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A poker hand consists of 5 cards.If the cards have distinct consecutive values and are not all of the same suit,we say that hand is straight.For instance a hand consisting of the five of spades,six of spades,seven of spades,eight of spades and nine of hearts is a straight.What is the probability that one is dealt is a straight?

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In a pack of cards there are 4 sets and 13 cards in each set $\rightarrow$ 1(ace),2,3,4,,5,6,7,8,9,10,J(11),Q(12),K(13).

The possible cases for straight hand can be (13,12,11,10,9) , (12,11,10,9,8) , (11,10,9,8,7) .....(5,4,3,2,1)

= ways of choosing $k$ consecutive integers from $n$ integers

=$(n-k+1)$

=$(13-5+1)$

=$9$ cases.

and Probability of selecting cards for each case = $\frac{4}{52} * \frac{4}{51} * \frac{4}{50} * \frac{4}{49} * \frac{4}{48}$

But this above cases also include the cases where all the $5$ cards can be from same slot and there are $4$ such cases possible

Also the order of selection of cards doesn't matter so we will multiply by $5!$

so now

Probability of selecting cards for each case = $\frac{4^{5}-4}{52*51*50*49*48}*5!$

$\therefore$ Required Probability

= $9*5!* \frac{4^{5}-4}{52*51*50*49*48}$

= $9*5!* \frac{1020}{52*51*50*49*48}$

= $\frac{1101600}{311875200}$

= $0.00354$

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