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From: jimg@diane.csg.mot.com (Jim Goedken)
Subject: Re: 3-prisoners problem
Reply-To: jimg@diane.csg.mot.com
Organization: Motorola Inc., Cordless
Date: Thu, 20 Oct 1994 15:18:27 GMT
Message-ID: <1994Oct20.151827.17915@schbbs.mot.com>
References: <MALONE.94Oct13164900@julia.asahi-kasei.co.jp>
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In article 94Oct13164900@julia.asahi-kasei.co.jp, malone@julia.asahi-kasei.co.jp (Declan Malone) writes:
 
> The second includes more knowledge about the outcomes, so it would be
> like saying
> 
>     A card is picked at random from a pack of cards. What is the
>     probability that the card is a queen, GIVEN that it is not the
>     Queen of Hearts?
> 
> This is obviously 3/52. In standard terms it would be written as

Actually the chances are 3/51.  

 
> 2 I've picked a card, then the dealer takes another card and shows it
>   to me: it's the Queen of Hearts. What's the probability, P(Y), that
>   I now have a Queen?
> 
>     At this point, the chance that I have a Queen in my hand is
> 
>     P(Y)=P(Queen\Not-queen-of-hearts) = 3/52
> 


Again it is 3/51.  

Imagine that the delear showed you ALL the other cards in the deck except 1 queen.
What the probability that you are holding a queen?  1/1.








