Four Aces

Lawrh

16 years ago

Algebra in the OT forum?

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KenThompson

16 years ago

Just taking a quick break from writing, Alexa?

Ken

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Bjarne Eldhuset 16 years ago

I'm no math expert, but why would it matter how many times you do this?

If there are 4 aces out of 52 cards, that means 1 of 13 cards is an ace. So I guess my answer is 13.

Or?
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KenThompson

16 years ago

I always hated probability and statistics problems in math. lol.

So, ok... the probability of drawing an ace out of a deck of 52
cards is 4/52 = 0.0769.

So the avg number of cards would 8, rounding up, before seeing
an ace. It doesn't matter if you do it 10 mil times because the
probability remains constant.

That's me out on a limb.

Ken

Ok, I realize you said off the top, and I said 'drawing.' Not sure how
that figures in. But since the frequency is 10 mil, then I'll venture
a guess that the two statements, off the top and drawing, will
converge since the frequency is so high.

That's a wag - wild ass guess.

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Bjarne Eldhuset

16 years ago

But if 1 card gives 8% probability, you'd need 13 cards to get 100%

Then again, may the answer is that your hands will fall off long before you manage to do this 10,000,000 times.

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KenThompson

16 years ago

Originally Posted by Bjarne Eldhuset

But if 1 card gives 8% probability, you'd need 13 cards to get 100%

Then again, may the answer is that your hands will fall off long before you manage to do this 10,000,000 times.

It's drawing 1 (out of possibly 4) from 52 total that gives that
probability.

Ken

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Alexa Smith

Banned 16 years ago

Originally Posted by Bjarne Eldhuset

why would it matter how many times you do this?

The sample size has to be substantial for the average to be realistic, no?

Originally Posted by Bjarne Eldhuset

I guess my answer is 13.

You're suggesting that if you just took one suit out of the cards (spades, say) and did the same thing with those 13 cards, the average number of cards you'd turn over would be 13? So the Ace would for some reason always find its way to the bottom, however you shuffled? :confused:

Remind me not to play cards with you!

In the "one-suit example", I'd expect the answer to be 7 (it has an equal chance of being any number from 1 to 13 inclusive, and the average of those numbers is 7). But the "four-suit example" is a very different question.

Originally Posted by KenThompson

the probability of drawing an ace out of a deck of 52
cards is 4/52 = 0.0769.

So the avg number of cards would 8, rounding up, before seeing an ace.

It's the "so" I don't understand, Ken: I think you're answering a different question from the one asked? Where is the question asked related to the probability of drawing an Ace out of a deck of 52 cards in one attempt (which I do agree is 1/13 or 4/52)?

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Bjarne Eldhuset

16 years ago

You're suggesting that if you just took one suit out of the cards (spades, say) and did the same thing with those 13 cards, the average number of cards you'd turn over would be 13? So the Ace would for some reason always find its way to the bottom, however you shuffled?

No, I meant to say that you had to turn over 13 cards to be pretty sure that there was 1 ace among those cards, but it would not necessarily be card number 13

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Bjarne Eldhuset 16 years ago

In the "one-suit example", I'd expect the answer to be 7 (it has an equal chance of being any number from 1 to 13 inclusive, and the average of those numbers is 7).

If the average is 7, and there are 4 aces, shouldn't you on average have found all aces after turning 28 cards?
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Alexa Smith

Banned 16 years ago

Originally Posted by Bjarne Eldhuset

No, I meant to say that you had to turn over 13 cards to be pretty sure that there was 1 ace among those cards

I'm with you now.

This is undeniable, but it wasn't the question: the question was "What's the average number of cards you'll turn over before seeing an Ace?"

Originally Posted by Bjarne Eldhuset

If the average is 7, and there are 4 aces, shouldn't you on average have found all aces after turning 28 cards?

Actually it's quite a bit more, but this, again, is a different question. The one I'm asking is easier.

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Bjarne Eldhuset

16 years ago

Well, you got me reading up on probability.

Check out the example on this page:
http://en.wikipedia.org/wiki/Event_(probability_theory)

So, I still think the answer is 13.

And I don't think there is any average number of cards, only a constant probability.

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Alexa Smith

Banned 16 years ago

Originally Posted by Bjarne Eldhuset

And I don't think there is any average number of cards

How do you mean?

