Problem 6 proof by cases a standard deck of 52 cards...

Problem 6. Proof by cases: A standard deck of 52 cards consists of 4 suites (hearts, diamonds, spades and clubs) each containing 13 different values (Ace, 2, 3, ..., 10, J, Q, K). If you draw some number of cards at random you might or might not have a pair (two cards with the same value) or three cards all of the same suit. However, if you draw enough cards, you will be guaranteed to have these. For each of the following, find the smallest number of cards you would need to draw to be guaranteed having the specified cards. Prove your answers. Three of a kind (for example, three 7's). A flush of five cards (for example, five hearts). Three cards that are either all the same suit or all different suits.

Transcript

00:01 Hi, so in the given problem we have to find the minimum number of cards that needs to be drawn such that all the three cards are of same kind so a same kind card so for that we have to draw a minimum number of card that mean that we need to find so here in order to solve we'll be using the pigeon hole principle so this is the pigeon hole principle which says that we say that if there's a pigeon hole principle that if there's a pigeon hole principle there are 13 holes or 13 types of card also here we'll be taking this as 13 types of card because these 13 cards will be so these are the 13 different kinds of card and the worst case scenario is 26 cards with all 13 having two cards each so now what we can do the 27 so the 27th card drawn the drawn card will go to one of the whole so hence making it carry three thus the answer is 27 so we can say that when the using the pigeon hole principle if we have 13 holes or here they we have 13 types of card the worst case scenario is 26 card with all 13 having two cards each then the 27th card will go to one of the hole so making it carry three so hence the answer would be 27 so the minimum number of card that we need to be drawn in the first case that all the three cards are of of same kind would be 27 now coming to the coming to the part b of this question we have to find the minimum number of card drawn such that the flush of five cards are there or for example we can say the flush all the five cards are hard hard so the flush of five cards so let's write it here now, so we'll be solving this by again taking a number of hole is four so here again we'll be using the pigeon hole principle now here the number of holes number of holes is that is five so holes are let's say four so that's that's four number of holes are four so here three holes getting filled or one of the whole carries three so so so the so we get the desired result so here what we have taking if if three holes are filled or one of the whole carries three we get the desired real result now suppose we draw three cards first so let's say we have taking out the three cars if all of them are of same suit so if three cars are drawn and all are of same same suit or all different so all or it could be all different then we get the desire real so the only desire the only or so the only remaining combination would be two plus one or we where two is one suit and one from the other suit so if we take two from one suit and one from the other suit we'll be getting so now after drawing fourth card it comes from the third card or the suit having two cards or can say that this is as let's start it again so here we'll be taking the whole number of holes as four so here this is of a different suit which is the spade so there are four different suits right so four suits which are spayed hurt diamond eclipse so against worst case is when all of them are having four cards so which makes four times four which is equal to 16 so the 17th card the 17th card will enter one of the whole making it carry five cards.

04:19 So hence we can say there are 17, we have to take minimum 17 cards such that we'll be getting the flush of the flush of five cards.

04:32 Now in the third part, we have to, so that's the solution for the second part...

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