Odds in tarot cards

[whqttt]

What are the odds of drawing the same seven cards from a 73 card deck of Tarot cards, twice?

[Marsh]

7/73 x 6/72 x 5/71 x 4/70 x 3/69 x 2/68 x 1/67

= 7! x 66! / 72!

But my calculator won't cope with working this out

[Marsh]

But my spreadsheet can:- 4.48032E-08

[ermintrude35]

..dunno but it happened to me the one and only ever time I had them read about 15 years ago..I can't remember much about the reading because it freaked out the reader so much....

[Hammond Egg]

Marsh, using your first calculation I made it 1 in 1,629,348,612 (or 6.13742E-10) which isn't the figure you got second time. I got the first figure doing everything the long way round, I think the second one is wrong.

[johnalex]

What are the odds of drawing them in the same order and how long would this take, turning one card per second? (I wouldn't have a clue, maybe 10,000 years?)

[whqttt]

That was the thing, the cards weren't in quite the same order, so they weren't simply being redrawn from the deck.Bizarre

[Marsh]

Hi everyone, I don't check in every day so this is a bit late.
Hammond Egg, I'm sure your'e right. I can't type a sentence right first time, let alone enter a long calculation.
Johnalex, i) 1/73 x 1/72 x... = 66!/73!

ii) 14 seconds + shuffling time, when it happens.

When does it happen? You can't say. The best you could do is to work how many draws gave a 50% chance of drawing the same 7 cards in two successive hands. To do this you have to calculate the chance of not getting two consecutive matches. For n draws the chance of not getting this is (72/73)^(n-1) [If n = 1 you've only drawn once and have nothing to match with]
The larger n gets the smaller the chance becomes until you eventually get below 0.5, giving a better-than-50% chance of two draws the same. I'll leave someone else to work it out.
If you want your last draw to match only your first draw the odds remain the same.
If you want your last draw to match any previous draw the chances of it not happening are (72/73 x 2x72/73 x 3x72/73�(n-1)72/73) = ((n-1)! )(72/73)^(n-1)

I think.

[johnalex]

Hi Marsh - Well done for those all those calculations, especially 14sec! Actually, that's a very good answer, as we humans tend to think that an event must occur after a certain number of repeats, which of course is not true. Same with the lottery - the only way to win is to cover all the combinations in the same week, not keep gambling on new games! Cheers, ja.