I'm just asking, if you did it a huge number of times, and saw each time how many cards you had to turn over before you saw an Ace, and wrote that number down each time, and at the end added them all up and divided by the number of times you'd done it, what would the average be?

There can't "not be" an average at all? :confused:

And the average, over enough samples, is surely going to be the same for everyone, isn't it?

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TimPhelan

16 years ago

My guess is the average has to be 13. I can't see it being any other number.

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Alexa Smith

Banned 16 years ago

Originally Posted by TimPhelan

My guess is the average has to be 13. I can't see it being any other number.

You don't think that there'll sometimes be an Ace among the first few cards, then? Just sometimes? That would surely bring the average down quite a bit?

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TimPhelan

16 years ago

Sure. Sometimes it might show up on the very first card. Or on the 49th card. But the other 12 cards would have the same probability. The average wouldn't come down if it shows in the first few cards, because there will be times it will take many more than 13 to get the first ace.

Originally Posted by Alexa Smith

You don't think that there'll sometimes be an Ace among the first few cards, then?

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Alexa Smith

Banned 16 years ago

Originally Posted by TimPhelan

Sometimes it might show up on the very first card. Or on the 49th card. But the other 12 cards would have the same probability. The average wouldn't come down if it shows in the first few cards, because there will be times it will take many more than 13 to get the first ace.

Yes, I see what you mean there. It isn't the answer, though, I promise ...

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TimPhelan 16 years ago

Hmm. OK, since the earliest ace possible is the first and the latest ace possible is the 49th, then perhaps you divide 49 by 4 instead of 52 by 4. That would equal 12.25. You are talking about a new deck each time right?

Originally Posted by Alexa Smith

Yes, I see what you mean there. It isn't the answer, though, I promise ...
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- Bjarne Eldhuset 16 years ago
  
  No offence taken, I just want to know the answer
  
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Bjarne Eldhuset 16 years ago

If the first card you draw is an ace, you have 51 cards left, and 3 of those
are aces.

The probability of getting an ace each time you draw a card after that now is 3/51 = around 5,8 percent.

(on a side note, I don't have a clue if using percents is "allowed")

When you start on the deck again, the chances are back to 4/52 = 1/13.
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- Alexa Smith Banned 16 years ago
  
  It's not a probability puzzle. It just looks like a probability puzzle. There's an easier solution. Not an "easy" solution, but much easier than delving into probability theory.
  
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Kay King 16 years ago

Of course sometimes an ace will be the first card - but it's also true that all four aces might be the last four cards.

All you have is probability and that's the best anyone could do without running 10,000 samples. Some might run two samples of 1000 tries each and average. Casinos do both on computers with new games - just to test the probability.

kay
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- Alexa Smith Banned 16 years ago
  
  Originally Posted by Kay King
  
  All you have is probability
  
  Er ...
  
  It's really not a probability puzzle.
  
  It's one of those "kick yourself" questions - to some extent - when you see the answer, because it becomes very clear that it's pretty unarguable. (And one can test and prove it experimentally too, more easily and quickly with a little computer program than with physical cards, of course, but the answer comes out exactly the same either way).
  
  No trickery, I promise - not a "trick question" in any sense.
  
  And not probability theory, either.
  
  My comments above, in response to the suggested answers, were aimed at being helpful. Sorry if they came across the wrong way.
  
  PS Why is everyone assuming that the answer's going to be a round number? "Averages" aren't normally whole numbers, are they?
  
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garyv

16 years ago

It depends on your definition of "shuffle well". If I shuffle a deck well, I can make one appear the first time every time.

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Bjarne Eldhuset 16 years ago

Originally Posted by garyv

It depends on your definition of "shuffle well". If I shuffle a deck well, I can make one appear the first time every time.

Haha, remind me not to play cards with you
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Frank Donovan

16 years ago

Originally Posted by Alexa Smith

If you do this 10,000,000 times,

Yikes! If it takes an average of 30 seconds to shuffle and draw the first ace, that's around nine and a half years of my life - with no sleep

I might lose count

Seriously, I wonder if the clue lies in the thread title. Four aces.

As there are four of every card in each pack, we are really only concerned with odds of 1 in 13, given that each card has a 1 in 13 chance of being drawn.

Therefore, I would expect the average for any card, including an ace, to be around 6.5.

Frank

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Alexa Smith Banned 16 years ago

Originally Posted by Frank Donovan

If it takes an average of 30 seconds to shuffle and draw the first ace, that's around nine and a half years of my life - with no sleep

Good point, and I'd hate to be responsible for that!

