Re: A card game probability





On Jan 26, 1:27 am, "Nick" <tulse04-ne...@xxxxxxxxxxx> wrote:
"Rob Johnson" <r...@xxxxxxxxxxxxxx> wrote in messagenews:20070125.165506@xxxxxxxxxxx





In article <UYqdnaf8o8kh0iTYnZ2dnUVZ8sikn...@xxxxxx>,
"Nick" <tulse04-ne...@xxxxxxxxxxx> wrote:
<jsepp...@xxxxxxxxx> wrote in message
news:1169756728.912704.141420@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Take a card deck with 52 cards. Pick cards one by one and compute the
cards by ace, two, three,..., jack, queen, king, ace,..., king, ace,
..., king, ace, ..., king. What is the probability that at least once
you turn a card of the same value as you say aloud? Here ace=1,
jack=11, queen=12, king=13.

Do you just call out any number from 1-13 at random. How are the numbers
that you call out at "random" chosen?

I think that it has been shown that people have a preference for
particular
numbers - so your brain wouldn't be a random number generator - hence the
need for random number tables/computers which do this.

It would depend whether there is an unlimited pool of numbers to call from
(1,..,13) or as someone else has said that the number generator is, in
fact,
a second pack of cards.

The numbers are not called randomly; they are called in sequence, ace
through king, four times.

Thanks

This is then the same as saying what is the probability (nth card =
remainder (n/4))

Huh?

.



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