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Cut the Cards
07-30-2020, 08:58 PM
Post: #2
RE: Cut the Cards
(07-30-2020 08:00 PM)David Hayden Wrote:  Choose a card at random from a deck of 52 and put the card back.
What is the average number of choices required to see all 52 cards?

First card picked must never been seen, thus only 1 try needed to get 1st card.
Probability of getting the next "new" card = 51/52, or averaged 52/51 tries to get 2nd card
...

We can get averaged picks either way:

XCas> sum(1/t, t=1 .. 52) * 52.0                 → 235.978285436
XCas> int((t^52-1)/(t-1), t=0 .. 1) * 52.0    → 235.978285436

→ averaged 236 picks to see all 52 cards
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Messages In This Thread
Cut the Cards - David Hayden - 07-30-2020, 08:00 PM
RE: Cut the Cards - Albert Chan - 07-30-2020 08:58 PM
RE: Cut the Cards - Albert Chan - 08-21-2020, 11:00 PM
RE: Cut the Cards - Jim Horn - 07-30-2020, 09:49 PM
RE: Cut the Cards - John Keith - 07-31-2020, 12:24 AM
RE: Cut the Cards - Gerson W. Barbosa - 08-24-2020, 01:57 PM
RE: Cut the Cards - Albert Chan - 08-25-2020, 06:14 PM
RE: Cut the Cards - Albert Chan - 07-30-2020, 10:21 PM
RE: Cut the Cards - pinkman - 08-24-2020, 09:49 PM
RE: Cut the Cards - Gerson W. Barbosa - 08-25-2020, 11:41 PM
RE: Cut the Cards - Albert Chan - 08-26-2020, 03:06 AM
RE: Cut the Cards - Gerson W. Barbosa - 08-26-2020, 08:23 AM
RE: Cut the Cards - Albert Chan - 08-26-2020, 02:13 PM
RE: Cut the Cards - Gerson W. Barbosa - 08-26-2020, 06:13 PM
RE: Cut the Cards - Gerson W. Barbosa - 08-27-2020, 10:07 PM
RE: Cut the Cards - Albert Chan - 08-28-2020, 09:26 PM
RE: Cut the Cards - Albert Chan - 08-29-2020, 04:02 PM
RE: Cut the Cards - Gerson W. Barbosa - 08-28-2020, 11:39 PM
RE: Cut the Cards - Albert Chan - 06-23-2021, 12:08 AM



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