[Guest]

I haven't been able to figure out the solution to this, so I thought I'd see if anyone out there can help!

Start with a standard deck of 52 playing cards, face down, in a random order. One at a time, turn the cards over. As you do so, count out loud in the following order: ace, 2, 3, 4, all the way up to jack, queen, king, ace, 2, 3, 4, all the way up to king...four times. That is, you count in order while you flip over cards.

The question is: What is the probability that you can make it all the way through the 52-card deck WITHOUT the flipped card matching the spoken card?

For example, let's say the first ten cards in the deck are K, 6, 8, Q, Q, J, 9, 8, 10, 3. You begin by flipping the first card, a king, and saying "Ace." The next card flipped is 6, and you say "two." Third card is 8, and you say "three." Fourth is queen, you say "four." Keep going. However, when you get to the 8th card, you flip over an 8 and say "eight." Since the flipped card and the spoken card are the same, you lose.

[Aaryan]

The chance of the flipped card BEING in that spot in the deck is 4/52. After that card is gone, the probably of the next card being in that spot is 3/51, then 2/50, then 1/49. That is true for every number of card. So the formula is Number of that numbered card still not taken from the deck/Number of cards in the deck.[/red]. Don't know where to go from there.

[Guest]

The probability of subsequent cards NOT matching what you're saying doesn't go down to 3/51, etc. When you're "looking" for an ace the first time, the probability of the card you turn over being an ace is 4/52. Which means that the probability of it NOT being an ace is 48/52. When you go to the second card, you're no longer looking for an ace -- now. you're looking for a two. So, the probability of THAT one being a two is 4/51, meaning that the probability of it NOT being a two is now 48/51. That's as far as I've thought it out so far. And, please tell me if my thinking is completely off. :-)

[Guest]

I like these starts. What's been tripping me up is the following. For the first card, the probability of it NOT being an ace is 48/52. For the second card, though, doesn't the probability depend on what the first card ended up being? For example, the probability of the second card NOT being a 2 is 48/51 UNLESS the first card flipped was a 2. Then the probability for the second card is 47/51. There's a lot going on here.

[BobbyGo]

Instead of flipping each card one at a time, what if we laid them down side by side all in a row. The chance of any specific card not being the "named" card (the one spoken aloud) would be 48/52, or about 92.308%. The probability of any two cards should then be (48/52) * (48/52), or about 85.207%. Continued for all 52 cards would give roughly a 1.557% chance of success.

Seems a little low to me, but that's where my thoughts are at right now.

[superprismatic]

I ran a simulation of 10⁹ trials and the

number of times the simulation got through

the whole deck was 16,226,616 which means

that the probability is near .01623.

[plainglazed]

am thinking the order is not relevent so consider if the first four cards are not an ace, the second four cards are not a two, etc. should be the same odds as BobbyGo came up with?

[Guest]

I like these starts. What's been tripping me up is the following. For the first card, the probability of it NOT being an ace is 48/52. For the second card, though, doesn't the probability depend on what the first card ended up being? For example, the probability of the second card NOT being a 2 is 48/51 UNLESS the first card flipped was a 2. Then the probability for the second card is 47/51. There's a lot going on here.

I was thinking this same thing, then my head started to hurt. So, I stopped thinking about it. Maybe later I'll track it out for each and every card.

[Guest]

haha very good question.

[Guest]

BobbyGo, I like your explanation. And the fact that it matches the probability from the simulation well is a good sign.

[Aaryan]

I like these starts. What's been tripping me up is the following. For the first card, the probability of it NOT being an ace is 48/52. For the second card, though, doesn't the probability depend on what the first card ended up being? For example, the probability of the second card NOT being a 2 is 48/51 UNLESS the first card flipped was a 2. Then the probability for the second card is 47/51. There's a lot going on here.

Since the Card you drew doesn't gonback in the deck, the next draw is a totally different probability, as the denominator(total cards) decreases.

I like BobbyGo's answer.

[Guest]

Instead of flipping each card one at a time, what if we laid them down side by side all in a row. The chance of any specific card not being the "named" card (the one spoken aloud) would be 48/52, or about 92.308%. The probability of any two cards should then be (48/52) * (48/52), or about 85.207%. Continued for all 52 cards would give roughly a 1.557% chance of success.

Seems a little low to me, but that's where my thoughts are at right now.

I agree most of your insights.

