1/30 “¬NE”1/3 0 “¬SE”1/3 0

Sleeping Beauty” = SB is in NWSWNESE Info=Info= “¬NW”01/3 “¬SW”1/30 “¬NE”1/3 0 “¬SE”1/3 0 / INFO=¬SE) = 2 × / INFO=¬SE) =

Judy Benjamin The informer checks SE and reports to JB whether she is there or not. –{“SE”, “¬SE”} | Info = “¬SE”) = 2/3 The informer checks all quadrants and informs JB of one quadrant where she is not. –{“¬NE”, “¬SE”, “¬NW”, “¬SW”} | 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”} – | Info = “E→¬S”) = 1/2

Judy Benjamin = SB is in NWSWNESE Info=Info= “¬NW”01/3 “¬SW”1/30 “¬NE”1/3 0 “¬SE”1/3 0 / INFO=¬SE) = 2 × / INFO=¬SE) =

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”} | 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”} – | Info = “E→¬S”) = 1/2

Judy Benjamin’ = JB is in NWNW SWSW NENE SESE Info=Info= “SE” 0001 “E→¬S” 1110 | Info=“E→¬S”) = 2 × | Info=“E→¬S”) =

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: = 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. = Ta | Info = “¬He-Tu”) = 2/3

Sleeping Beauty = JB is in NWNW SWSW NENE SESE Info=Info= “E→¬N” ½½01 “E→¬S” ½½10 | Info=“E→¬S”) = 2 × | Info=“E→¬S”) =

Igor Douven and Jan-Willem Romeijn ‘A New Resolution to the Judy Benjamin Problem’

‘If it Rains, then no Sd’ ¬RainRain ¬Sd¼¼ Sd¼¼ ¬RainRain ¬Sd¼½ Sd¼0

‘If he robbed him, then he shot him’ ¬RoRo Sh.1 ¬Sh.1.7 ¬RoRo Sh.1 ¬Sh.80

‘If he robbed him, then he shot him’ ¬RoRo Sh¼¼ ¬Sh¼¼ ¬RoRo Sh¼¼ ¬Sh½0

Modus Ponens ¬RainRain ¬Sd¼½ Sd¼0 ‘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 ¬RoRo Sh¼¼ ¬Sh½0 ‘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/HeR/Ta Q/ Mo ¼¼ +1/8 S/ Tu ¼ +1/8 0 B/HeR/Ta Q/ Mo 1/3 S/ Tu 1/30

Questions Epistemic entrenchment and probabilism? –Beta-functions? Epistemic entrenchment in the JB? Epistemic entrenchment in the SB? If not, then what?