Before Becoming A Card Shark, You Need To Know This Method - Part 1

Bob will solve it indeed first. Case 1: Alice She goes through 52 rounds through the remaining set of cards to sort them out.

Case 2: Bob He goes through one round to sort the cards as per denomination (heart etc.) Then he goes through total of 13 rounds for each denomination but the difference from the case of Alice is that his size of set to search will go from 13,12,11 ..... ,1 in the similar manner four times. But in the case of Alice it goes from 52,51, ......, 13,12,11 ....,1 , she has to go through many more number of cards than Bob.

We can use Asymptotic Analysis to figure out which one is quicker. In this method we try to come up with and assign a cost to the worst case scenario

Let v = # times we view a card.

CASE 1: In Alice's case, she looks at each card in the deck and determines if it's the one she is looking for. This is an example of Selection Sort.

Worst case scenario, she ends up looking through all the cards before she finds the correct one. In the initial step, she ends up looking through 52 cards. In subsequent steps, the number of cards she looks through is decreased by 1, since each time 1 card will be removed from the list.

$v = 52 + 51 + \cdots + 2 + 1 = 1378$

CASE 2: In Bob's case, he first sorts the deck into 4 piles by suit, then he performs the same operation as Alice did on each pile. The first part is an example of Bucket Sort. He is not concerned with order, he looks at each card and places it in the correct pile. In this case he will end up looking at each card just once to determine the pile it belongs in.

During the second part, like Alice did on the deck, he performs a Selection Sort on each pile. Each pile will have 13 cards. Worst case scenario he will have to look through all cards in their respective piles to find the correct card. And he will repeat that 4 times for each pile.

$v_{Bucket} = 52 \quad v_{Pile} = 13 + 12 + \cdots + 2 + 1 = 91 \\ v = v_{Bucket} + 4 \cdot v_{Pile} = 52 + 4 \cdot 91 = 416$

Bob's method results in a smaller number of views. So theoretically, it is faster.

U can use logic.Bob's method is more efficient than Alice's

Alice goes through the deck systematically, and so the probability she finds each card she is looking for is approximately $\frac{1}{52!}$ . This comes out to approximately $10^{-68}$ .

Bob, discounting his initial sorting, takes approximately $4(\frac{1}{13!}$ ) which comes out at approximately $10^{-10}$ .

Since there are significantly fewer permutations for Bob, his sorting must be quicker.

(Ok, ignore the math, but I stand by my logical reasoning xD)

Its just simple logic. It takes less time to go through the pile instead of going through entire deck.

For Alice : total time

                                 = 52 + 51 + .....+ 2 + 1 = 26 * 53

For Bob: total time = 52 + forAllColor(13 + 12 + 11 +...+ 2 + 1)

                                 = 52 + 4 * (13 * (13 + 1)/2)

                                 = 52 + 4 * (13 * 7) 

                                 = 52 + 28 * 13

So bobTime < AliceTime

Actually , what Bob did , is little bit similar to the concept of divide and conquer algorithm . At first when Alice started sorting , he worked with all 52 cards whereas Bob divided that sorting problem into 4 smaller parts and solved them individually . Then , he merged 4 smaller problems into 1 to find the final solution . You can find out the worst case complexity of the procedure of both Bob and Alice . And then you will understand why Bob needed fewer time than Alice.

Step 1:

Suppose , there are n cards . So , if we want to work with all n cards , we may need to at most n cards if we want to find the desired card (it's the worst case) . And once we find the card with least value , next stage our worst case will be 1 less than n , n-1 . And next it will be n-2_then n-3 then n-4 ......... and so on . So, in that case at most c1 = $\frac {n*(n + 1)}{2}$ number of checking will be needed .

Step 2:

(i) At first we want to separate 4 different decks it is enough to check all n cards once . Here , complexity : $n$

(ii) Then , We can sort them individually by using Step 1: 4 times . Here , complexity : $4*(\frac { \frac { n }{ 4 } *(\frac { n }{ 4 } +1) }{ 2 } )$

(iii) At last , we can merge these $4$ decks by only 4 operation .

So , that procedure will need at most c2 = n + $(4*(\frac { \frac { n }{ 4 } *(\frac { n }{ 4 } +1) }{ 2 } ))$ + 4 number of checking .

And if you calculate c1 and c2 , you will find that c1 will be always bigger than c2 for any n . So, Bob will need fewer time than Alice . That's how divide and conquer work's . And sorry for poor English :)

Lets consider the worst case: Lets say while going through the cards, each time a card is considered it consumes a 1 unit time. Therefore Alice will consume time 52+52+50+....1 = 1352. While Bob will sort the cards in suits in one 52 steps there after he needs to sort each suit in numbers which will take 13+12+11...1 time there his total time is 52+ 4(13+12+11...) =416 which is much smaller than Alice's.

No need to use calculation, the logic for this question is simply easy.

Well I think it goes this ways: Alice has to search for one card in the deck of 52 cards, then one card in the deck of 51 cards and so on.. This way the average number of steps are: 52/2 + 51/2 + 50/2 + ... = 689. Now Bob has do the same thing but in a different way. Initially Bob has to do 52 steps compulsorily to separate the deck suit wise. Now he has to find one card in suit of 13. The same way as with Alice avg. number of steps = 13/2 + 12/2 + 11/2 + ... That is 45.5 But he has to repeat this for all the suits therefore total of 45.5*4 = 182 times. Total steps for Bob = 52 + 182 = 234 Which is way lesser then that of Alice.

Its just that it takes less time to go through a suit than the whole deck.

Bob,obvious.I have used probability to solve this. Alice's prob to get ace of hearts in first time=1/52 then 2 of hearts = 1/51..likewise prob of her to solve this=1/52 * 1/51 ...=1/52! On the other hand, Bob's prob to sort each one of them=13/52 * 13/39 * 13/26 * 13/13=1/24. Then to solve this=(1/13 * 1/12 *..) 1/4 * 1/24 which is effectively greater than 1/52!. Hence Bob should be the first to complete.