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Topic: Turning Cards (Read 1038 times) |
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ThudnBlunder
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Turning Cards
« on: Nov 5th, 2003, 7:26am » |
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A standard 52-card deck is well shuffled and the cards are turned over one-by-one.
What is the expected number of cards needed to be turned over for the following to appear (in whichever order)?
a) A '2'
b) A '2' and a '3'
c) A '2', a '3', and a '4'
d) One of each of the 13 denominations (2, 3, 4, 5,....J, Q, K, and A)
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| « Last Edit: Nov 5th, 2003, 10:11pm by ThudnBlunder » |
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Noke Lieu
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Re: Turning Cards
« Reply #1 on: Mar 23rd, 2004, 10:16pm » |
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Oooh, this one went unnoticed. Or maybe everyone ran away from it.
expected number? as in probabity has risen to an acceptably high level?
Part One would presumably be about 48... 
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towr
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Re: Turning Cards
« Reply #2 on: Mar 24th, 2004, 12:50am » |
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a)
[sum]49i=1 i * [48!/(49-i)!]/[52!/(53-i)!] * 4/(53-i)=53/5
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kellys
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Re: Turning Cards
« Reply #3 on: Mar 24th, 2004, 1:29am » |
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Noke Lieu: If P(X=k) is the probability of a random variable X being equal to k, then the expected value of X is defined to be [sum] k*P(k), where the sum is taken over all possible values k of X.
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Noke Lieu
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Re: Turning Cards
« Reply #4 on: Mar 24th, 2004, 9:49pm » |
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Thanks.
Live and learn- probability and I haven't been on speaking terms for some time now....
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Barukh
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Re: Turning Cards
« Reply #5 on: Mar 25th, 2004, 5:28am » |
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Wonderful puzzle, THUD&BLUNDER!
It's tempting to use random indicator variables in such kind of puzzles. Trying this, I've got for a) the same answer as towr's, which also correlates well with the simulations.
However, for b) I've got [smiley=square.gif]13.711[smiley=square.gif], which is way off from the simalated value [smiley=blacksquare.gif]~15.3[smiley=blacksquare.gif]
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towr
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Re: Turning Cards
« Reply #6 on: Mar 25th, 2004, 7:29am » |
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for b)
:: 689/45 ~= 15.3111
pretty much same method as above
first wait till you have one of the eight cards, then untill you have one of the four you still need.
::
for c), similarwise, I get
:: 3551/195 ~= 18.2103 ::
It's a bit bothersome to extend this to 13 though, and I assume there is a better way.. (Which, however, I do not yet know)
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Barukh
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Re: Turning Cards
« Reply #7 on: Mar 25th, 2004, 9:35am » |
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on Mar 25th, 2004, 7:29am, towr wrote:
Incredible! Can you please explain how did you manage to simplify the expressions so nicely? When I tried to write down several terms of your formula for a), I've got a rather complicated series...
BTW, after fixing the bug in my reasoning in b), I've got [smiley=blacksquare.gif]4823/315 = 15.311...[smiley=blacksquare.gif] - the same as yours. 
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towr
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Re: Turning Cards
« Reply #8 on: Mar 25th, 2004, 10:26am » |
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I let Derive5 help me in simplifying the formulas..
But the first one can at least be simplified to
[sum](i(50 - i)(i - 51)(i - 52)/1624350, i, 1, 49)
And that's 'just' the sum of a few sums of the form a[sum]ip (p=0..4) which afaik all have some closed form (which I don't know from the top of my head)
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| « Last Edit: Mar 25th, 2004, 10:30am by towr » |
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aeongirrl
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Re: Turning Cards
« Reply #9 on: Mar 26th, 2004, 7:03pm » |
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nine
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Barukh
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Re: Turning Cards
« Reply #10 on: Mar 27th, 2004, 9:36am » |
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Here's how I approached this problem.
[smiley=blacksquare.gif]
a) Let X be a random variable equal to the number of cards turned before the first '2' appeared. Clearly, these may be 48 cards other than '2'-s. Associate with every card ci ( i = 1,…48 ) a random variable (indicator) Xi, which equals 1 if ci appears before the first '2', and 0 otherwise. Then, X = [sum]i=148 Xi, so the expected value E(X) = [sum] E(Xi) = 48E(Xi). But E(Xi) = Prob{Xi = 1}, and this equals 1/5, because there 5! ways to order five cards (ci and four '2'-s), and in 4! cases ci comes first.
Thus, the sought expected value is E(X) + 1 = 53/5.
b) In a similar manner, but here things get more complicated… First, determine the number X of 44 cards other than '2'-s and '3'-s that are turned before the pair appeared. In this case, Prob{Xi = 1} = (8! + 8! + 2C42!6!2 + 3C43!5!2 + 4!4!2) / 9! = 91/315(do you see why?). To this, we need to add an expected number of cards of one denomination (e.g. '2') before another denomination ('3') appears, and this equals 3/5.
Thus, the sought expected value is 44x111/315 + 3/5 + 2 = 4823/315.
[smiley=blacksquare.gif]
Unfortunately, this approach is not "scalable": even for 3 denominations it's too clumsy, I am certain to get all the 13 one needs another method… 
I would like THUD&BLUNDER to comment on this.
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SWF
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Re: Turning Cards
« Reply #11 on: Mar 29th, 2004, 6:47pm » |
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Barukh, I like your "random indicator variable" method. I never heard of that before. For the final case, here is how I did it with a spreadsheet. Towr can probably use his Derive5 program to find the exact fraction:
Define p(i,j)= probability of having i different denominations after having dealt the first j cards.
Start with p(0,0)=1, and when j>0 p(0,j)=0.
After j cards are dealt, there are two ways of having i different denominations having had appeared:
1) There were already i denominations after card number j-1, and of the (53-j) possible values for card j, one of the 4*i+1-j cards of the already drawn denominations was drawn.
2) There were i-1 denominations after card number j-1, and of the (53-j) possible values for card j, one of the 4*(14-i) cards of a new denomination were drawn.
Combining those possiblities gives:
p(i,j) = p(i,j-1)*(4*i+1-j)/(53-j) + p(i-1,j-1)*4*(14-i)/(53-j)
With a spreadsheet a 14 by 52 table can be filled by using the starting values and copying that formula into the cells. Probability of the 13th and final denomination being drawn on card number N is p(13,N)-p(13,N-1), and that allows expected value to be computed. It turns out to be about 27.9971.
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Barukh
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Re: Turning Cards
« Reply #12 on: Mar 30th, 2004, 5:27am » |
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SWF, that's a nice solution! It is in fact iterative. I was also thinking about iterations, but didn't arrive at anythng satisfactory...
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