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lynnl
07-11-2010, 11:45 AM
In view of how we like to butt heads and debate endlessly, issues that don't really matter (:D), I thought some of you might find this interesting.

The following logic problem was in today's Parade Magazine with the Sunday paper. It was following the "Ask Marilyn" column, so I assume it was part of that. At any rate here's the problem:

I shuffle a deck of playing cards, and then place them face down. Then I ask you this question: "If I draw one card at a time from the top, which card drawn is most likely to be the first red queen? Is it the first card? The second? Third? And so on, to the 52nd card?"

Of course it's highly unlikely that you'll be right. But which position has the BEST CHANCE? Or maybe none of them do.

What say ye?

(I'll wait a couple of hours or so before posting the answer given, to permit any ingenious solutions from the gang.)

Evan
07-11-2010, 12:17 PM
There are some unstated conditions in your question. It depends on the initial order and the type of shuffle used. A new deck is in a set and known order. Eight perfect "out" riffle shuffles put the deck back in order. Prestidigitators practise the riffle for hours to take advantage of this. I can usually get it pretty close as I practised card tricks for many hours. One of my first jobs as a youngster was running a Heidelberg press for a magic shop one summer.

If we assume truly random order then all positions have the same probability initially. As cards are removed from the deck that are not a red queen then the chance of picking a red queen next increases but the probability of any particular card position in the deck remains the same.

lynnl
07-11-2010, 12:45 PM
Well those were the only conditions stated. So let's say it was a fairly new deck, but not "brand spanking new", and they'd been shuffled some unknown # of times previously.
Regardless, that's all irrevelant, insofar as the answer given is concerned. (Which I'm not sure I agree with. Certainly not the rationale given.)


...ah what the heck, I'll just go ahead and give the answer they provided:

ANSWER: (verbatim)
<quote>
The first red queen is more likely to occupy the top spot than any other place in the deck. Wrap your brain around this readers: The later the position of the first red queen, the more it confines the area that can be occupied by the second red queen. The second red queen cannot come first. So the chances of the first red queen occupying the 51st position are almost nil.
And, of course, the first red queen can't occupy the last position. <end quote>

aostling
07-11-2010, 12:45 PM
Evan a card sharp? That's the most surprising news of the day, so far.

If we assume that we stop removing cards once the first red queen is uncovered, then the 51st card has a 100% chance of being the first red queen, since none of the prior 50 cards were a red queen. Is that the answer?

[edit] On reflection, my answer is nonsense. Should learn to hold my tongue!

lynnl
07-11-2010, 12:48 PM
No, the question isn't "what the probability is any specific position", but what is the "most likely" position.

...also, that would answer the question "...at or before.."

Evan
07-11-2010, 12:58 PM
In a randomly ordered deck all card positions and sequences are equally likely.

The answer given is inapplicable to the question posed.

IdahoJim
07-11-2010, 01:10 PM
In view of how we like to butt heads and debate endlessly, issues that don't really matter (:D), I thought some of you might find this interesting.

The following logic problem was in today's Parade Magazine with the Sunday paper. It was following the "Ask Marilyn" column, so I assume it was part of that. At any rate here's the problem:

I shuffle a deck of playing cards, and then place them face down. Then I ask you this question: "If I draw one card at a time from the top, which card drawn is most likely to be the first red queen? Is it the first card? The second? Third? And so on, to the 52nd card?"

Of course it's highly unlikely that you'll be right. But which position has the BEST CHANCE? Or maybe none of them do.


What say ye?

(I'll wait a couple of hours or so before posting the answer given, to permit any ingenious solutions from the gang.)

The position with the best chance is obviously 51. Assuming you'd drawn all the others without finding the 1st red queen, the 51st position chances are 100%.
Jim

RancherBill
07-11-2010, 01:12 PM
Without doing the math, the graph of the two probability curves meet in the middle. One is the curve of probability of a miss and the other is the probability of a hit.

