What would be the odds, and how would they be calculated if you had 13 spectators each select a card, and all 13 of them were from the same suit?

Assuming you mean a 52 card standard deck:

This is 13/52*12/51*11/50*...*1/40. Probability = 1.211361171114698E-12.

Of course this is the probability, not the odds, but its easy from there when you check a book out of your local library.

Sorry, I haven't time to give you a better explanation, I'm sure someone more patient will.

M

Chop off the first factor (13/52 = 1/4) and you've got it.
Remember: Suit was not specified.

Stan Alger

I would have to disagree with that assumption. There has to be specific odds that 13 spectators would each and only select all 13 cards of one suit.

No, Stanalger's right. What is mean by this statement, is that the probability doesn't matter on the first card, since you don't care which suit is chosen (only that all of them are the same after that). So my calculation gives you the probability of choosing a specific suit, say spades, while knocking off the first fraction gives you the probability of choosing all 13 of the same (any) suit. I should know better than to answer maths questions just before bed.

So by that reasoning, we have a new answer:

Probability = 3.820446770438662E-12

There you go.

M

Hi,

my PC says 12/51*11/50....*1/40 = 6.299E-12

but it doesn't really matter, 1 magician -assuming he needs 1 minute for the trick- has to work for more than 300,000 years (whitout a single coffeebreak!) to find it happening once...

On the other hand: try, maybe it's tomorrow!

Heinz

You want odds? I'll give you odds.
The odds AGAINST all 13 cards being of the same suit are
158,753,389,899 to 1.

(I agree with Heinz Weber's figure for the probability of all 13 being of the
same suit.)

Stan Alger

After checking my program for flaws (missed a card), I concur. Thanks for the correction guys, now I get:

6.299078089796431E-12

I'll let someone who got it right show you how to convert odds to probability and back again.

M

So if a presentation were set up having 13 different spectators select a card each, then revealed that they were all the same suit . . . how would you tell the spectators in laymen's terms what the odds are of all of them selecting the same suit?

Uh...have another look at the third post above this one or the one above that...I think that about wraps it up. Enjoy!