A little computer program's much faster, for an "experimental proof", for someone who knows "Visual Basic" or whatever people use for things like this (I'm way too incompetent and technophobic to know!).

Originally Posted by Frank Donovan

Seriously, I wonder if the clue lies in the thread title. Four aces.

No, sorry - not trying to help/mislead anyone with the title: I just couldn't think what to put in the subject-line! Ignore it.

Originally Posted by Frank Donovan

Therefore, I would expect the average for any card, including an ace, to be around 6.5.

This was my instinctive guess, too. It actually turns out to be higher, though. Strangely.

I was wondering if we have Warriors with a math degree here (I'm sure I remember one or two saying so in that huge long thread where people were saying what they did at college?), who would solve it in 3 seconds ...

Originally Posted by KenThompson

But right now... I'm going out for coffee.

I'm going out for a while, too ... see you later.
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KenThompson

16 years ago

I always hated test questions that included the comment, "It is
self evident." lol

And I wondered if I was straying with the mention of 'average' and
then my over-engineering with probability.

But right now... I'm going out for coffee. I'm having one of those days
when you have a ton of work to do, and you just can't seem to get
started. One of those days...

I'll give this more thought when I return.

Ken

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Kurt

16 years ago

To be totally accurate, you'd use Baye's formula for probability, which is the same one used by casinos, insurance companies, NASA, etc:
Bayesian probability - Wikipedia, the free encyclopedia

People are correct that there are 4/52 chance for the first card, and as each card is dealt, the odds increase.

If there were only 13 cards, one of each suit, then the answer would be (1/13/)/2 or 6.5 cards. However, you can't apply this ratio for the situation given.

Now here's where we get in trouble using straight ratios and not Baye...There's virtually an unlimited number of combinations for the distribution of Aces, so it's really impossible to calculate without Baye's.

By the way, as a card counter it isn't all that rare to go 13 cards without seeing an ace. Counters keep track of these things.

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TimPhelan 16 years ago

OK, I'll give it another try since I guess 12.25 isn't the answer.

All 13 types of cards will have the same average obviously. So, what you do is take 100% and divide it by 13 and you get 7.6923076.... :-)

Edit: I realized this was flawed after I posted it and went to the gym. Right during my bar dips I said to myself "That's BS"
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KimW

16 years ago

You might want to ask this guy:

Persi Diaconis - Wikipedia, the free encyclopedia

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Bill Farnham

16 years ago

I'm reserving this spot so that after the answer is given I can come back and edit this post to include the right answer thus giving me the appearence that I was the first to get the answer right and so therefore I must have miraculous powers of deduction only granted to a few chosen humans by the very Gods that created all knowledge in the first place.

Especially the "deck 'o cards" Gods...

~Bill

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Kurt

16 years ago

Originally Posted by Bill Farnham

I'm reserving this spot so that after the answer is given I can come back and edit this post to include the right answer thus giving me the appearence that I was the first to get the answer right and so therefore I must have miraculous powers of deduction only granted to a few chosen humans by the very Gods that created all knowledge in the first place.

Especially the "deck 'o cards" Gods...

~Bill

Go for it Bill!

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John Henderson

16 years ago

If there were only one ace in Alexa's pack of 52 cards, there would be an equal chance of that ace being in any position from card 1 to card 52. The average of these over 10 million shuffles would be position 26; half way thru the pack.

If there were two aces in the pack of 52 cards, there's now an increased chance of me finding one of them sooner. Over 10 million shuffles, I'm not going to have to wait to get 1/2 way thru the pack -- I suspect that it might average out at 1/3 way thru the pack.

So if this line of reasoning is correct... with four aces in the pack, the average number of cards that will have to be turned over to find an ace is 1/(4+1) or 1/5 of the pack: 52/5 = 10.4 cards.

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Alexa Smith

Banned 16 years ago

Originally Posted by John Henderson

If there were only one ace in Alexa's pack of 52 cards, there would be an equal chance of that ace being in any position from card 1 to card 52. The average of these over 10 million shuffles would be position 26; half way thru the pack.

Indeed.

Originally Posted by John Henderson

If there were two aces in the pack of 52 cards, there's now an increased chance of me finding one of them sooner. Over 10 million shuffles, I'm not going to have to wait to get 1/2 way thru the pack -- I suspect that it might average out at 1/3 way thru the pack.

