Sowhat I'm askingmyselfis, howmanystickerswill I needtobuyuntil I getallofActually, thisissomethingthatpeoplehavetriedtodobefore.
Themostrecentstorythatcameouttalkedabouthowitwouldcostabout $1000 tofillthealbum, and I seewheretheygetthat.
But I want I wanttogiveus a differentwayofthinkingaboutthis, becausetheproblemisthatthatestimateover $1000 assumesthatyou'rejustbuyingandbuyingandbuyingandbuyingstickers.
Let's saytheyweretryingtofindhisnumberandwhichisthenumberofstickers I expecttobuybefore I geteverystickeratleastonce.
Thisisthelonelystickercollectorversion.
I mean, I'm I'm inthe u.
S.
I havenobodytotradewith.
Andso I'm justgoingtoimaginethat I'm that I'm buyingandbuyingstickersuntil I getLetmetrytodefineAnd I Toby, howmanystickersittookmetogetbetweentheIastnewstickeronthat I plusonemistake.
Okay, So, forexample, andzerowillbehowmanystickersUntil I getthefirst, Nooneandonewillbehowmanydoesn't get a second.
Soyoucanimagine, forexample, whatisand 200.
Thatmeansthat I'vebought a lotofpacks.
I have 200 theplayers, andnow I reallywanttogetthenextone I'm lookingforforsomenewonethat I don't haveinmyalbum.
Andthen I keep I keepbuyingandbuyingandbuyingstickersuntil I getonethat I don't haveyetandhowevermanystickersittakesme, I'm goingtocallthatnumberand I thetotalnumberofstickersisgoingtobebasically, youknow, howlongdidittakemetogetthefirstnewone?
Plus, howlongdidittakemetogetthesecondnewoneandsoonuntil I getthelastnewone?
Thisishowlongit's goingtotakemetofindtheverylast.
Andsoif I wanttofindthisnumberandhowmanystickers I needtobuy, what I'm goingtodois I'm goingtoactuallytrytofindeachoneofthesenumbers.
Andsowhat I wanttodonowisGeneralGeneralEyesanddothisforanyoneofthesenumbers.
And I thinkthekeythingthatweshouldrealizehereiswheredidthis 482 comefrom?
Well, yousawwhathappened.
Itwas 682 minus 200.
And I, whichishowmanystickers I needtobuybetweentheicenewstickerand I placefirstnewstickerisgoingtobe 682 dividedby 682 minus I.
So, if I combinethesetwo, what I'm goingtogetisthatinitsequalto 682 overseasattorneyto 680 toover 681 682 over 680.
Youcanseewhenyouopenyourfirstpack, you'redefinitelygoingtoget a stickyYoudidn't haveexactly.
Sothisis, thisisthenumberendsupzero.
Howlongisitgonnatakemetogetthefirstnewstickerforsure, Nomatterwhatthe 1st 1 I getisthat's gonnajustgonnacomerightaway.
Andasyougoon, itgetsharderandharderandharderandhardertoget a newwhatisthelastone.
So I get 692 dividedby 6 32 minutes 6 81 whichisequaltoonethatmakessenseaswell, becausewhenyouwinthisonestickinginaid, you'vegot a onein 682 chanceofgettingandit's sohardtogetit.
And I thinkit's worthdoublingonthispoint a littlebit.
Here I getoneplus X plus X squaredtosay, exceptforone X squaredoveronetimestwo x cubedoveronetimes, twotimesthreeandsoonuntilthenumberoffriendsminusonetakeallofthatand I dividedbyeithertheex.
I warnedyou, Thisismessy.
Nowyoutake a parenthesishereandyoutakethiswholethingandraiseittotheend, whichisthenumberofstickers, andthenyoucloseparenthesisandthensenatorintegralwithvariable X.
Sointegralhere, Dex.
It's a monster.
It's a monster.
I thinkit's beautiful, butyouhavetohave a tasteforthesekindsofthings.
When I thinkit's beautiful, isnotthat I seethisand I thinkit's reallybeautiful.
What I thinkisbeautifulisthat I justneedtopluginandforthenumberofstickers f forthenumberofpeople.
And I'm goingtogettheanswerforhowmanystickersperpersonweexpecttobe.
Howmuchof a savingisittobe a stickerswapper?
Sowhatweneedtodoisfigureouthowtopluginthisthing.
Onething I willsayisthatifthenumberendandthenumber F areprettysmall, thenthisexpressionisactuallynottoobad.
Andifyou'vetaken a courseinintegralcalculus, youwillhavethetoolstocomputethisforan f small.
Buttheproblemisthatwe'relookingat N isequalto 682.
Wewanttodothisfor a burialvaluesof F.
Andsoreally, we'regonnahavetoask a computerforhelphere.
Andsowe'rejustgoingtopluginto a computer.
And I didthisaheadoftimeand I'llshowyousomeoftheresults.
So I gotthewonderfulhelpofmystudent, BibianaMarcusand I wenttogivehercreditbecauseshe's muchbetterwithcomputersandme.
Sothat's lookingmuchbetter, I think.
What's thisheadingtowards, Youknow, if I liveinColombia, whereeveryone's obsessedwiththesestickersandhas a populationinthemillionslikehowlowcanthisguyjustputhimoutanywhere, S SoSothere's somethingcalledthelawofLargeNumberswhichsaysthatbasically, onceweoncewehaveenoughpeople, thenit's gonnacosteachofusessentiallythecostofanalbum 682.
Soif I multiplythatby 20 cents, I get 136.
Ofcourse, itcanbelowerthanthatbecause I needtobuyallthestickers.
Uh, andsoyouknow, basically, however, whateverpriceyou'rewillingtopay, that's kindofhowyouneedtofigureouthowmuchyouneedtotrade.
And I thinkthisis a goodargumentforreallyusingthepaniniasthesocialexperiencethatitisandactuallygoingoutandseekingourfriends, andandthemorethatyoudowiththelowerthepricecanbe.
Anditcouldbedowntoprettyclosetothat 136 1 thingthat I wantedtodoisthat I don't knowaboutyou, Brady, but I findthisformulaabsolutelyamazing.
Youknow, I satdownand I understoodit, and I readtheproofonit.
Itmakessensetome, andyet I havetroublebelievingit.
It's somagical.
Andsoonethingthat I didisthat I reachedouttobeViennamarketsagain, and I askedher, Hey, you'reamazingwithprogramming, canyoujustrunsomesimulationsthatwilltake 10 friendsandwe'llbuystickersforTemperanceandfigureouthowmanystickerstheyspentandrunit 100 timessothatwecangetsomekindofaverageofoffwhereitwillbe, andso I'llshowyoutheVienna's numbers.
I thinkthat's good, thatnotonlydowehave a proofofthisthatweunderstand, I thinksometimesit's goodtojustrunsomeexperimentstomakesurethattheformulaalsoholdsupinreality.
Anditreallydoes, justlike I don't knowwhichstickersairinsidethispackonmyopenthisin a minute.
I alsoneverknowwhattheproblemoftheweekisgonnabeover a brilliantdog, Solet's have a looktoday.