However, do you think probability of 1st card and 2nd card independent? P(AB)=P(A)*P(B)

[Guest]

I think this right: (12 / 13)^52 = 0.0155729351

There is a 1/13 chance of any card being a match. You do it 52 times.

Edited June 1, 2011 by NoMinorChords

[EventHorizon]

I think this right: (12 / 13)^52 = 0.0155729351

There is a 1/13 chance of any card being a match. You do it 52 times.

That's equivalent to BobbyGo's approximation. It's a pretty good approximation based on superprismatic's simulation, but not exact.

On a more interesting note, thinking about this problem resulted in me rediscovering a multidimensional extension of Pascal's triangle. Too bad it was discovered at least 20 years ago (and the idea is probably a lot older than that).

[Guest]

I will go with BobbyGo's approach. In fact the two events - "The cards are dealt and flipped successively", and "Cards are first dealt and then turned up" are the same. Lets see mathematically. Find the probability that the second card is "2":

Case-1 After the 1st card is dealt, then the 2nd: there are 2 possibilities

(i) 1st card is "2" and the 2nd card is also "2" - probability = (4/52)*(3/51)

(ii) 1st card is not "2" and 2nd card is "2" - probability = (48/52)*(4/51)

Therefore, probability of a "2" in the 2nd position = (4/52)*(3/51) + (48/52)*(4/51) = 4/52

Case-2 All cards are distributed. Irrespective of the 1st and other cards, probability of the 2nd card being a "2" = 4/52.

You see, both cases are identical

[Guest]

(48/52)^52 = 0.015573

See the argument above.

[Guest]

Shouldn't the probability depend upon the cards not being on the specific spot?

So, an Ace shouldn't be in the spots 1, 14, 27, 40. Similarly, a two shouldn't be in the spots 2, 15, 28, 41; and so on.

Now, if we were to place the cards in order:

For each Ace the possible arrangements are: ⁴⁸P₄

For each two the possible arrangements are: ⁴⁴P₄

...

For each King the possible arrangements are: ⁴P₄

And we know that the possible arrangements for a deck are: 52!

So, overall probability should be: (⁴⁸P₄ * ⁴⁴P₄ * ... * ⁴P₄) / 52!

= 48! / 52! = 1 / ⁵²P₄ = 1 / 6497400 = 0.00002%

[Guest]

Mukul Verma, thanks for going through that calculation. I never expected both probabilities to be the same, regardless of whether the first card is a two or isn't a two. I'm convinced!

[Guest]

Since the Card you drew doesn't gonback in the deck, the next draw is a totally different probability, as the denominator(total cards) decreases.

I like BobbyGo's answer.

You could look at it that way, or you could consider it without conditional probability, such as what is the probability that the first card will be an ace, or the 2nd card will be a two etc., and as you are flipping through the deck you are just "checking" not eliminating cards.

Then the probability would be something like 4/52 to the 52 power?

[Guest]

BobbyGo is correct.

[superprismatic]

BobbyGo's explanation doesn't take into account

some dependencies, For example, if the first 4

kings are drawn in the first 12 draws, we are

guaranteed never to get a match with a king.

So the factor of 48/52 should only be used

when we face a king in the case where no king

has been drawn yet.

Here's a small example with a 4-long deck:

1,2,3,4. Now, using BobbyGo's method, we get

a probability of not matching of (3/4)⁴ which

is approximately .32. But, actually listing

all 24 ways that the cards could be shuffled,

we can see that 10 of these do not produce

a match. So, the actual probability is 10/24

which is about .42, a bit bigger than .32.

So, an exact calculation requires one to consider

all dependencies. I can't figure out how to do

this yet, but it might be able to be done with

a recursive calculation.

[Guest]

(12 / 13)^52 = 0.0155729351 This is because there is a 1 in 13 chance of a match on any card and you try it 52 times. Which is about what Bobby Go got.

[Guest]

(4^52)/(52!) or roughly 2.5 x 10 ^ -37) - not very good odds

[Guest]

quote name='psycho' date='01 June 2011 - 06:53 PM' timestamp='1306950825' post='284440']

(4^52)/(52!) or roughly 2.5 x 10 ^ -37) - not very good odds

[Guest]

The answer is quite simple - the probability of the card matching is 100%. To prove it, I just opened a brand new deck of cards. Don't waste time shuffling. Sure 'nuff, they came up in the correct order!