So if this line of reasoning is correct... with four aces in the pack, the average number of cards that will have to be turned over to find an ace is 1/(4+1) or 1/5 of the pack: 52/5 = 10.4 cards.

You win the prize for the closest answer by far, John, and for your method of solving it, which is basically right. (The prize is a deck of cards, Aces removed, which I'll deliver next time I'm down in Sussex).

I'll explain the answer. Please excuse a long-winded explanation but I think it makes it clear, and it's the one that was offered to me by my math lecturer neighour who originally tried this puzzle out on me.

Forget cards; forget probability.

I have a long, narrow glass rod which is 4-feet long and I drop it on the floor and it breaks cleanly at 4 different points into 5 pieces. What's the average length of the pieces? (This is actually the same problem as the cards, above).

Clearly their average length is 9.6 inches, because 48 inches have been cut up into 5 pieces, and 48/5 = 9.6.

So, the 48 inches of the glass rod are the 48 remaining non-Ace cards in a deck of cards (52-4) after the Aces have been removed. If you remove the 4 Aces and then insert them randomly ("shuffled") that's mathematically (and physically, in practice) the same as inserting 4 breaks into the glass rod, and it divides what's left into 5 segments.

Each of those segments has an average length of "9.6 cards".

The number of cards you turn over, on average, before seeing an Ace is the whole of first segment (9.6 cards) plus one more card (the Ace in its average position - you have to turn it over to see it, of course), so the answer is therefore 10.6.

And if you produce a little computer program that does this randomly and run it a million times and ask it to work out the average, it will indeed be 10.6. And if you do it just 100 times on your own, it will also be 10.6 (perhaps give or take a fraction for the error of a small sample-size).

So there we are: nothing to do with probability at all, actually. Just arithmetic.

Many thanks to those who had a go at it!

PS, we posted at nearly the same time (I stopped for a while to reply to an email while I was typing this) and I didn't actually see your totally correct answer before posting mine, Gary, apologies and very well solved!! I think you share the prize of 24 cards each ... and you both did better with it than I did ...

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CDarklock

16 years ago

Originally Posted by Alexa Smith

I have a long, narrow glass rod which is 4-feet long and I drop it on the floor and it breaks cleanly at 4 different points into 5 pieces.

I solved this slightly differently.

The key to this problem is to understand that the four aces do not divide the deck into four pieces, but into five.

The average size of each piece is therefore 52/5 = 10.4 cards.

However, the last piece does not contain an ace, and to compensate for this inaccuracy we imagine a hypothetical 53rd card.

If the final card in the deck is always an ace, the probabilities for the rest of the deck don't change, so the actual answer is 53/5 = 10.6

Same answer, different method, no Bayes.

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Alexa Smith

Banned 16 years ago

Originally Posted by CDarklock

The key to this problem is to understand that the four aces do not divide the deck into four pieces, but into five.

Exactly so.

Originally Posted by CDarklock

The average size of each piece is therefore 52/5 = 10.4 cards.

However, the last piece does not contain an ace, and to compensate for this inaccuracy we imagine a hypothetical 53rd card.

If the final card in the deck is always an ace, the probabilities for the rest of the deck don't change, so the actual answer is 53/5 = 10.6

Yes, I see exactly what you mean - interesting ... well done!

Originally Posted by CDarklock

no Bayes.

Indeed.

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Jagged

16 years ago

Originally Posted by John Henderson

If there were only one ace in Alexa's pack of 52 cards, there would be an equal chance of that ace being in any position from card 1 to card 52. The average of these over 10 million shuffles would be position 26; half way thru the pack.

If there were two aces in the pack of 52 cards, there's now an increased chance of me finding one of them sooner. Over 10 million shuffles, I'm not going to have to wait to get 1/2 way thru the pack -- I suspect that it might average out at 1/3 way thru the pack.

So if this line of reasoning is correct... with four aces in the pack, the average number of cards that will have to be turned over to find an ace is 1/(4+1) or 1/5 of the pack: 52/5 = 10.4 cards.

Wouldn't it be more like this...

The pack of cards with only one ace would only be a 49 card deck...with 3 aces removed...(if it was still 52 cards..what are the other 3 cards?)
The pack with only 2 aces would be a 50 card deck & so on....use your method to break it down to this ratio's & see if it works out even closer...

