Probability of Playing Cards?

Question

Can anyone tell me how to work out the following:

A shuffled pack of 52 normal playing cards, each card is dealt face up in order, how long (on average) do you have to wait before (say) the Ace of Clubs is dealt?

Of course it is less than or equal to 52 because it has to come up, but how long on average?

Thanks

Also if anyone knows any good books that deal with such question s (and others that I have asked) I would be grateful for a recommendation!

blahb31: Thanks for the answer, how would you work it out for (say) any ace?

Answer 1

The ace of clubs is equally likely to be in any of the 52 positions in the deck.

(1+2+3+...+52)/52 = (52*53/2)/52 (using 1+2+...+n = n*(n+1)/2)
= 53/2
= 26.5

So, on average, you would expect to go through 26.5 cards to get to the ace of clubs, or any other card for that matter.

edit: Well, you would have to find the probability of getting the first ace in the first, second, third, ..., all the way up to the 49th position.

Probability it is 1st is 4/52.
2nd is (48/52)(4/51).
3rd is (48/52)(47/51)(4/50).
4th is (48/52)(47/51)(46/50)(4/49).
and so on...

Then multiply each probability with each postion and add up. I'm sure it's around 13. There's probably an easier way to do it. I'm not familiar enough with the subject of waiting times to know it.

edit (again): Running this through R, I get 10.6 as the expected number of cards using my suggestion. This is likely less than 13 due to the fact that the first ace can't occur in the 50, 51, or 52 positions.

Answer 2

26

Answer 3

26

Answer 4

we've 52/4 = 13 "suitable" playing cards and 39 "incorrect" ones. We %. 4 out of the 13 suitable playing cards and 0 out of the 39 incorrect playing cards and in entire we picked 4 out of 52 playing cards: (13nCr4 * 39nCr0) / 52nCr4 = 0.00264 With case in point 5nCr3 = 5! / (3!*2!) = 10 this isn't a similar as your answer, it quite is 0.01056 i don't understand ways you receive that... you won't be able to apply powers right here via the indisputable fact that is without replace, so that you each and every time you draw a million card the full volume is one a lot less so so that you could apply facculties (!)

Answer 5

Havent time to work it out, but it's a very long equation as the probabilty increases with every subsequent card, ie 1/52, then 1/51, 1/50 all the way down to 1/1 for the last card. I'll leave it to others to work out the answer! A good book though is ...

PROBABILITY GUIDE TO GAMBLING: The Mathematics of Dice, Slots, Roulette, Baccarat, Blackjack, Poker, Lottery and Sport Bets

http://www.amazon.com/PROBABILITY-GUIDE-GAMBLING-Mathematics-Blackjack/dp/9738752035/sr=8-6/qid=1160293432/ref=sr_1_6/102-2504526-3715345?ie=UTF8&s=books

Answer 6

26 for any card on average

Answer 7

26 for any card on average. The wait could be longer if you have a slow dealer. lol