# Conditional Probability on Cards of the Same Suit

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: $$\text{Probability}=P(\text{Bob has all hearts} |\text{Ann has all hearts})+P(\text{Bob has all clubs/spades/diamonds}|\text{Ann has all hearts})$$

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

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

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) =$$\frac{^{13}C_4\cdot ^3C_1+^9C_4}{^{52}C_4}$$

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

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

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

• To obtain $P(B \mid A)$, type $P(B \mid A)$. Also, check your last line. Commented Nov 19, 2018 at 10:40
• @N.F.Taussig thanks for that, but whats wrong in last line? Commented Nov 19, 2018 at 11:02
• Your probability is greater than $1$. Commented Nov 19, 2018 at 11:04
• @N.F.Taussig Yeah...$C(52,8)$. And i later added $C(4,1)$ to select a suite-heart; but we dont need to it seems... Commented Nov 19, 2018 at 12:20
• @N.F.Taussig And Thanks, for taking me through this. Got to learn something new. Commented Nov 19, 2018 at 12:41

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.

$$P(X|H)=\frac{P(X\cap H)}{P(H)}$$

where

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

and

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

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