Ross's First Course on Probability 10th edition Chapter 2, question 22

Question

I am currently self-studying Sheldon Ross' Intro to Probability and I am stuck on this problem. I tried myself and got an answer. Check whether I'm right, if not then whats wrong with my approach.

Each of 52 people are given a deck of cards, which they are asked to shuffle independent of each other. What is the probability that (a) the order of the cards in each shuffled deck is unique? (b) there is exactly one card that occupies the same position in the shuffled decks received from all 52 persons? (c) all cards occupy the same position in all the shuffled decks?

My approach:

1. $$ \frac{52!*(52! - 1)*(52! - 2)*....*(52! - 51)}{(52!)^{52}} $$ Clearly, there are 52! ways to arrange these 52 cards of one person. And this is same for all the 52 persons. Hence the denominator. Now, in the numerator the first person can arrange his cards in 52! ways, and then the next person, inorder to have a unique order, can arrange his cards in 52! ways except for that one particular arrangement that the first person has used. So, the second person has 52!-1. Similarly, the 3rd person has 52!-2 and so on till the last(52nd) person has 52!-51.

2. $$ \frac{\binom{52}{1}*\binom{52}{1}*51!*(51!-1)*...*(51!-50)}{(52!)^{52}} $$

For exactly one card to occupies the same position, we first choose a position from the 52 positions. Then we choose a specific card from the possible 52 cards for that position. Then, we are left with 51 cards for each person. Rest can be done in a similar fashion as above.

3. $$ \frac{52!*1*....*1}{(52!)^{52}} = \frac{1}{(52!)^{51}} $$

The first person can arrange his cards in 52! ways. And in order for all cards occupy the same position in all the shuffled decks. All other 51 people should have the exact same arrangement as that of the first person.

Answer 1

Now that I'm sober (but hungover) I will give it another try :)

The answers to 1. and 3. are correct, 2. is incorrect. Let $f(c,p)$ be the number of ways to distribute the same deck of $c$ different cards to each of the $p$ players without any card occupying the same position for all players. We now have $$c!^p=\sum_{k=0}^ck!{c\choose k}^2f(c-k,p)$$ where $f(0,p)=1$. The formula comes from the partition of the $c!^p$ possible configurations without restriction into those where no card occupies the same position, those where exactly one card occupies the same position and so on.

The $k$-th term inside the sum is the number of configurations where exactly $k$ cards occupy the same position for all players, there are ${c\choose k}$ number of ways to choose those cards, ${c\choose k}$ to choose those positions and $k!$ ways to permute the let's say cards (permuting the cards or positions is equivalent). The rest of the $c-k$ cards have to be arranged such that non of those cards occupy the same position for all players and there are $f(c-k,p)$ such arrangements.