So my final answer is 25 or 26(I told you I didn't do the math/graph):)

lynnl
07-11-2010, 01:16 PM
I haven't fully decided if I agree with the top spot.
But I'm with you Evan, I don't think the rationale addresses the question right.

That sentence "So the chances of the first red queen occupying the 51st position are almost nil. ", for example: the "So" implies the sentence is somehow predicated on some preceding argument. But the fact that the 2nd one can't be in spot 1, has nothing to do with the first one being in spot 51.

I can see that postion 51 would require two dependent conditions: 1) the red queen in question be there, and 2) the other red queen follow it. And of course that can occur in two ways. But that still has nothing to do with spot 1.

The only logic that supports it to me is: Ok, position 1 can be occupied by any of 52 cards. And of those, there are two posibilities for a 1st red queen.
For any subsequent positions, there is a diminished possibility for a FIRST red queen, since the other red queen may have preceded it. (I'm open to criticism here...)

Barrington
07-11-2010, 01:19 PM
Let's reduce the problem to something more manageable: five cards of which two are queens.

For this small number we can easily list all the possibile arrangements:-

Q Q x x x
Q x Q x x
Q x x Q x
Q x x x Q
x Q Q x x
x Q x Q x
x Q x x Q
x x Q Q x
x x Q x Q
x x x Q Q

Now we can see that the first queen occurs in the 1st position 4 times, the second position 3 times, the third position 2 times and the 4th position once.

i.e. the first position is the most likely...!

Cheers

.

lynnl
07-11-2010, 01:25 PM
The position with the best chance is obviously 51. Assuming you'd drawn all the others without finding the 1st red queen, the 51st position chances are 100%.
Jim

No, that isn't addressing the question asked. That is addressing the question "what position has the highest probability the 1st red queen will be found 'AT OR BEFORE'.

In other words, I'm gonna offer you 10 bucks if you can correctly predict which spot will hold the FIRST red queen if we start pulling them off the top.
You, naturally want to maximize your chances for scarfing up my money. If you're not right on the correct spot, then you don't get the $10. Of course you realize going in, that the odds are greatly against you, but after all, it isn't costing you anything to try.

Paul Alciatore
07-11-2010, 01:34 PM
No, that isn't addressing the question asked. That is addressing the question "what position has the highest probability the 1st red queen will be found 'AT OR BEFORE'.

In other words, I'm gonna offer you 10 bucks if you can correctly predict which spot will hold the FIRST red queen if we start pulling them off the top.
You, naturally want to maximize your chances for scarfing up my money. If you're not right on the correct spot, then you don't get the $10. Of course you realize going in, that the odds are greatly against you, but after all, it isn't costing you anything to try.

Excuse me, but you didn't say "or before" in the original post. This makes a difference. With "or before" then the answer is the 51th card as by the time you get there the chances are 100% that you will have gotten to the first red queen (assuming 52 cards - no jokers). But for the original statement of the problem, then the answer you give from the magazine and Barrington's analysis holds and it is the first card that has the highest probability of being the "first red queen".

aostling
07-11-2010, 01:34 PM
Now we can see that the first queen occurs in the 1st position 4 times, the second position 3 times, the third position 2 times and the 4th position once.
i.e. the first position is the most likely...!


Impeccable logic, and easily extended to the full deck. Well done!

lynnl
07-11-2010, 01:43 PM
Excuse me, but you didn't say "or before" in the original post. This makes a difference. With "or before" then the answer is the 51th card as by the time you get there the chances are 100% that you will have gotten to the first red queen (assuming 52 cards - no jokers). But for the original statement of the problem, then the answer you give from the magazine and Barrington's analysis holds and it is the first card that has the highest probability of being the "first red queen".


No I didn't. Because that's NOT the problem as stated. My reply to someone else was simply intended to point that out.

