Magic trick probability

What is the probability that a King will be adjacent to an Ace in a well shuffled deck of 52 cards ?

#Combinatorics #Probability #Math

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

The problem has been solved here

The required probability is: $\displaystyle 1- \dfrac{44!\cdot \sum_{k=1}^{3} \binom{3}{k-1}(4!) \binom{45}{k}\, (45-k)\, (46-k)\, (47-k)\, (48-k)}{52!} = \dfrac{284622747}{585307450} \approx 0.486279043603494$

gopinath no - 7 years, 8 months ago

First , you should clarify it. Do you mean a particular king will be adjacent to a particular ace or at least one king would be adjacent to an adjacent to at least one ace or a particular king would be adjacent to at least one ace or at least one king will be adjacent to a particular ace ?

jatin yadav - 7 years, 8 months ago

Probably, any king would be adjacent to any ace.

Tim Vermeulen - 7 years, 8 months ago

18 but jatin is wright

Parag Tipre - 7 years, 8 months ago

Any king adjacent to any ace.

Shubham Raj - 7 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$