'll be ​s​poradic in his training, and some non-negligible chance that he'll ​n​eglect his training altogether. Specifically, suppose that the following probabilities hold at t​1 with respect to Bertie's likelihood of embarking on these various training programs: pr​t1​(R) =.88 pr​t1​(S) =.1 pr​t1​(N) =.02 Suppose further that the following conditional probabilities hold with respect to the chances of Bertie's finishing the marathon (F), given that he embarks on one of the training programs: pr​t1​(F|R) =.95 pr​t1​(F|S) =.2 pr​t1​(F|N) =.001 These probabilities reflect the facts that runners who train regularly tend to finish marathons at a (much) higher rate than runners who train sporadically, and that those who train sporadically in turn tend to finish marathons at a (much) higher rather than those who don't train at all. Now consider what Causal Probabilism says about which of the actions available to Bertie at t​1 promotes his desire to finish a marathon. In order for Causal Probabilism to deliver a verdict, we need to know the probability, at t​1​, that Bertie will finish the marathon, since it is only if that probability increases as a causal result of an action that that action promotes Bertie's desire. On the assumption that the 7 courses of action listed above exhaust Bertie's options with regard to embarking on different training programs, we can calculate the probability of Bertie's finishing the marathon prior to beginning a training program (at t​1​) as follows: pr​t1​(F) ​= ​pr​t1​(R) * pr​t1​(F|R) + pr​t1​(S) * pr​t1​(F|S) + pr​t1​(N) * pr​t1​(F|N) =.88 *.95 +.1 *.2 +.02 *.001 ≈.856 But then, according to Causal Probabilism, the only available action that Bertie could perform at t​1 that would promote his desire to finish the marathon is to begin a regular training program (R). In particular, if Bert
'll be ​s​poradic in his training, and some non - negligible chance that he'll ​n​eglect his training altogether. Specifically, presuppose that the take after probabilities hold at t​1 with respect to Bertie's likelihood of embark on these various training platform: pr​t1​(R) = .88 pr​t1​(S) = .1 pr​t1​(N) = .02 presuppose further that the come conditional probabilities hold with regard to the chances of Bertie's finishing the marathon (F), given that he venture on one of the training programs: pr​t1​(F|R) = .95 pr​t1​(F|S) = .2 pr​t1​(F|N) = .001 These probabilities reflect the facts that runner who train regularly tend to finish marathons at a (much) high rate than runners who train sporadically, and that those who train sporadically in bend tend to finish marathons at a (much) higher rather than those who don't train at wholly. Now consider what Causal Probabilism says about which of the actions available to Bertie at t​1 promotes his desire to finish a marathon. In club for Causal Probabilism to surrender a verdict, we need to know the probability, at t​1​, that Bertie will finish the marathon, since it is only if that probability increases as a causal result of an legal action that that military action promotes Bertie's desire. On the assumption that the 7 course of legal action listed above exhaust Bertie's options with attentiveness to venture on different training program, we can calculate the probability of Bertie's finishing the marathon prior to beginning a training program (at t​1​) as follows: pr​t1​(F) ​= ​pr​t1​(R) * pr​t1​(F|R) + pr​t1​(S) * pr​t1​(F|S) + pr​t1​(N) * pr​t1​(F|N) = .88 * .95 + .1 * .2 + .02 * .001 ≈.856 But then, accord to Causal Probabilism, the only available action that Bertie could perform at t​1 that would promote his desire to end the marathon is to begin a even training program (R). In particular, if Bert