Yep, Barrington offers the simplest way of looking at it.
I think my logic .... <quote>The only logic that supports it to me is: Ok, position 1 can be occupied by any of 52 cards. And of those, there are two posibilities for a 1st red queen.
For any subsequent positions, there is a diminished possibility for a FIRST red queen, since the other red queen may have preceded it. (I'm open to criticism here...) <end quote>
... is the same thing, but Barrington's is much clearer.

dp
07-11-2010, 01:50 PM
In view of how we like to butt heads and debate endlessly, issues that don't really matter

This pretty much nailed it.

aostling
07-11-2010, 02:40 PM
The following was going around the internet a few weeks ago, after the death of Martin Gardner. It is the Two Sons problem, with a twist: http://www.sciencenews.org/view/generic/id/60598/title/When_intuition_and_math_probably_look_wrong

Black_Moons
07-11-2010, 02:45 PM
Its the same probability for each spot...
Because I play with a 4 color deck: http://www.push-poker.com/shop/art/cards/Modiano4Color.jpg
(There is no second red queen)

Ray Sidell
07-11-2010, 04:01 PM
[QUOTE=Barrington]Let's reduce the problem to something more manageable: five cards of which two are queens.

For this small number we can easily list all the possibile arrangements:-

Q Q x x x
Q x Q x x
Q x x Q x
Q x x x Q
x Q Q x x
x Q x Q x
x Q x x Q
x x Q Q x
x x Q x Q
x x x Q Q

What if you list the same possiblities as below?

xxxQQ
xxQxQ
xQxxQ
QxxxQ
xQQxx
xQxQx
xQxxQ
xxQQx
xxQxQ
xxxQQ

Evan
07-11-2010, 04:10 PM
Barrington's logic is flawed because he used an odd number of cards. Of course it will produce a bias. The real deck has an even number of cards.

Note also that if your turn the deck over the last position is equally likely.

2 red cards=0

2 black cards=1

1100
0011
0101
1010
1001
0110

aostling
07-11-2010, 05:18 PM
What if you list the same possiblities as below?

xxxQQ
xxQxQ
xQxxQ
QxxxQ
xQQxx
xQxQx
xQxxQ
xxQQx
xxQxQ
xxxQQ

That is a bogus list, because 1st = 10th, 2nd = 9th, 3rd = 7th. Each combination must be unique.

aostling
07-11-2010, 05:22 PM
Barrington's logic is flawed because he used an odd number of cards. Of course it will produce a bias.

Write out Barrington's combinations, for two red queens in six cards. It shows that a red Queen will be the first card 5 times, the second card 4 times, the third card 3 times, the four card 2 times, the fifth card once. So the logic holds, with an even number of cards in the deck.

Barrington
07-11-2010, 05:25 PM
Barrington's logic is flawed because he used an odd number of cards. Of course it will produce a bias.Here's a 'deck' of 8 cards - the argument is no different:-

Position ->>
1 2 3 4 5 6 7 8

Q Q x x x x x x
Q x Q x x x x x
Q x x Q x x x x
Q x x x Q x x x
Q x x x x Q x x
Q x x x x x Q x
Q x x x x x x Q
x Q Q x x x x x
x Q x Q x x x x
x Q x x Q x x x
x Q x x x Q x x
x Q x x x x Q x
x Q x x x x x Q
x x Q Q x x x x
x x Q x Q x x x
x x Q x x Q x x
x x Q x x x Q x
x x Q x x x x Q
x x x Q Q x x x
x x x Q x Q x x
x x x Q x x Q x
x x x Q x x x Q
x x x x Q Q x x
x x x x Q x Q x
x x x x Q x x Q
x x x x x Q Q x
x x x x x Q x Q
x x x x x x Q Q

7 in position 1, 6 in position 2, 5 in position 3 etc.


Note also that if your turn the deck over the last position is equally likely. - But if you turn the deck over the 'last position' is the one you encounter first !

