A blind man doing a visual trick?
0 views
Skip to first unread message
Mickey
May 4, 2010, 7:58:02 AM5/4/10
to nextgen_engg
A blind man is handed a deck of 52 cards and told that exactly 10 of
these cards are facing up. How can he divide the cards into two piles,
not necessarily of equal size, with each pile having the same number
of cards facing up?
It is upto the blind man to create the pile, all you get is two piles
with the said property. He does not take anyone else's help and there
is no way for him to find the orientation of any card.
these cards are facing up. How can he divide the cards into two piles,
not necessarily of equal size, with each pile having the same number
of cards facing up?
It is upto the blind man to create the pile, all you get is two piles
with the said property. He does not take anyone else's help and there
is no way for him to find the orientation of any card.
sachithanandam karthikselvan
May 4, 2010, 9:29:01 PM5/4/10
to nextge...@googlegroups.com
Still I am learning probability so there can be gaps in my reasoning.
You can entirely reject my reasoning or close the gaps. 10 facing
cards can be divided in to two parts. In this case each pile will
contain 5 facing up cards.
The prob. to select a first card which is facing up for pile 1= 10/52
The prob. to select a 2nd card which is facing up for pile 1= 10/52 *
9/51 (Probability without replacement)
The prob. to select a 3rd card which is facing up for pile 1= 10/52 *
9/51 * 8/50
The prob. to select a 4th card which is facing up for pile 1= 10/52 *
9/51 * 8/50 * 7/49
The prob. to select a 5th card which is facing up for pile 1= 10/52 *
9/51 * 8/50 * 7/49 * 6/48 = x
Now the blind man can stop adding cards for pile 1 since there is no
need to have equal cards for each pile.
Hence the prob. for adding zero cards for pile 1 after adding 5 face
up cards = x * (0/(47-5) 'Cause other 5 face up cards is allocated for
pile 2.
I used intuition here. Multiplaction will make the prob. zero. so I am
using addition.
So the prob. for adding zero cards for pile 1 after adding 5 face up
cards = x + (0/42)
The prob. for adding 1 card for pile 1 after adding 5 face up cards =
x * (0/42 + 1/42)
The prob. for adding upto 42 card for pile 1 after adding 5 face up
cards = x * (0/42 +1/42 + 2/42 + 3 /42 + ... +42/42)
There is a mistake in my reasoning. I hope someone will correct my
mistakes with better reasoning;-)
Final prob. = (10/52 * 9/51 * 8/50 * 7/49 * 6/48) * (0/42 +1/42 +
2/42 + 3 /42 + ... +42/42)
Thanks
Karthikselvan
Trivial comment: You guys give answer without proper reasoning. Please
give the answer with proper reasoning in English statements.
You can entirely reject my reasoning or close the gaps. 10 facing
cards can be divided in to two parts. In this case each pile will
contain 5 facing up cards.
The prob. to select a first card which is facing up for pile 1= 10/52
The prob. to select a 2nd card which is facing up for pile 1= 10/52 *
9/51 (Probability without replacement)
The prob. to select a 3rd card which is facing up for pile 1= 10/52 *
9/51 * 8/50
The prob. to select a 4th card which is facing up for pile 1= 10/52 *
9/51 * 8/50 * 7/49
The prob. to select a 5th card which is facing up for pile 1= 10/52 *
9/51 * 8/50 * 7/49 * 6/48 = x
Now the blind man can stop adding cards for pile 1 since there is no
need to have equal cards for each pile.
Hence the prob. for adding zero cards for pile 1 after adding 5 face
up cards = x * (0/(47-5) 'Cause other 5 face up cards is allocated for
pile 2.
I used intuition here. Multiplaction will make the prob. zero. so I am
using addition.
So the prob. for adding zero cards for pile 1 after adding 5 face up
cards = x + (0/42)
The prob. for adding 1 card for pile 1 after adding 5 face up cards =
x * (0/42 + 1/42)
The prob. for adding upto 42 card for pile 1 after adding 5 face up
cards = x * (0/42 +1/42 + 2/42 + 3 /42 + ... +42/42)
There is a mistake in my reasoning. I hope someone will correct my
mistakes with better reasoning;-)
Final prob. = (10/52 * 9/51 * 8/50 * 7/49 * 6/48) * (0/42 +1/42 +
2/42 + 3 /42 + ... +42/42)
Thanks
Karthikselvan
Trivial comment: You guys give answer without proper reasoning. Please
give the answer with proper reasoning in English statements.
