[shevek]

Got asked the chance of a player being dealt one of each pip (A,K .....2)
I thought ....

Not so rare if I got it right, about 1 in 9,462

Shuffle the cards and deal out 13 to a player.
First can be anything, so 52/52
Second can be 48/51. 51 cards left, can’t be one of the 3 other (aces, say)
52/52 x 48/51 x 44/50 ……. 4/40. (Come the last card, there are 40 left, need to be dealt one of the four (2s, say)

= 0.000010568. 1/x = 9462.

Nick

[smerriman]

Looks good to me. An alternative calculation is 4^13 / (52C13) which gives the same result.

Edit - incidentally, I just discovered you posted this exact same thing 4 years ago.