Game Theory (a) Find the number of distinct hands in Straight Poker. (A deck of

Question

Game Theory

(a) Find the number of distinct hands in Straight Poker. (A deck of cards contains 52 cards. A Straight Poker hand contains 5 cards. You are therefore asked how many ways there are of selecting five cards from 52 cards when the order in which they are selected is irrelevant.) (b) What is the probability of being dealt a royal flush in Straight Poker? (A royal flush consists of the A, K, Q, J, and 10 of the same suit.) (e) Suppose that you are dealt the A, K, Q and 10 of hearts and the 2 of clubs. In Draw Poker you get to change some of your cards after the first round of betting. If you discard the 2 of clubs, hoping to draw the J of heart, what is the probability that you will be successful? (d) What is the probability of drawing a straight? (Any J will suffice for this purpose.)

Explanation / Answer

Solution

Back-up Theory

Number of ways of selecting r things out of n things, when order is not important, is given by: nCr = (n!)/{(r!)(n – r)!}………………………………………………………………(1)

Probability of an event, E = P(E) = n/N, ……………………………………………..(2)

where N = total number of possible outcomes and n = number of possible outcomes favourable to E.

Part (a)

Number of distinct hands in Straight Poker

= Number of ways of selecting 5 cards out of 52 cards = 52C5 = (52!)/{(5!)(47!)}

= (52 x 51 x 50 x 49 x 48)/120 = 2598960 ANSWER

Part (b)

A royal flush => a hand of 5 cards AKQJ10 of the same suit. So, this can happen in just 4 ways – AKQJ10 of Clubs, Hearts, Spades or Diamonds.

So, vide (2) under Back-up Theory and answer of Part (a), probability of a royal flush

= 4/2598960 = 0.000001539 ANSWER

Part (c)

Having got AKQ10 of Hearts as hand and discarded 2 of Clubs, there are 47 cards left in the pack of which J of Hearts is to be selected. This can be done in only one way. Hence, vide (2) under Back-up Theory, the requires probability = 1/47 = 0.0213 ANSWER

Part (d)

For a straight, any J would suffice. Since there are only 4 J’s, number of ways of selecting a straight = number of ways of selecting a J out of 4 and selecting the remaining 4 cards from the remaining 48 cards = 4 x 48C4 = 4 x (48!)/{(4!)(44!)}

= (48 x 47 x 46 x 45)/24 = 778320. Hence, vide (2) under Back-up Theory and answer of Part (a), the required probability = 778320/2598960 = 0.2995 ANSWER

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