<<- Position
8 7 6 5 4 3 2 1

Q Q x x x x x x
Q x Q x x x x x
Q x x Q x x x x
Q x x x Q x x x
Q x x x x Q x x
Q x x x x x Q x
Q x x x x x x Q
x Q Q x x x x x
x Q x Q x x x x
x Q x x Q x x x
x Q x x x Q x x
x Q x x x x Q x
x Q x x x x x Q
x x Q Q x x x x
x x Q x Q x x x
x x Q x x Q x x
x x Q x x x Q x
x x Q x x x x Q
x x x Q Q x x x
x x x Q x Q x x
x x x Q x x Q x
x x x Q x x x Q
x x x x Q Q x x
x x x x Q x Q x
x x x x Q x x Q
x x x x x Q Q x
x x x x x Q x Q
x x x x x x Q Q

The bold Q is the first Q encountered going 'backward through the deck.
Same frequencies as above - 7 in the first position encountered, 6 in the second etc.

Cheers

.edit: Sorry, cross posted...

IdahoJim
07-11-2010, 05:29 PM
No, that isn't addressing the question asked. That is addressing the question "what position has the highest probability the 1st red queen will be found 'AT OR BEFORE'.

In other words, I'm gonna offer you 10 bucks if you can correctly predict which spot will hold the FIRST red queen if we start pulling them off the top.
You, naturally want to maximize your chances for scarfing up my money. If you're not right on the correct spot, then you don't get the $10. Of course you realize going in, that the odds are greatly against you, but after all, it isn't costing you anything to try.

I think what threw me off was the "one card at a time". If you are asking which card, given that you can only draw once, it has to be the top card. That card is the only one with a 100% chance of being above the 2nd red queen.
Jim

lynnl
07-11-2010, 07:23 PM
Look at it this way:

If you consider any of the 51 positions (the 1st red Q can't be in #52), then for all positions OTHER THAN #1, two (2) conditions must be met: 1) a red Q must be there, and 2) the other red Q must be later or deeper in the stack.
For postion #1, only the condition 1) must be met, the other condition is automatically satisfied.

lynnl
07-11-2010, 07:30 PM
I think what threw me off was the "one card at a time". If you are asking which card, given that you can only draw once, it has to be the top card. That card is the only one with a 100% chance of being above the 2nd red queen.
Jim

Yeahbut..., that's true, but you aren't asked to pick a position with 100%. Only to pick the most likely. Even tho it is the only spot with, as you say, 100% chance of being above the 2nd red Q, it still only has a 1/52 chance of being a red Q. Any other spot also has a 1/52 chance of having a red Q, but to meet the stated requirement, that 1/52 is further reduced by the additional stipulation that it be FOLLOWED by the other red Q.

lynnl
07-11-2010, 07:39 PM
The following was going around the internet a few weeks ago, after the death of Martin Gardner. It is the Two Sons problem, with a twist: http://www.sciencenews.org/view/generic/id/60598/title/When_intuition_and_math_probably_look_wrong


I looked at that discussion. I think some people are full of horse hockey.
If someone tells me he has two children, and that one is a son, that leaves two, and only two, possibilities for the other child. And, ignoring any slight statistical M/F birth rate difference, the probability the other child in question is either M or F is 50%. Doesn't matter what day either was born on.
The probability of multiple independent events is the product of their individual probabilities, so in the case: 100% X 50%. And in this neck of the woods that computes to be 50%.

ieezitin
07-11-2010, 07:54 PM
who cares?

lynnl
07-11-2010, 08:01 PM
who cares?

Obviously not everyone. But some do. Some enjoy puzzles and problems, even though there's no practical purpose served.

That's true of lots of topics. Not everyone cares.

Carld
07-11-2010, 08:14 PM
The probability of it being probable is probably not as probable as you probably would think it probable so it's not probable at all as you observe the probabilities of it.:rolleyes:

aostling
07-11-2010, 09:30 PM
I think some people are full of horse hockey.