Mickey
May 4, 2010, 11:08:18 PM5/4/10
to nextgen_engg
Thinking along these lines, I think the calculation of the probability
should stop at 10/52 *9/51 * 8/50 * 7/49 * 6/48. The blind guy has
made up two piles of 5 and 47 cards respectively. Taking out any more
cards will only decrease the probability.
I cannot think of any other way of breaking the deck that can ensure
any higher probability.
On the other hand this question has an answer other than what we can
gather from probability, going over the question once again:
"with each pile having the same number of cards facing up."
Let us think he can flip the cards. Of course he doesn't know if it
was facing up or down but he can flip. And this gives him the power to
exactly satisfy the above property. He can make two piles from the
given deck that satisfies the above property with certainty. So how
does he form the piles now and when does he apply this flip operation?
Regards,
Jyoti
On May 5, 6:29 am, sachithanandam karthikselvan
should stop at 10/52 *9/51 * 8/50 * 7/49 * 6/48. The blind guy has
made up two piles of 5 and 47 cards respectively. Taking out any more
cards will only decrease the probability.
I cannot think of any other way of breaking the deck that can ensure
any higher probability.
On the other hand this question has an answer other than what we can
gather from probability, going over the question once again:
"It is upto the blind man to create the pile, all you get is two piles
with the said property. He does not take anyone else's help and there
is no way for him to find the orientation of any card."
And the property is:
with the said property. He does not take anyone else's help and there
is no way for him to find the orientation of any card."
"with each pile having the same number of cards facing up."
Let us think he can flip the cards. Of course he doesn't know if it
was facing up or down but he can flip. And this gives him the power to
exactly satisfy the above property. He can make two piles from the
given deck that satisfies the above property with certainty. So how
does he form the piles now and when does he apply this flip operation?
Regards,
Jyoti
On May 5, 6:29 am, sachithanandam karthikselvan
SUDHEER DURUSOJU
May 5, 2010, 10:49:23 AM5/5/10
to nextge...@googlegroups.com
Just to complete the probability calculation. I can think of like this.
10/52 *9/51 * 8/50 * 7/49 * 6/48 is to pick up 5 cards facing up to pile 1. Now we need to make sure that remaining 5 facing up cards are not picked up for pile 1.
Probability to not picking up first card facing up for pile 1 : (47-5)/47 = 42/47
Probability to not picking up second card facing up for pile 1 : (46-4)/46 = 42/47*42/46
Probability to not picking up third card facing up for pile 1 : (45-3)/45 = 42/47*42/46*42/45
Probability to not picking up forth card facing up for pile 1 : (44-2)/44 = 42/47*42/46* 42/45*42/44
Probability to not picking up fifth card facing up for pile 1 : (43-1)/43 = 42/47*42/46*42/45* 42/44*42/43
Now the total probability satisfying the property is
(Probability to pick up first 5 cards facing up) * (Probability to not picking up remaining 5 cards facing up)
(10/52 *9/51 * 8/50 * 7/49 * 6/48) * (42/47*42/46*42/45* 42/44*42/43)
-Sudheer
10/52 *9/51 * 8/50 * 7/49 * 6/48 is to pick up 5 cards facing up to pile 1. Now we need to make sure that remaining 5 facing up cards are not picked up for pile 1.
Probability to not picking up first card facing up for pile 1 : (47-5)/47 = 42/47
Probability to not picking up second card facing up for pile 1 : (46-4)/46 = 42/47*42/46
Probability to not picking up third card facing up for pile 1 : (45-3)/45 = 42/47*42/46*42/45
Probability to not picking up forth card facing up for pile 1 : (44-2)/44 = 42/47*42/46* 42/45*42/44
Probability to not picking up fifth card facing up for pile 1 : (43-1)/43 = 42/47*42/46*42/45* 42/44*42/43
Now the total probability satisfying the property is
(Probability to pick up first 5 cards facing up) * (Probability to not picking up remaining 5 cards facing up)
(10/52 *9/51 * 8/50 * 7/49 * 6/48) * (42/47*42/46*42/45* 42/44*42/43)
-Sudheer
Mickey
May 5, 2010, 1:38:57 PM5/5/10
to nextgen_engg
This would be when he is planning to make two piles of 10 and 42 first
five terms ensure that five cards that face up are included and the
next five terms ensure that those cards are not facing up. If he is
making pile of 5 and 47 the probability should be (10/52 *9/51 * 8/50
* 7/49 * 6/48) only.
Regards,
Jyoti
With the flipping operation permitted he can achieve the final
constraint "How can he divide the cards into two piles, not
necessarily of equal size, with each pile having the same number of
cards facing up?" Exactly. With a probability of 1.
