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Question Number 43805 by Penguin last updated on 15/Sep/18

In a properly shuffled deck, what is the probability  that the last 10 cards are in order:  A♥, 2♥, 3♥, 4♥, 5♥, 9♠, 10♠, J♠, K♠, Q♠

$$\mathrm{In}\:\mathrm{a}\:\mathrm{properly}\:\mathrm{shuffled}\:\mathrm{deck},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{last}\:\mathrm{10}\:\mathrm{cards}\:\mathrm{are}\:\mathrm{in}\:\mathrm{order}: \\ $$$${A}\heartsuit,\:\mathrm{2}\heartsuit,\:\mathrm{3}\heartsuit,\:\mathrm{4}\heartsuit,\:\mathrm{5}\heartsuit,\:\mathrm{9}\spadesuit,\:\mathrm{10}\spadesuit,\:\mathrm{J}\spadesuit,\:\mathrm{K}\spadesuit,\:\mathrm{Q}\spadesuit \\ $$

Commented by Penguin last updated on 15/Sep/18

This just happened to a friend in a deck that was  VERY well shuffled (literally thrown into  the air).

$$\mathrm{This}\:\mathrm{just}\:\mathrm{happened}\:\mathrm{to}\:\mathrm{a}\:\mathrm{friend}\:\mathrm{in}\:\mathrm{a}\:\mathrm{deck}\:\mathrm{that}\:\mathrm{was} \\ $$$$\boldsymbol{{VERY}}\:\mathrm{well}\:\mathrm{shuffled}\:\left(\mathrm{literally}\:\mathrm{thrown}\:\mathrm{into}\right. \\ $$$$\left.\mathrm{the}\:\mathrm{air}\right). \\ $$

Answered by MJS last updated on 16/Sep/18

if it′s a set of 4×13=52 cards  (1/(52×51×50×49×48×47×46×45×44×43))≈1.74×10^(−17)

$$\mathrm{if}\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{set}\:\mathrm{of}\:\mathrm{4}×\mathrm{13}=\mathrm{52}\:\mathrm{cards} \\ $$$$\frac{\mathrm{1}}{\mathrm{52}×\mathrm{51}×\mathrm{50}×\mathrm{49}×\mathrm{48}×\mathrm{47}×\mathrm{46}×\mathrm{45}×\mathrm{44}×\mathrm{43}}\approx\mathrm{1}.\mathrm{74}×\mathrm{10}^{−\mathrm{17}} \\ $$

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