"What you have told us is rubbish. The world is really a flat plate supported on the back of a giant tortoise." The scientist gave a superior smile before replying, "What is the tortoise standing on?" "You're very clever, young man, very clever", said the old lady. "But it's turtles all the way down!"

SmoggyTurnip
07-12-2010, 10:32 AM
The probability that the first red queen appears as the X'th card is:

(52 - X)/1326.

aostling
07-12-2010, 12:17 PM
The probability that the first red queen appears as the X'th card is:

(52 - X)/1326.

That passes first-order sanity checks. How did you figure it out?

lost_cause
07-12-2010, 12:26 PM
i can't answer this question, as i'm not playing with a full deck.

SmoggyTurnip
07-13-2010, 09:57 AM
That passes first-order sanity checks. How did you figure it out?
In order to get the first red queen on the X'th draw you first need to draw (X-1) cards that are not queens.
There are 50 cards that are not red queens and 52 cards in the deck.
The probability that the first card is not a red queen is (50/52).
for the next card there are only 51 cards left of which 49 are not red queens now so the probability that is not a red queen is (49/51)
for the next card there are only 50 cards left of which 48 are not red queens now so the probability that is not a red queen is (48/50)
etc...
Then we need to draw a red queen. Since there are 2 red queens and (53-X) cards left we have a probability of 2/(53-X) for this event.
So the probability of these events happening in succession is the product of the probabilities of each individual event.
This gives:

(50/52)*(49/51)*(48/50)*(47/49)*...*((52-X)/(54-X)) * 2/(53-X)

50 * 49 * 48 * 47 * ... *(53-X) * (52-X) * 2
= --------------------------------------------------------------------
52 * 51 * 50 * 49 * 48 * 47 * ... * (54-X) *(53-X)


(52-X) * 2
= -----------
52 * 51

=(52-X)/(1326)

Tony Ennis
07-13-2010, 12:26 PM
Without reading any other replies...

The trick is that the cards are revealed sequentially and there are two equivalent queens.

The 51st card is most likely. The reason is that either queen has an equal probability of being in any of the 52 positions. But once you hack your way down to the 51st position, 51 *must* be a queen. And so must 52. But since we'll never turn 52 over, we know 51 is the winnah.

I totally made that up. :D

edit - I agree that my answer is shyte :-D

Another way to think about it, or perhaps a restatement of what has been said...

Chance of Q1 and Q2 being shuffled at 51 and 52 is 1/1326. (Note I said "shuffled" not "discovered by turning cards over." Once we start turning cards over, we change the odds. That's what's confusing us. This is similar to the "goat" problem* in that respect.)

Chance of Q1 being shuffled at 1: 2/52. Chance of Q2 being shuffled below it: 51/51. Chance of red Q on top = 1/26.

Here's the full table of probabilities. The first column is the card position, the next is the number of combinations where that card is the first red queen, and the percentage of combinations where the first queen is in this position.

1 51 3.85
2 50 3.77
3 49 3.70
4 48 3.62
5 47 3.54
6 46 3.47
7 45 3.39
8 44 3.32
9 43 3.24
10 42 3.17
11 41 3.09
12 40 3.02
13 39 2.94
14 38 2.87
15 37 2.79
16 36 2.71
17 35 2.64
18 34 2.56
19 33 2.49
20 32 2.41
21 31 2.34
22 30 2.26
23 29 2.19
24 28 2.11
25 27 2.04
26 26 1.96
27 25 1.89
28 24 1.81
29 23 1.73
30 22 1.66
31 21 1.58
32 20 1.51
33 19 1.43
34 18 1.36
35 17 1.28
36 16 1.21
37 15 1.13
38 14 1.06
39 13 0.98
40 12 0.90
41 11 0.83
42 10 0.75
43 9 0.68
44 8 0.60
45 7 0.53
46 6 0.45
47 5 0.38
48 4 0.30
49 3 0.23
50 2 0.15
51 1 0.08
52 0 0.00