On May 5, 7:49 pm, SUDHEER DURUSOJU <sdurus...@gmail.com> wrote:
> Just to complete the probability calculation. I can think of like this.
>
> 10/52 *9/51 * 8/50 * 7/49 * 6/48 is to pick up 5 cards facing up to pile 1.
> Now we need to make sure that remaining 5 facing up cards are not picked up
> for pile 1.
>
> Probability to not picking up first card facing up for pile 1 : (47-5)/47 =
> 42/47
>
> Probability to not picking up second card facing up for pile 1 : (46-4)/46 =
> 42/47*42/46
>
> Probability to not picking up third card facing up for pile 1 : (45-3)/45 =
> 42/47*42/46*42/45
>
> Probability to not picking up forth card facing up for pile 1 : (44-2)/44 =
> 42/47*42/46* 42/45*42/44
>
> Probability to not picking up fifth card facing up for pile 1 : (43-1)/43 =
> 42/47*42/46*42/45* 42/44*42/43
>
> Now the total probability satisfying the property is
>
> (Probability to pick up first 5 cards facing up) * (Probability to not
> picking up remaining 5 cards facing up)
>
> (10/52 *9/51 * 8/50 * 7/49 * 6/48) * (42/47*42/46*42/45* 42/44*42/43)
>
> -Sudheer
>
five terms ensure that five cards that face up are included and the
next five terms ensure that those cards are not facing up. If he is
making pile of 5 and 47 the probability should be (10/52 *9/51 * 8/50
* 7/49 * 6/48) only.
Regards,
Jyoti
With the flipping operation permitted he can achieve the final
constraint "How can he divide the cards into two piles, not
necessarily of equal size, with each pile having the same number of
cards facing up?" Exactly. With a probability of 1.
On May 5, 7:49 pm, SUDHEER DURUSOJU <sdurus...@gmail.com> wrote:
> Just to complete the probability calculation. I can think of like this.
>
> 10/52 *9/51 * 8/50 * 7/49 * 6/48 is to pick up 5 cards facing up to pile 1.
> Now we need to make sure that remaining 5 facing up cards are not picked up
> for pile 1.
>
> Probability to not picking up first card facing up for pile 1 : (47-5)/47 =
> 42/47
>
> Probability to not picking up second card facing up for pile 1 : (46-4)/46 =
> 42/47*42/46
>
> Probability to not picking up third card facing up for pile 1 : (45-3)/45 =
> 42/47*42/46*42/45
>
> Probability to not picking up forth card facing up for pile 1 : (44-2)/44 =
> 42/47*42/46* 42/45*42/44
>
> Probability to not picking up fifth card facing up for pile 1 : (43-1)/43 =
> 42/47*42/46*42/45* 42/44*42/43
>
> Now the total probability satisfying the property is
>
> (Probability to pick up first 5 cards facing up) * (Probability to not
> picking up remaining 5 cards facing up)
>
> (10/52 *9/51 * 8/50 * 7/49 * 6/48) * (42/47*42/46*42/45* 42/44*42/43)
>
> -Sudheer
>
SUDHEER DURUSOJU
May 5, 2010, 1:58:43 PM5/5/10
to nextge...@googlegroups.com
Not just 10 and 42. The probability i gave will work for any number more than 10 cards in
Pile 1. Anyone can correct me if I am thinking wrong !
-Sudheer.
Pile 1. Anyone can correct me if I am thinking wrong !
-Sudheer.
SUDHEER DURUSOJU
May 5, 2010, 3:52:38 PM5/5/10
to nextge...@googlegroups.com
Thinking more about the blind man's situation ....he can solve the problem like this.
He can flip every other card while making up the 2 piles and also flip the piles while adding cards.
1 card straight to pile 1
2nd card flipped to pile 2
3rd card straight to pile 2
4th flipped to pile 1
5th straight to pile1
6th flipped to pile 2
........
----
As we have 10 cards ( even number ) facing up in the total cards 52 ( even number ) we can get the cards that are facing up equally distributed between pile 1 and 2.
Each pile will get 26 cards. 13 flipped and 13 straight.
In a set of fully closed set of cards, we will get 13 cards facing up in each pile. But as we have 10 cards initially faced up the no of cards that are facing up will change depending on their position in the original set. A bit lazy to calculate the original number....Definitely each pile will have same no.of cards facing up!!
-Sudheer
He can flip every other card while making up the 2 piles and also flip the piles while adding cards.
