31:T565,Where this gets really difficult though, is in determining the probability that _**ANY**_ of the flavors will run out of ice cream. **We need to know the probability that the local news runs a smear campaign on this ice cream shop!** The clever reader may suggest: >"Well, isn't the probability of _**ANY**_ flavor running out, simply 100% minus the probability-that-none-run-out, which can be calculated by multiplying each individual wont-run-out probability together?" Unfortunately, the answer here is "No - we can't do that" The reason being, **these selections are not independent**. For example, running out of Chocolate and Vanilla ice-cream provides information to me that it's going to be less likely to have run out of Strawberry, since a certain number of potential-Strawberry-enthusiasts must have chosen Chocolate and Vanilla. There are certainly ways to handle complex non-independent probabilities like this, using combinatorics, enumerating the various combinations and permutations of flavor selections. In fact [Mark McClure](https://observablehq.com/@mcmcclur) has [shown](https://marksmath.org/temp/MattsIceCream.html) that this solution involves the sum of 😱 **84 Million terms** 😱 for 10 flavors, 8-scoop-tubs and 30 visitors. In this case, it turns out that the most practical way to get a "close enough" answer here is to simply simulate it.32:Tf63,function factorial(x){ const a = [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, 6227020800, 87178291200, 1307674368000, 20922789888000, 355687428096000, 6402373705728000, 121645100408832000, 2432902008176640000, 51090942171709440000, 1124000727777607680000, 25852016738884976640000, 620448401733239439360000, 15511210043330985984000000, 403291461126605635584000000, 10888869450418352160768000000, 304888344611713860501504000000, 8841761993739701954543616000000, 265252859812191058636308480000000, 8222838654177922817725562880000000, 263130836933693530167218012160000000, 8683317618811886495518194401280000000, 295232799039604140847618609643520000000, 10333147966386144929666651337523200000000, 371993326789901217467999448150835200000000, 13763753091226345046315979581580902400000000, 523022617466601111760007224100074291200000000, 20397882081197443358640281739902897356800000000, 815915283247897734345611269596115894272000000000, 33452526613163807108170062053440751665152000000000, 1405006117752879898543142606244511569936384000000000, 60415263063373835637355132068513997507264512000000000, 2658271574788448768043625811014615890319638528000000000, 119622220865480194561963161495657715064383733760000000000, 5502622159812088949850305428800254892961651752960000000000, 258623241511168180642964355153611979969197632389120000000000, 12413915592536072670862289047373375038521486354677760000000000, 608281864034267560872252163321295376887552831379210240000000000, 30414093201713378043612608166064768844377641568960512000000000000, 1551118753287382280224243016469303211063259720016986112000000000000, 80658175170943878571660636856403766975289505440883277824000000000000, 4274883284060025564298013753389399649690343788366813724672000000000000, 230843697339241380472092742683027581083278564571807941132288000000000000, 12696403353658275925965100847566516959580321051449436762275840000000000000, 710998587804863451854045647463724949736497978881168458687447040000000000000, 40526919504877216755680601905432322134980384796226602145184481280000000000000, 2350561331282878571829474910515074683828862318181142924420699914240000000000000, 138683118545689835737939019720389406345902876772687432540821294940160000000000000, 8320987112741390144276341183223364380754172606361245952449277696409600000000000000, 507580213877224798800856812176625227226004528988036003099405939480985600000000000000, 31469973260387937525653122354950764088012280797258232192163168247821107200000000000000, 1982608315404440064116146708361898137544773690227268628106279599612729753600000000000000, 126886932185884164103433389335161480802865516174545192198801894375214704230400000000000000, 8247650592082470666723170306785496252186258551345437492922123134388955774976000000000000000, 544344939077443064003729240247842752644293064388798874532860126869671081148416000000000000000, 36471110918188685288249859096605464427167635314049524593701628500267962436943872000000000000000, 2480035542436830599600990418569171581047399201355367672371710738018221445712183296000000000000000, 171122452428141311372468338881272839092270544893520369393648040923257279754140647424000000000000000, 11978571669969891796072783721689098736458938142546425857555362864628009582789845319680000000000000000, 850478588567862317521167644239926010288584608120796235886430763388588680378079017697280000000000000000, 61234458376886086861524070385274672740778091784697328983823014963978384987221689274204160000000000000000, 4470115461512684340891257138125051110076800700282905015819080092370422104067183317016903680000000000000000, 330788544151938641225953028221253782145683251820934971170611926835411235700971565459250872320000000000000000, 24809140811395398091946477116594033660926243886570122837795894512655842677572867409443815424000000000000000000, 1885494701666050254987932260861146558230394535379329335672487982961844043495537923117729972224000000000000000000]; return(a[x]); }33:[] 2f:{"title":"🍨 Probability of Running-Out of Ice Cream","cells":[{"id":0,"value":"# 🍨 Probability of Running-Out of Ice Cream\n\nLet's imagine a scenario where there are ${N_cats} flavors of ice cream in the universe. The local ice cream shop has 1 tub of ice cream for each of the ${N_cats} flavors, and each tub can only serve ${N_overflow-1} guests.