-=-=-=-

* goat problem: You're playing a game show. There are three curtains. The host tells you that there is a car behind one curtain and goats behind the other 2. Let's assume you want the car, not the goats :D
Q1: You choose a curtain. What are your chances of selecting the curtain hiding the car?
Q2: After you make your choice, but before your curtain is opened, the host reveals a goat behind one of the other curtains. Now there are two curtains, your curtain and the other. He asks you if you'd like to change your mind and choose the other curtain instead. Do you?

aostling
07-13-2010, 01:01 PM
In order to get the first red queen on the X'th draw you first need to draw (X-1) cards that are not queens.


Turnip,

I often find combinatorics tough. But your exposition is easy to follow, and puts the whole problem in perspective.

Are you a mathematician, as well as machinist?

SmoggyTurnip
07-13-2010, 01:12 PM
Turnip,

I often find combinatorics tough. But your exposition is easy to follow, and puts the whole problem in perspective.

Are you a mathematician, as well as machinist?

No I am a computer programmer and electronic tech. Math was once a hobby of mine, now machining is. Maybe someday I will work on spelling. I am guessing you have a bit of experience with math since you had the patience to read through my explaination.

lynnl
07-13-2010, 01:19 PM
I follow the first part, and the decrementing process. But I'm not following the introduction of the numbers 53 and 54. Where did that come from?

aostling
07-13-2010, 01:37 PM
I follow the first part, and the decrementing process. But I'm not following the introduction of the numbers 53 and 54. Where did that come from?

Write Turnip's products of fractions, first for X=3, then for X=4. You will see that most numerators cancel the denominators. The numbers 53 and 54 in the formula will be apparent.

SmoggyTurnip
07-13-2010, 01:47 PM
I follow the first part, and the decrementing process. But I'm not following the introduction of the numbers 53 and 54. Where did that come from?

(53-X) is the denominator in the last term from step 1 (drawing (X-1) cards with out getting a red queen.
So if you want to hit a red queen on the 1st card the denominator is 53-1 or 52. on the second card it is 53-2 or 51 etc.

(54-X) is the denominator in the second last term from step 1.
(55-X) is the denominator in the third last term from step 1.
etc.

for example if X is 5

we have:

50*49*48*47*46 *2
----------------------------
52*51*50*49*48 *47

= 46/1326.

The probability that the first card is the first red queen is 51/1326
The probability that the second card is the first red queen is 50/1326
The probability that the third card is the first red queen is 49/1326
.
.
.
The probability that the 50th card is the first red queen is 2/1326
The probability that the 51st card is the first red queen is 1/1326
The probability that the 52st card is the first red queen is 0/1326

lynnl
07-13-2010, 01:56 PM
Yeah, I picked up on the canx'ing of numerators and denominators right away, but the 53 didn't jump right out at me. (still doesn't)

I majored in math, and was very good at it (member of the math honor society), but that was a Loooonnng time ago, and other than further class requirements in the early years, my math skills have mostly lain dormant.

RancherBill
06-07-2012, 08:35 PM
The following logic problem was in today's Parade Magazine with the Sunday paper. It was following the "Ask Marilyn" column, so I assume it was part of that. At any rate here's the problem:

I shuffle a deck of playing cards, and then place them face down. Then I ask you this question: "If I draw one card at a time from the top, which card drawn is most likely to be the first red queen? Is it the first card? The second? Third? And so on, to the 52nd card?"

It is the 51st card. The probability of the first card is 3.846% (2/52), the second is 3.921% (2/51) etc. On the the fiftyfirst first draw there are only red queens left so on that draw the probability is 100% .

Fasttrack
06-08-2012, 01:00 AM
:) This problem was on my "Probability and Statistics for Scientists and Engineers" first exam several years ago. Barrington and SmoggyTurnips nailed it. I can fish out my exam or my text book and scan it for anyone who doubts...

