Four Aces

by Banned 48 replies
55
Shuffle a standard deck of 52 playing cards well, and then deal cards off the top, one at a time, turning each one face-up.

If you do this 10,000,000 times, what's the average number of cards you'll turn over before seeing an Ace?
#off topic forum #aces
  • Algebra in the OT forum?
  • Just taking a quick break from writing, Alexa?


    Ken
    • [1] reply
    • 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?
  • 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.
    • [2] replies
    • 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.
      • [1] reply
    • Banned
      The sample size has to be substantial for the average to be realistic, no?

      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.

      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)?
      • [1] reply
  • It depends on your definition of "shuffle well". If I shuffle a deck well, I can make one appear the first time every time.
    • [1] reply
  • 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
    • [1] reply
    • Banned
      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!).

      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.

      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 ...

      I'm going out for a while, too ... see you later.
  • 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
  • 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.
    • [ 1 ] Thanks
    • [1] reply
    • 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"
  • You might want to ask this guy:

    Persi Diaconis - Wikipedia, the free encyclopedia
    • [1] reply
  • 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.
    • [ 1 ] Thanks
    • [2] replies
    • Banned
      Indeed.

      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 ...
      • [ 2 ] Thanks
      • [1] reply

    • 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...
      • [1] reply
  • 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
    • [ 1 ] Thanks
  • Woo hoo - And I did stay at a Holiday Inn last night...
    • [1] reply
    • 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
      • [1] reply
  • if I had to figure all this out to do marketing I would be one poor mofo.
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    Four Aces

    Shuffle a standard deck of 52 playing cards well, and then deal cards off the top, one at a time, turning each one face-up. If you do this 10,000,000 times, what's the average number of cards you'll turn over before seeing an Ace?