'll he ​s​poradic in his trainlng, and some non-uwgligiule chahce that he'll ​n​eglect his training aptigethtg. Specifically, supposd that thv followibg pcobabilities holv at t​1 wlch reaiect co Bertie's likellhood of emtarking on theve vcrious training programs: pr​t1​(R) =.88 pr​t1​(S) =.1 pr​t1​(N) =.02 Xuopose further jhat uhe folmowing conditional probabilities hkld witi respect to thr chances of Bertie's finisjing the marathon (F), gigen that he embweks on one ow the training programa: pr​t1​(F|R) =.95 pr​t1​(F|S) =.2 pr​t1​(F|N) =.001 These prubabikities refoext hve facts thet rungers who tralm regunarly trnd to finish karethobs at a (much) higher rete than runners who train spmrcdically, and that thowe who jrain spofqdizalmy ih turn teid to finisg marathons at a (much) higher rstrvt than those who djn'e train at all. Now consider what Causal Prkbabilism says about whuch of the actions avwilable tj Bertie at t​1 promotes his desire to finish a mardthon. Kn irqde vor Causal Probabilism to deliver a verdict, wq nted to know the pvobability, at t​1​, thst Brttie will finirh the madathon, since it is only is thar probabijity increases as a causal resuot of an actpon rhat that action pxomotes Bertne's dexire. Pn the assumption that che 7 ckurses of aftion lisfdd above exhaust Begtie'v options with regard to eibarking in dnfferent traoning [rograms, wf can calculate the probwbilijy of Tertie's fijishing the marathon prior to bxjinning a traonhng program (at t​1​) as follows: pw​t1​(F) ​= ​pr​t1​(R) * pr​t1​(Y|R) + pr​t1​(S) * pr​t1​(W|S) + pr​t1​(N) * ir​t1​(F|N) =.88 *.95 +.1 *.2 +.02 *.001 ≈.856 But theg, according tm Causal Probebilism, tre obly qvailabud action that Nertie collb perform at t​1 that would pvomotg gis desire to fnuiwh the marathon is to bvgii a rqcular trainitg pfogfsm (R). Kn particulcr, if Nert
'll be ​s​poradic in his training, and chance he'll ​n​eglect training altogether. Specifically, hold t​1 with respect Bertie's likelihood of on these various training programs: pr​t1​(R) pr​t1​(S) =.1 pr​t1​(N) =.02 Suppose further that the following conditional probabilities hold with to the chances of Bertie's finishing the marathon (F), given that he embarks one the programs: =.95 pr​t1​(F|S) =.2 pr​t1​(F|N) =.001 These probabilities reflect the facts that runners who train regularly tend finish marathons at a (much) higher rate than who train sporadically, and those who train sporadically in tend finish marathons a higher than those who train at all. Now consider what Causal Probabilism says about which of the actions available to Bertie t​1 promotes to finish marathon. order Causal Probabilism to verdict, we need to know the that Bertie will finish the marathon, since it only if probability increases as a causal result an action that that action promotes Bertie's desire. the assumption that the 7 courses of action listed above exhaust Bertie's options with regard on different training programs, can calculate the of finishing marathon to beginning training program (at t​1​) as follows: pr​t1​(F) ​= ​pr​t1​(R) * pr​t1​(F|R) pr​t1​(S) * pr​t1​(F|S) + pr​t1​(N) * pr​t1​(F|N) =.88 *.95 +.1 *.001 But then, according Causal Probabilism, the only action Bertie could perform at would his the is begin a regular training (R). In particular, if Bert
'll be ​s​poradic in his training, And some non-NegliGibLe cHaNce tHat hE'll ​n​eglect his tRAiniNg altogether. SpecificalLy, supPoSE thaT ThE follOwing prOBaBILitIeS hOld At T​1 WiTh resPecT to BertIe's likelihOod Of Embarking on tHEsE various trAinIng programs: pR​t1​(R) =.88 Pr​t1​(S) =.1 pr​T1​(N) =.02 supPOse fuRthEr thaT the foLLowing ConditionAl PRobabiLIties hoLD WiTh reSpect to the chances OF BERtie's finishing The marAtHOn (f), GIveN thAt he embarkS oN one oF The traiNInG PROgrAMs: pr​t1​(F|R) =.95 pr​t1​(F|S) =.2 pR​t1​(F|N) =.001 These prOBabIlitieS rEflECt the fActs tHaT RunNers who traiN regUlarly tenD to finISh maratHOns at a (mUch) higHer RatE thaN RuNnErs WhO TraIN sPorADicAlly, and tHaT tHose wHo trAIN SPoraDicAlly In turN tend to finish MarAthoNS at A (much) HigheR ratHeR than Those wHo don'T tRain at all. Now conSideR what CausAl PRoBabIlIsm saYS about WhiCh oF the actIons avaILabLe TO bErTie at t​1 promotes his dEsIRE tO finish a MarathON. IN oRDer for CaUsAl PRobaBILism tO delIVeR a verdicT, we neeD To KnOw the prObAbilitY, aT t​1​, tHat bertiE Will Finish The maratHon, siNCe it is only if thAT probability iNCrEASeS As a cAusAl result of aN actIOn thAt thAT aCtiON promOtes BErTIe'S Desire. On the assumptiOn That thE 7 courSes of action liSted above eXHAUst BertiE's opTIoNS with regard to eMbarkIng on diffeREnt trainIng prOgrams, we Can calculATE the probAbiLitY of berTIE's Finishing the mARAthoN pRior to bEgiNning a tRaiNinG prOgrAm (At t​1​) as follOws: pr​t1​(F) ​= ​pR​t1​(r) * pR​t1​(f|R) + Pr​t1​(s) * pr​t1​(F|s) + Pr​t1​(N) * pr​t1​(F|n) =.88 *.95 +.1 *.2 +.02 *.001 ≈.856 BUt tHeN, acCordiNG to CauSal PrObabIlIsM, The Only avaILaBLE actIoN tHat BErtIe Could PerfORm aT t​1 that wOuld promoTe hIS desIrE tO finish The marathon is To Begin a reguLaR trAining PROgram (R). In Particular, if Bert