For those of you that say it is the 51st position ... you have reduced the problem to one of only two cards, both of which are red queens. Thus, the first card will be the first red queen with 100% probability. That is not the question! The question is which position is most likely given an entire deck. As the original solution stated, a queen in the first position is the least constrained arrangement and thus the most likely. It requires only that one card be in one location. In the case so many of you have argued for (i.e. the 51st position) we make a statement about TWO cards, requiring that these have specific locations relative to each other AND that the doublet has a specific location relative to the deck.


A more concise version of what I just wrote: The problem with the 51st card argument is that you building a specific scenario. If I just give you a random deck of cards, you have no way of knowing where any cards are. At that point, the first card has the highest probability of being the first red queen. Once you draw one card, the problem changes and now the second card becomes the most likely - in this way you continue drawing cards and changing the problem until you either draw the first red queen or you end up at the situation where you know with absolute certainty that the first red queen is next. However, that is no longer the 51st position but rather the first position of two cards. Understand? Instead of thinking about this problem as if you were actually drawing cards, imagine that you take the deck and spread it out so that you have a line of cards, all face down and ordered 1 through 52. Now you simultaneously flip over all the cards. Record the position of the first red queen. Now does this an infinite number of times and you'll find that the statistics converge to the probabilty given by SmoggyTurnip. If you do the experiment a more realistic number of times ... say 100, you'll find that first red queen appears in the first position most often, but the statistics will be slightly off from the probability given by SmoggyTurnip.

Fasttrack
06-08-2012, 01:09 AM
Q2: After you make your choice, but before your curtain is opened, the host reveals a goat behind one of the other curtains. Now there are two curtains, your curtain and the other. He asks you if you'd like to change your mind and choose the other curtain instead. Do you?

Yes! Absolutely.

A.K. Boomer
06-08-2012, 09:53 AM
I don't think there is any true answer to this,,,

Im just not seeing it, But --- here's my hillbilly way of thinking for what it's worth and this is what I would try,

Lets just say there's only one red queen in the entire 52 card deck --- where is it going to lie in the order of things?, I don't know what to tell you other than your possibility of hitting it right off is just as good as not, so split the difference and the number is the 26th card...

Now add two red queens - this doesn't mean you cut the 26 number in half - it means you divide 52 by 3 and the magical card # is 17.333333333333333333333333

so in rounding that out to the closest possibility I chose card 17:D

that's my story and im sticking to it..

SmoggyTurnip
06-08-2012, 12:09 PM
Without reading any other replies...

* goat problem: You're playing a game show. There are three curtains. The host tells you that there is a car behind one curtain and goats behind the other 2. Let's assume you want the car, not the goats :D
Q1: You choose a curtain. What are your chances of selecting the curtain hiding the car?
Q2: After you make your choice, but before your curtain is opened, the host reveals a goat behind one of the other curtains. Now there are two curtains, your curtain and the other. He asks you if you'd like to change your mind and choose the other curtain instead. Do you?

Yes of course you change your mind because the ONLY way you can lose is if you picked the car on your first guess which has a probability of 1/3.

george4657
06-08-2012, 03:54 PM
The probability first card is a red queen is 2/52 = 3.845%
If card is a red queen then the probability it is the first red queen is 100%
So net probability first card is red queen is 1/52 = 3.845%

The probability second card is a red queen is also 2/52 = 3.845%
However there is a chance it is the second red queen as the first could also be a red queen.
The chance of the first 2 being red queens is 2/52 * 1/51 = .075%
This is a net of 3.77% second card is first red queen.

The chance for third card being first red queen is reduced even further as first or second could also be red queen.

The chance of each card being the first red queen reduces for every card till you get to the last card where the probability is Zero of being first red queen.

In answer to question first card has the highest probability. But remember the probability for any card is the same.

George