Jtdylktuy

A normal deck of 52 playing cards is well shuffled and then 4 cards are dealt to Ann and 4 cards ar dealt to Bob. Ann looks at her c that all 4 of them are from the same suit, that suit being hearts. She is interested in the probability that Bob also has 4 cards that belong to a single suit, allowing for her knowledge of the cards she holds.

Please correct my answer if it is incorrect:
$$text{Probability}=P(text{Bob has all hearts} |text{Ann has all hearts})+P(text{Bob has all clubs/spades/diamonds}|text{Ann has all hearts})$$

$$=frac{binom{9}{4}}{binom{48}{4}}+frac{binom{3}{1}timesbinom{13}{4}}{binom{48}{4}}$$

$$=frac{757}{64860}$$

I am aware that this is a conditional probability, but I was unsure how to apply Bayes' Theorem in this scenario.

$P(A)$ = P(Anne has all hearts) =$$frac{ ^{13}C_4}{^{52}C_4}$$
$P(B)$ = P(Bod has same suite) =$$frac{^{13}C_4cdot ^3C_1+^9C_4}{^{52}C_4}$$

Using:
$$P(Bmid A)=frac{P( A ∩ B)}{P(A)}$$

P(A ∩ B)=$$frac{(^{13}C_4 cdot 3+ ^{9}C_4)cdot ^{13}C_4}{^{52}C_8}$$

You get: $$P(Bmid A)=(frac {(^{13}C_4 cdot 3+ ^{9}C_4)cdot ^{13}C_4}{^{52}C_8 })cdot(frac{^{52}C_4}{^{13}C_4})$$

Letting $X$ be the event where Bob has $4$ cards of the same suit and letting $H$ be the event where Ann has four hearts.

$$P(X|H)=frac{P(Xcap H)}{P(H)}$$

where

$$P(H)=frac{binom{13}{4}}{binom{52}{4}}$$

and

$$P(X cap H)=frac{binom{13}{4}times left(binom{3}{1}timesbinom{13}{4}+binom{9}{4} right)}{binom{52}{4}timesbinom{48}{4}}$$

Substituing into the conditional probability gets the result that I originally posted in my solution.