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February 12th, 2012, 08:31 PM  #1 
Newbie Joined: Feb 2012 Posts: 2 Thanks: 0  Card Game  settle an argument
While playing cards with friends, with 5 players each receiving 1 card each, the question arose, "what are the chances that no player will receive a trump". The trumps being determined by turning over the top card of the undealt deck. (Trump means any card of a particular suit  so if the 5 of clubs was turned over to be the trump suit, the remaining 12 clubs card are known as trumps). Player A said; Its fairly straightforward. After seeing the trump, there are then 51 unknown cards, and 39 of them are nontrumps, so the chance of the first players card not being a trump, is 39/51. And then we have 50 unknown cards, 38 of which are now known to be not trumps, so the second players odds of not getting a trump are 38/50.. and so forth so that we get the chances of none of the 5 players getting a trump as (39x38x37x36x35) / (51x50x49x48x47) Player B said; No, it seems at first like it should be as player A said, but 'Bayes Theorem' shows that things aren't this simple. I cannot remember how it works, but in such a case we can't work out the probability as simply as Player A thinks. So, who, if either is correct? If neither, what is the correct answer. 
February 12th, 2012, 11:27 PM  #2 
Senior Member Joined: Apr 2011 From: USA Posts: 782 Thanks: 1  Re: Card Game  settle an argument
I'm not exactly believing this was a real argument. How many card players know Baye's Theorum? So I'm assuming it's homework and wanting to know what did you try? A is right, and yes, it's that simple. However, I would have done it using combinations: You'll get the same answer doing A's. I don't really know Baye's Theorum. Seems to me when I read it, it was nothing but conditionals  like I solved the example without even reading it but had never put a name to it. I don't know if it encompasses more than I read, but as to conditionals, I don't in any way see how that would fit here. This is just a basic card probability and no different really than figuring out, say, the probability of getting a 5spade hand. So we threw out a card first  just means we start with one card missing and 12 of that suit left, and it makes no difference which is trump. 
February 13th, 2012, 04:57 AM  #3 
Newbie Joined: Feb 2012 Posts: 2 Thanks: 0  Re: Card Game  settle an argument
Thanks a ton for your answer. Actually, yes it was a real argument. We are university educated people in science, but none of us in Maths, and we like to play cards :P My friend (Player B) has clearly heard of Bayes theorem, but doesn't quite know exactly what it is nor when it applies. I was players A, and only did a little probability at Pure Maths Alevel. But I can see that my expression; (39x38x37x36x35) x (51x50x49x48x47) , which i deliberately left crude, is essentially the same as yours expressed in factorials. And I left the explaination posted here the same as the way i described it to other players in the pub. (ie tried to explain it in common sense terms like, 'we can now choose 1 of 39 in a deck of 51 total'. Though I basically know 'how to do it' the way you've mentioned, I'd forgotten if its combinatronics. I also have no idea what Bayes Theorem is, but having heard of more sophisticated theorems can I guess be a little dangerous sometimes. Incidentally the game we were playing is know as "nomination(or estimation) whist" where we have to bid how many tricks we win. In the middle we play a "blind" round, where the players have only 1 card each again, and this time hold it face out on their foreheads for everyone else to see, so now all the other players know what your card is, but you do not know yourself. In this round several of the players seem to believe that if, say, there are 4 players total, and they can see the other 3 do NOT have a trump, then given that out of 4 random cards there will be, on average, 1 trump out (adjusted slightly for the 1 already known trump, but basically about right), so they believe that they themselves MUST have THE trump, or at least, are far more likely to have a trump. Whereas in reality I would see that they still only have .. erm.. 12/48 chance. (12 remaining unseen trumps, and (52  3 player cards seen  1 card turned over for trumps)). These are the people I am dealing with! :P But i wasnt really able to argue well when I'd never even heard of Bayes Theorem. If anyone has a fairly simple explaination of what Bayes Theorem applies to, and hence why it doesnt apply here, that would also be great. 
February 13th, 2012, 07:31 AM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond  Re: Card Game  settle an argument
Perhaps it is a matter of perspective: If someone watching that game, assuming no player is allowed to reveal their card, is asked to give the probability any particular player holds a nontrump, they should answer 39/51. So, from that perspective, the probability in question is (39/51)^5. If a player, Q, is asked to give the probability, again assuming no player is allowed to reveal their card, that any other player holds a nontrump they should answer 39/50 or 38/50, the former if Q holds a trump and the latter if Q doesn't hold a trump. From that perspective, the probability is (38/50)^4 that no other player holds a trump, without any players revealing their cards and given that Q holds a nontrump. So the probability, from Q's perspective, depends on the probability of Q holding a nontrump.

February 13th, 2012, 07:04 PM  #5  
Senior Member Joined: Apr 2011 From: USA Posts: 782 Thanks: 1  Re: Card Game  settle an argument Quote:
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