# Shuffling Probability

Integers 1 through 53 are written on cards, one per card. The stack is thoroughly shuffled. Five cards are drawn. What is the probability that the cards are drawn in their natural order the smallest first, and the rest in increasing order of magnitude?

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Copyright © 1996-2017 Alexander Bogomolny

1/120. To better grasp the solution start with drawing just two cards. What is the probability that the first out is smaller than the second one?

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Copyright © 1996-2017 Alexander Bogomolny

You have five cards which may have been drawn in any of 120 possible ways. Only one of these is the natural increasing order. The probability of this happening is 1/120.

The specific number of cards is a *red herring*. Whatever the number you start with, as long as you draw 5 cards there are 5! = 120 variants. Each comes with probability of 1/120.

### References

- C. W. Trigg,
*Mathematical Quickies*, Dover, 1985, #91.

### What Is Red Herring

- On the Difference of Areas
- Area of the Union of Two Squares
- Circle through the Incenter
- Circle through the Incenter And Antiparallels
- Circle through the Circumcenter
- Inequality with Logarithms
- Breaking Chocolate Bars
- Circles through the Orthocenter
- 100 Grasshoppers on a Triangular Board
- Simultaneous Diameters in Concurrent Circles
- An Inequality from the 2015 Romanian TST
- Schur's Inequality
- Further Properties of Peculiar Circles
- Inequality with Csc And Sin
- Area Inequality in Trapezoid
- Triangles on HO
- From Angle Bisector to 120 degrees Angle
- A Case of Divergence
- An Inequality for the Cevians through Spieker Point via Brocard Angle
- An Inequality In Triangle and Without
- Problem 3 from the EGMO2017
- Mickey Might Be a Red Herring in the Mickey Mouse Theorem
- A Cyclic Inequality from the 6th IMO, 1964

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Copyright © 1996-2017 Alexander Bogomolny

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