How Likely Is A Flush?

Probability Level 1

Kyle draws 5 cards from a standard deck of cards . The probability that he has a flush (including straight and royal flushes) is $\frac{ 4 \times { 13 \choose 5 } } { 52 \choose 5 } = \frac{ 33} { 16660 }.$ From the remaining 47 cards, Linh draws 5 cards. Given that Kyle has a flush, what is the probability that Linh has a flush?

Kyle hosts a DataSkeptic podcast, which explores topics in statistics, machine learning, big data, artificial intelligence, and data science. He focuses on scientific skepticism, misconceptions, and misinformation in these topics.

Equal to

\frac{33}{16660}

Less than

\frac{33}{16660}

Greater than

\frac{33}{16660}

2 solutions

Andy Hayes
Oct 11, 2017

Relevant wiki: Math of Poker - Basics

After Kyle draws his flush, there are now 47 cards left in the deck: 3 suits of 13 cards and 1 suit of 8 cards. So the probability that Linh has a flush is:

$\frac{3 \times \binom{13}{5}+1\times \binom{8}{5}}{\binom{47}{5}}$

Without calculating this probability exactly, we can compare it to the original probability of drawing a flush: $\dfrac{4 \times \binom{13}{5}}{\binom{52}{5}}$ . The numerator is (relatively speaking) not much less than the original numerator. Meanwhile, the denominator is significantly less than the original denominator. We can reasonably conclude that this probability is greater than the original probability.

The numerical probabilities are below:

$\begin{aligned} \frac{4 \times \binom{13}{5}}{\binom{52}{5}} &= \frac{33}{16660} \approx 0.00198 \\ \frac{3 \times \binom{13}{5}+1\times \binom{8}{5}}{\binom{47}{5}} &= \frac{3917}{1533939} \approx 0.00255 \end{aligned}$

Nice solution! But there is a typo in this problem. The denominator is 16660 instead of 16600.

Judy Gu - 3 years, 7 months ago

Thanks! Fixed.

Andrew Hayes Staff - 3 years, 7 months ago

Maybe I dont understand the english syntax but "Given that Kyle has a flush, what is the probability that Linh has a flush?" does this not imply that the resulting probability is the product of these probabilities?

Jakub Zajac - 3 years, 7 months ago

"Given that Kyle has a flush" means that Kyle has already drawn his cards and already has his flush. That event has already happened. Kyle having a flush only matters for determining what cards are left in the deck.

The product would be the probability that both events happen. If you had a shuffled 52 card deck, and both Linh and Kyle are about to draw 5 cards, then the probability that they both get flushes is the product:

$\frac{33}{16660} \times \frac{3917}{1533939}.$

Andy Hayes - 3 years, 7 months ago

Linh's probability is greater than Linh's probability of drawing a flush in a DIFFERENT suit, which is $$3×{13 \choose 5} \over {47 \choose 5}$$ To compare it with Kyle's probability without difficult calculations, we can first drop $13\choose 5$ in both and note that $${x \choose 5} × 5! = x(x-1)(x-2)(x-3)(x-4)$$ So here we go: $${48\over43}>{49\over44}>{50\over45}>{51\over46}>{52\over47}>{13\over12}$$ $${13^2\over12^2}>{7\over6}$$ $${13^4\over12^4}>{7^2\over6^2}>{4\over3}$$ Therefore $${52×51×50×49×48\over47×46×45×44×43}>{13^5\over12^5}>{13^4\over12^4}>{4\over3}$$ That means $${47×46×45×44×43\over3}<{52×51×50×49×48\over4}$$ Divide both sides by 5! $${{47 \choose 5} \over 3}<{{52 \choose 5} \over 4}$$ $${3\over{47 \choose 5}}>{4\over{52 \choose 5}}$$

Tomaž Cedilnik - 1 year, 11 months ago

Oops, my TeX formulae didn't render properly....

Tomaž Cedilnik - 1 year, 11 months ago

Ariijit Dey
Oct 17, 2017

After Kyle has his hand, there are exactly 3 suits kept intact. Also the 4th suit from which Kyle draws out his cards, has 8 cards remaining. The probability that Linh also has a flush must be greater than drawing a a flush out of the remaining 3 intact suits. This turns out to be 0.0025... .

Lol kyle and Linh! They're the hosts of Data Skeptic! Hahaha

Phillip Deon - 3 years, 7 months ago

How Likely Is A Flush?

2 solutions

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