\n\nIn this scenario we'll be inviting ${N_samples} people to the store, to order their favorite ice cream, and we'll wonder:\n\nWhat is the probability that **_any_** of the flavors gets ordered ${N_overflow} or more times _(meaning we wont be able to serve them)_? If this happens, the local news will find out and it will become a whole thing.","pinned":false,"hidden":false,"mode":"md"},{"id":29,"value":"viewof N_cats = Inputs.range([3, 14], {value: 10, step: 1, label: \"# of Flavors\"})","pinned":false,"hidden":false,"mode":"ojs"},{"id":30,"value":"viewof N_samples = Inputs.range([10, 50], {value: 30, step: 1, label: \"# of Samples\"})","pinned":false,"hidden":false,"mode":"ojs"},{"id":31,"value":"viewof N_overflow = Inputs.range([2, N_samples], {value: 9, step: 1, label: \"Overflow Level\"})","pinned":false,"hidden":false,"mode":"ojs"},{"id":97,"value":"## 🍦 Not all Flavors are Created Equal\nTo make things interesting, let's allow for people to select their flavor non-uniformly.\nAfter all, some flavors just aren't as good as others.\n\nuse the sliders below to determine \"how good\" each flavor is. \n\n#### arbitrary unitless flavor popularity:","pinned":false,"hidden":false,"mode":"md"},{"id":81,"value":"viewof prefernce_sliders = Inputs.form(d3.range(N_cats).map((x,ii) => Inputs.range([0, 100], {step: 1, value: Math.round(100/(ii+0.2)), label: flavors[ii]})))","pinned":false,"hidden":false,"mode":"ojs"},{"id":109,"value":"_Probability of selection 👇 will then be a normalized version of the above 👆 popularities_","pinned":false,"hidden":false,"mode":"md"},{"id":136,"value":"{\n const mdtable = await dataToMarkdown(p_vals.map((x,ii) => {return({flavor: flavors[ii], probability_selected: Math.round(x*1000)/10+\"%\"})}));\n yield(md`${mdtable}`)\n}","pinned":false,"hidden":false,"mode":"ojs"},{"id":113,"value":"## 🍨 Determining the Probability of Disaster\n\nDetermining the likelihood that each individual flavor will run out is fairly straight-forward. This is a type of problem often called a [Bernoulli Trial](https://en.wikipedia.org/wiki/Bernoulli_trial) and we can use the [Binomial Distribution](https://en.wikipedia.org/wiki/Binomial_distribution) to determine the probability that something with probability `p` will happen at least `k` times given `N` chances.\n\nThe table below lists those probabilities of **_any individual flavor_** getting more than ${N_overflow-1} requests among our ${N_samples} guests.","pinned":false,"hidden":false,"mode":"md"},{"id":163,"value":"{\n const mdtable = await dataToMarkdown(p_vals.map((x,ii) => {return({flavor: flavors[ii], \n probability_selected: Math.round(x*1000)/10+\"%\", \n probability_runout: Math.round(p_each_overflow[ii]*1000)/10+\"%\"})}));\n yield(md`${mdtable}`)\n}","pinned":false,"hidden":false,"mode":"ojs"},{"id":170,"value":"$31","pinned":false,"hidden":false,"mode":"md"},{"id":195,"value":"{\n const mdtable = await dataToMarkdown([{Naive_Independent_Approx: Math.round(p_any_overflow_naive*1000)/10+\"%\", \n Simulation_Result: Math.round(simulation.any_overflows*1000)/10+\"%\"}]);\n yield(md`${mdtable}`)\n}","pinned":false,"hidden":false,"mode":"ojs"},{"id":200,"value":"Note that the values between the Naive Independent Approximation, and our Simulation differ particularly when there isn't a single flavor that dominates the popularity charts.\n\nNow that you've made it to the end of the boring words. Please, go mess around with all the values yourself. 🍦 ","pinned":false,"hidden":false,"mode":"md"},{"id":20,"value":"simulation = {\n let N = 5e3;\n let c = d3.cumsum(p_vals);\n\n // initialize my counts to zero\n let bin_overflows = d3.range(N_cats).map(x => 0);\n let any_overflows = 0;\n\n // perform the very simple simulation\n for(let ii=0; ii d3.randomUniform()()).map(r => N_cats-d3.sum(c.map(cc => r v.length, d => d).map(x => {return({bin: x[0], count: x[1]})})\n let sim_bin_overflow = sim_bin_count_map.filter(x => x.count>=N_overflow);\n if(sim_bin_overflow.length>0){\n any_overflows+=1;\n sim_bin_overflow.forEach(x => bin_overflows[x.bin]+=1)\n }\n }\n \n // return the results\n return({bin_overflows:bin_overflows.map(x => x/N), \n any_overflows:any_overflows/N})\n}","pinned":true,"hidden":false,"mode":"ojs"},{"id":227,"value":"## Supporting Calculations","pinned":false,"hidden":false,"mode":"md"},{"id":17,"value":"p_vals = {\n let s = d3.sum(prefernce_sliders);\n return prefernce_sliders.map(x => x/s)\n}","pinned":false,"hidden":false,"mode":"ojs"},{"id":18,"value":"p_each_overflow = p_vals.map(x => p_at_least_k(x, N_samples, N_overflow))","pinned":false,"hidden":false,"mode":"ojs"},{"id":19,"value":"p_any_overflow_naive = {\n let p = 1;\n for(let ii=0; ii