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John Henderson

16 years ago

Originally Posted by Jagged

Wouldn't it be more like this...

The pack of cards with only one ace would only be a 49 card deck...with 3 aces removed...(if it was still 52 cards..what are the other 3 cards?)
The pack with only 2 aces would be a 50 card deck & so on....use your method to break it down to this ratio's & see if it works out even closer...

Hi Ken,

I varied the number of aces but kept the total amount of cards the same (52). It was just a hypothetical pack that allowed me to show a pattern emerging as I added aces to my 52 card pack. The other cards in the pack could have been blank -- the important thing was the ratio of aces to non-aces.

But yeah, if it had been a real pack, you'd be absolutely correct.

John H.

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garyv

16 years ago

If there are 4 aces, then there would actually be 5 sections of cards that would need to be averaged. One section on each side of each ace. So the answer would be 52 cards - 4 aces (the dividing points) divided by 5 sections of cards + 1 to turn over the next card (the ace) 48/5 +1 = 10.6

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garyv

16 years ago

Woo hoo - And I did stay at a Holiday Inn last night...

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Kay King

16 years ago

If I were betting I wouldn't raise my bet on that average. I've talked to players who use a similar formula to make decisions when playing blackjack - I took their money

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garyv

16 years ago

Originally Posted by Kay King

If I were betting I wouldn't raise my bet on that average. I've talked to players who use a similar formula to make decisions when playing blackjack - I took their money

You were the dealer?

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Alexa Smith

Banned 16 years ago

Remind me not to play cards with any of you ...

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Kurt

16 years ago

Originally Posted by Kurt

To be totally accurate, you'd use Baye's formula for probability, which is the same one used by casinos, insurance companies, NASA, etc:
Bayesian probability - Wikipedia, the free encyclopedia

People are correct that there are 4/52 chance for the first card, and as each card is dealt, the odds increase.

If there were only 13 cards, one of each suit, then the answer would be (1/13/)/2 or 6.5 cards. However, you can't apply this ratio for the situation given.

Now here's where we get in trouble using straight ratios and not Baye...There's virtually an unlimited number of combinations for the distribution of Aces, so it's really impossible to calculate without Baye's.

By the way, as a card counter it isn't all that rare to go 13 cards without seeing an ace. Counters keep track of these things.

Originally Posted by Alexa Smith

Remind me not to play cards with any of you ...

I posted the correct answer, although I didn't do the math. And if your answer does NOT use Baye's, then you are incorrect. I say that will 100% probability.

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Kay King 16 years ago

You were the dealer?

Yup - it was like my fourth career. Had never been in a casino till I moved to Mississippi and my son used his connections to set up an interview for a new casino. They hired me for my people experience in sales/management and trained me.

Been dealing since 94 - full time for years but now I work weekends only.
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garyv

16 years ago

Originally Posted by Kurt

I posted the correct answer, although I didn't do the math. And if your answer does NOT use Baye's, then you are incorrect. I say that will 100% probability.

The question was one of averages, based upon dealing the cards 10 million times. Your answer deals not w/ an average but a probability, which does not deal w/ averaging 10 million shuffles, but what will "probably" happen on this shuffle.

Which is why Kay is also right. A good card counter does not use an averages formula when playing black-jack. They use a formula closer to yours that deals w/ probability. And that probability formula changes w/ each card that is dealt. It is the house that uses the averages formula, which is why after they've dealt millions of hands, they always come out ahead.

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John Henderson

16 years ago

Originally Posted by Alexa Smith

You win the prize for the closest answer by far, John, and for your method of solving it, which is basically right. (The prize is a deck of cards, Aces removed, which I'll deliver next time I'm down in Sussex).

Woo Hoo! I'm the winner!

Originally Posted by Alexa Smith

Gary, apologies and very well solved!! I think you share the prize of 24 cards each ... and you both did better with it than I did ...

I've got to share the prize???

Originally Posted by Kurt

I posted the correct answer, although I didn't do the math. And if your answer does NOT use Baye's, then you are incorrect. I say that will 100% probability.

Oh, now Kurt's muscling in on my prize...

Originally Posted by garyv

The question was one of averages, based upon dealing the cards 10 million times. Your answer deals not w/ an average but a probability, which does not deal w/ averaging 10 million shuffles, but what will "probably" happen on this shuffle.

Phew! That was a close call! :rolleyes:

P.S. Thanks Alexa, that was fun.