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#1
04-14-2005, 12:50 AM
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Re: once the cards are shuffled, arent the odds just the same
[ QUOTE ]
0%, because once you've removed the A [img]/images/graemlins/spade.gif[/img] there is no A [img]/images/graemlins/spade.gif[/img] left in the deck. [/ QUOTE ] i'm pretty sure you're A) missing the question and B) wrong... first of all, read the original question and you'll see that this is obviously with replacement... secondly, and pay attention to this one, if you draw the A[img]/images/graemlins/spade.gif[/img] and replace it, and truly shuffle the deck to the point that it is random again and independent of the previous draw / shuffle, the odds of drawing the A[img]/images/graemlins/spade.gif[/img] is the same every time... it should theoretically work the same as independent trials... 
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#2
04-14-2005, 01:04 AM
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Re: once the cards are shuffled, arent the odds just the same
[ QUOTE ]
[ QUOTE ] 0%, because once you've removed the A [img]/images/graemlins/spade.gif[/img] there is no A [img]/images/graemlins/spade.gif[/img] left in the deck. [/ QUOTE ] i'm pretty sure you're A) missing the question and B) wrong... first of all, read the original question and you'll see that this is obviously with replacement... secondly, and pay attention to this one, if you draw the A[img]/images/graemlins/spade.gif[/img] and replace it, and truly shuffle the deck to the point that it is random again and independent of the previous draw / shuffle, the odds of drawing the A[img]/images/graemlins/spade.gif[/img] is the same every time... it should theoretically work the same as independent trials... [/ QUOTE ] im prety sure youre both wrong regarding terrys question
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#3
04-14-2005, 04:14 AM
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Re: once the cards are shuffled, arent the odds just the same
[ QUOTE ]
[ QUOTE ] 0%, because once you've removed the A [img]/images/graemlins/spade.gif[/img] there is no A [img]/images/graemlins/spade.gif[/img] left in the deck. [/ QUOTE ] i'm pretty sure you're A) missing the question and B) wrong... [/ QUOTE ] Just a wild guess but from the tone of the posts I think he actually might be C) joking :P
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#4
04-14-2005, 08:50 AM
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Re: once the cards are shuffled, arent the odds just the same
sorry i wasnt specific.. once the hand is OVER and the cards are RESHUFFLED.. i shuffle the deck, the top card is A [img]/images/graemlins/spade.gif[/img], i put it back INTO the deck, RESHUFFLE, the odds are still 1 in 52... im coming from the angle of i get QQ one hand, the hand is over, the next deal is again, QQ
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#5
04-14-2005, 10:08 AM
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Re: once the cards are shuffled, arent the odds just the same
Assuming the shuffle randomises the deck, the odds are the same. Why would they be any different..?
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#6
04-14-2005, 12:05 PM
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Re: once the cards are shuffled, arent the odds just the same
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Assuming the shuffle randomises the deck, the odds are the same. Why would they be any different..? [/ QUOTE ] I'm guessing he got into an argument with someone who doesn't understand what random means, and thinks there's an even distribution - i.e., if you get AA one hand, you shouldn't see it again for about another 220 hands. Which is of course, silly. So yeah, if you get AA ten hands in a row, the odds of getting it on the 11th hand are still 220-1. However, the odds of getting AA 11 times in a row (before you've seen even 1 AA come) are 58431830141132800000000000 to 1. Approximately. [img]/images/graemlins/smile.gif[/img]
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#7
04-14-2005, 10:35 PM
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you are right
You are absolutely right. Once you have the A [img]/images/graemlins/spade.gif[/img], for the next game it's exactly 1 in 52. Only if you want to know the probability of getting it two times in a row, without knowing what will happen the first time, than you have to calculate 52 x 52 = 2704.
All speculations about so-called "emperical probabilities" are nothing but wishful thinking, otherwise all casinos would have already gone bancrupt!
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#8
04-14-2005, 11:05 PM
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Re: you are right
[ QUOTE ]
You are absolutely right. Once you have the A [img]/images/graemlins/spade.gif[/img], for the next game it's exactly 1 in 52. Only if you want to know the probability of getting it two times in a row, without knowing what will happen the first time, than you have to calculate 52 x 52 = 2704. All speculations about so-called "emperical probabilities" are nothing but wishful thinking, otherwise all casinos would have already gone bancrupt! [/ QUOTE ] Whoops, you are right. I added them instead of multiplying (52+52=104 whereas 52x52=2704) Thus 1/2704=.0003, which means you'd have to take a card from the deck and reshuffle after each time 3 million times before you'd pull out one twice in a row. [img]/images/graemlins/tongue.gif[/img]
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#10
04-15-2005, 01:35 AM
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Re: you are right
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Almost there. What are the odds? Not the probability [percent of times it will happen] – what are the odds? [/ QUOTE ] what are the odds that no matter what anyone responds there'll be someone who doesn't agree? hmmm....
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