This formula leads to a recursion which with a modern laptop evaluates instantly for the numbers considered. The answer for 2. will then be $$\frac{1!{52\choose 1}^2f(52-1,52)}{52!^{52}}=\frac{52^2f(51,52)}{52!^{52}}.$$ In fact we have that $f(51,52)$ is equal to $$8194878626029100213158341829015278440600410342343524058985214361778896\ 1138920807102375825646393673626741006680903784355152047966280697649972\ 1023209464875728007862790607510001168826592553873527922707758072430954\ 1696699562802756695403432702679293904466008347931258410005602665917241\ 0003243589234395423435763314488593404794831861542206092386710078960491\ 8215650609351490697649308355734906917508353989018681605372404633549963\ 6529208502782598222644817536453556294970544304616427278324904069489516\ 6499612172944741768520833424192415139526102703939524391803888625375711\ 7596965144314495312795616956469847049689089190668641186786511460449422\ 8054652503232700030717904816845649618210588162716797596763249192873503\ 9766127767182984408361192038145333780222300168001028967557244212117461\ 6369125071857013055369877018210272887675484488104639144621556875261623\ 4684458246788657410861804826107270728012981849828582887674817275351499\ 4033176027734397267943466512618674418868833532261565350238605531131848\ 2006589787774091971697136315639454263299476413087387879594466255689409\ 8413797518711950948289804004376376900642813294429066471960616770310282\ 8318337939515404076394841213522450056990856679840686703047844215484236\ 1361450985127583734113869810504853919541728412506735658662968829818927\ 0013253907974420479325128668885723505800906221045138137039683699123788\ 1188692771558171615244961259257923888441516771093843247056748208462229\ 9896775497146356915426929947767162130444724218285644356759999349054696\ 5056625758184903892376584187076585042949805918852842722298883502845542\ 6795407399462241696700106522410255361618077536360175007723565457067552\ 3795896025552082305238876489431625058948310666940497973389275609221668\ 4719857889400305856996138361076471919151315467282368985698255620622291\ 1513867096582979740710305599548980113506895886786129535239493578089959\ 2458170312920131518793947714340413614968927983705322757004923505481094\ 9214687601013756435814849167651417850145770508266165937149555650999559\ 9983102097045662937406222327705793326109099027796347408796663478020833\ 4758048849315339055347192085791465265730473655979680606582962143517442\ 1575151273175679164536842330294738874406208574334949477309765874595608\ 9193549513895627964562731468453398361835722130651011673890552217373115\ 6811337587266395214594432901648522461485826941064460589257903717916012\ 1080123328019707650268106235771807735207981774487242638428328972172798\ 2276025085569079552734304752479255244271972829084520326583862405575581\ 0059200451239668912494447923187170342252595644440152476342160569390584\ 9789316809137856607443552590758770827619939451122178670521448566168852\ 0099454615898623146667715610248581063387199794986728656924719900683263\ 4800013734475136564825014540906925950289877229405461646644213518311646\ 8545166648618114914106177718558519856478561434502577075879425340157610\ 2954220496015196953114519114343858950715907356813618656529969145693261\ 0915940558468138137748931663383232689546089456987135451845347940886806\ 7325430837795459331759204745786262716133951576572656009578307105270752\ 1493336386278103099455302828575213288845827291998181522184608582756364\ 5224801768118331122633061645690002229240472908612820856287713659407850\ 8199981149588619637067023800484933584239655205280534609205184300785393\ 9151018248545177703554711692687464180807007246481453917362757285761888\ 8649094250973054979256029129478869675466532787720308885278984438584425\ 5604611626758559357261524753696433432050812002748924323869936215654400\ 000000000000$$ if I haven't done any off by one errors.

Answer 2

Your answer for questions 1 and 3 are correct.

For question 2, underflow's answer makes it seems like you need to use methods which were not mentioned in the book to solve this book's problem. My answer shows the way that Ross likely intended to solve this problem; I will use the principle of inclusion-exclusion, which was an earlier topic in the chapter.

First, let us find the probability that the $\spadesuit A$ is in the first position in everyone's decks, but no other card is in the same position in everyone's decks. The actual answer will be $52^2$ times the answer to this simpler problem, since there are $52^2$ analogous cases ($52$ choices for the card, $52$ choices for the position).

The probability that the $\spadesuit A$ is first card of everyone's decks is $(1/52)^{52}$. Conditional on this occurring, each person has $51$ cards besides $\spadesuit A$ to shuffle, and we require none of these cards to be in the same position for everyone. To this end, define the event $E_i$, for each $i\in \{1,\dots,51\}$, by $$ E_i=\{\text{$i^\text{th}$ card is at the same position in everyone's decks}\} $$ We want to find the probability $\mathbb P(E_1^c\cap \dots \cap E_{51}^c)$. Inclusion-exclusion is perfectly suited for this. $$ \begin{align} \mathbb P(E_1^c\cap \dots \cap E_{51}^c) &= \sum_{k=0}^{51}(-1)^k\sum_{1\le i_1<\dots<i_k\le 51 }\mathbb P(E_{i_1}\cap \dots \cap E_{i_k}) \\&=\sum_{k=0}^{51}(-1)^k\binom{51}k \mathbb P(E_1\cap \dots \cap E_k) \end{align} $$ Now we just need to ask, when $52$ friends shuffle decks of $51$ cards, what is $\mathbb P(E_1\cap \dots \cap E_k)$, the probability that cards numbered $1$ to $k$ all end up in the same place in everyone's decks? I claim that $$ \mathbb P(E_1\cap \dots \cap E_k)=\left(\frac{(51-k)!}{51!}\right)^{51} $$ To see this, consider the first person's deck. Cards numbered $1$ to $k$ will end up at some positions $p_1,p_2,\dots,p_k$ in the first person's deck. Then, we only succeed if every person's deck has cards $1$ to $k$ at these positions. Since the probability that $k$ particular cards land at $k$ particular positions in a particular deck is $\frac{(51-k)!}{51!}$, and since the $51$ other decks are independent, the result follows.

Putting this altogether, the final answer is $$ \mathbb P(\text{exactly one card at same position})=\frac{1}{52^{50}}\sum_{k=0}^{51}(-1)^k\binom{51}k \left(\frac{(51-k)!}{51!}\right)^{51}. $$ My answer numerically agrees with underflow's answer exactly.