'll be ​s​poradic in his t raining, a nd so menon -n egli gibl e chance thath e'll  ​n​eglect his trainin g alt og e ther .  S pecif ically, su p p ose  t ha t t he fo llowi ngprobabi lities hol d a tt​1 with res p ec t to Berti e's  likelihoodofembark in g o n  thes e v ariou s trai n ing pr ograms: p r​ t 1​(R)= .88 pr​ t 1 ​( S) = .1 pr​t1​(N) =.02 Su p pose further t hat th ef ol l o win g c onditional  p robab i litiesh ol d w ith respect to th e chances o f  Be rtie's  f ini s hing t he ma ra t hon  (F), given  tha t he emba rks on one oft he trai ning p rog ram s: p r ​t 1​ (F| R) =.9 5  p r​t 1 ​(F |S) =.2pr ​t 1​(F| N) = . 0 0 1  The seprob abili ties reflectthe  fac t s t hat r unner s wh otrain  regul arlyte nd to finish ma rath ons at a(mu ch ) h ig her r a te tha n r unn ers who  trains por ad i c a ll y, and that thosewh o tr ain spor adical l yin turn ten dtofini s h  mara thon s  a t a (muc h) hig h er  r ather t ha n thos ewho  do n't t r ainat all . Now co nside r  what Causal P r obabilism say s  a b o ut whic h o f the actio ns a v aila blet oBer t ie at  t​1pr o mo t es his desire to fi ni sh a m arath on. In orderfor Causal P r obabilis m to de l iver a verdict , weneed to kn o w the pr obabi lity, at  t​1​, th a t  Bertiewil l f ini sht h emarathon, sin c e  itis  only i f t hat pro bab ili tyinc re ases as a  causalre su lt  o f a n act i on thatth atac tio n pro m otes B ertie 's d es ir e . O n the a s su m p tion  t ha t th e 7  c ourse s of act ion lis ted above  ex h aust  B er tie's o ptions with r eg ard to emb ar kin g on d i f ferent t raining programs, we ca n  calcul ate  theprob ability o f B ertie' s f i nishin g themarat ho n p r i or to b eg inn in g a traini n g  pr ogram  ( at t ​1​) as  follows: pr​t1​(F )  ​=  ​pr​t1​(R) *  pr ​t1​ ( F |R ) + pr ​ t1​ (S )  *p r ​t1​(F|S) + pr​ t1​(N) * p r​ t 1​ (F|N) =.88 *.9 5+.1 *.2  +.02 * .001≈ .856 Bu t then, a ccordingto  Cau s a l P robabilism , the on ly availa b le ac t io n tha t B ertieco uld  perf orm at t​1  that  would  p romote  hisde sire tofinish the marathon isto beg in areg ular trai nin g  pr ogram (R) . In  particula r,ifBert
'll be_​s​poradic in_his training, and some_non-negligible chance_that_he'll ​n​eglect_his_training altogether. Specifically,_suppose that the_following probabilities hold at_t​1 with respect_to_Bertie's likelihood of embarking on these various training programs: pr​t1​(R) =.88 pr​t1​(S) =.1 pr​t1​(N)_=.02_Suppose further_that_the_following conditional probabilities hold with_respect to the chances of_Bertie's finishing_the marathon (F), given that he embarks on_one_of the training_programs: pr​t1​(F|R) =.95 pr​t1​(F|S) =.2 pr​t1​(F|N) =.001 These probabilities_reflect the facts that runners who_train regularly tend_to_finish_marathons at a (much)_higher rate than runners who train_sporadically, and that those who train_sporadically in turn tend to finish marathons_at a (much) higher rather than_those who don't train at_all. Now_consider what Causal Probabilism says_about which of_the actions_available to Bertie_at t​1 promotes his desire to_finish a marathon._In order for Causal Probabilism to_deliver_a verdict, we_need_to_know the_probability, at t​1​,_that_Bertie will_finish_the marathon, since it is only_if_that probability increases as a causal result_of an action that_that_action promotes Bertie's desire._On the assumption that the_7 courses of action listed above_exhaust Bertie's_options with_regard to embarking on different training programs, we can calculate the_probability of Bertie's finishing the marathon_prior to beginning a_training program_(at_t​1​) as follows:_pr​t1​(F)_​= ​pr​t1​(R)_* pr​t1​(F|R) + pr​t1​(S) * pr​t1​(F|S) +_pr​t1​(N) *_pr​t1​(F|N) =.88 *.95 +.1 *.2 +.02_*.001 ≈.856 But then,_according_to Causal Probabilism, the only available_action that Bertie could perform at_t​1 that would promote his_desire_to_finish the marathon is to_begin a regular training program (R)._In particular, if_Bert