1 card straight to pile 1
2nd card flipped to pile 2
3rd card straight to pile 2
4th flipped to pile 1
5th straight to pile1
6th flipped to pile 2
........
----
As we have 10 cards ( even number ) facing up in the total cards 52 ( even number ) we can get the cards that are facing up equally distributed between pile 1 and 2.
Each pile will get 26 cards. 13 flipped and 13 straight.
In a set of fully closed set of cards, we will get 13 cards facing up in each pile. But as we have 10 cards initially faced up the no of cards that are facing up will change depending on their position in the original set. A bit lazy to calculate the original number....Definitely each pile will have same no.of cards facing up!!
-Sudheer
Mickey
May 6, 2010, 1:51:20 AM5/6/10
to nextgen_engg
That is quite involved and i couldn't quite get it work for me. Does
flipping piles mean switching them or flipping all the cards present
in that pile? What if I all the up cards at position 1, 5, ... They
all will land up in pile 1, right?
How about this:
Divide the deck in two piles. One of 10 and the other of 42. Now flip
all the cards in pile of 10 cards. done.
Regards,
Jyoti
On May 6, 12:52 am, SUDHEER DURUSOJU <sdurus...@gmail.com> wrote:
> Thinking more about the blind man's situation ....he can solve the problem
> like this.
>
> He can flip every other card while making up the 2 piles and also flip the
> piles while adding cards.
>
> 1 card straight to pile 1
> 2nd card flipped to pile 2
> 3rd card straight to pile 2
> 4th flipped to pile 1
> 5th straight to pile1
> 6th flipped to pile 2
> ........
> ----
>
> As we have 10 cards ( even number ) facing up in the total cards 52 ( even
> number ) we can get the cards that are facing up equally distributed between
> pile 1 and 2.
>
> Each pile will get 26 cards. 13 flipped and 13 straight.
>
> In a set of fully closed set of cards, we will get 13 cards facing up in
> each pile. But as we have 10 cards initially faced up the no of cards that
> are facing up will change depending on their position in the original set. A
> bit lazy to calculate the original number....Definitely each pile will have
> same no.of cards facing up!!
>
> -Sudheer
>
flipping piles mean switching them or flipping all the cards present
in that pile? What if I all the up cards at position 1, 5, ... They
all will land up in pile 1, right?
How about this:
Divide the deck in two piles. One of 10 and the other of 42. Now flip
all the cards in pile of 10 cards. done.
Regards,
Jyoti
On May 6, 12:52 am, SUDHEER DURUSOJU <sdurus...@gmail.com> wrote:
> Thinking more about the blind man's situation ....he can solve the problem
> like this.
>
> He can flip every other card while making up the 2 piles and also flip the
> piles while adding cards.
>
> 1 card straight to pile 1
> 2nd card flipped to pile 2
> 3rd card straight to pile 2
> 4th flipped to pile 1
> 5th straight to pile1
> 6th flipped to pile 2
> ........
> ----
>
> As we have 10 cards ( even number ) facing up in the total cards 52 ( even
> number ) we can get the cards that are facing up equally distributed between
> pile 1 and 2.
>
> Each pile will get 26 cards. 13 flipped and 13 straight.
>
> In a set of fully closed set of cards, we will get 13 cards facing up in
> each pile. But as we have 10 cards initially faced up the no of cards that
> are facing up will change depending on their position in the original set. A
> bit lazy to calculate the original number....Definitely each pile will have
> same no.of cards facing up!!
>
> -Sudheer
>
> On Wed, May 5, 2010 at 1:58 PM, SUDHEER DURUSOJU <sdurus...@gmail.com>wrote:
>
> > Not just 10 and 42. The probability i gave will work for any number more
> > than 10 cards in
> > Pile 1. Anyone can correct me if I am thinking wrong !
>
> > -Sudheer.
>
>
> > Not just 10 and 42. The probability i gave will work for any number more
> > than 10 cards in
> > Pile 1. Anyone can correct me if I am thinking wrong !
>
> > -Sudheer.
>
SUDHEER DURUSOJU
May 6, 2010, 10:59:31 AM5/6/10
to nextge...@googlegroups.com
That's a simple and perfect solution.
I spent lot of time thinking to solve this. Flipping piles means switching them. The method gave by me felt like working. But I was not able to do imagination with 52 cards. I tried now and it won't work.
--Sudheer
I spent lot of time thinking to solve this. Flipping piles means switching them. The method gave by me felt like working. But I was not able to do imagination with 52 cards. I tried now and it won't work.
--Sudheer
Reply all
Reply to author
Forward
0 new messages