A little math fun....

EazzyEazzy Red Chipper Posts: 983 ✭✭✭✭
You have a deck of cards ...you turn over cards till an Ace comes.....
whats the odds that the next card will be another Ace....

Comments

  • NTD12NTD12 Red Chipper Posts: 73 ✭✭
    How many cards did you turn over?
  • NinjahNinjah Red Chipper Posts: 1,138 ✭✭✭✭
    edited November 8
    The odds for the first draw would be 1/13. I'm probably missing something here but I think the sequence for each undrawn card would be 4/(n-1) after each card gets drawn that is not an ace.
  • CrodigglyCrodiggly SF Bay AreaRed Chipper Posts: 34 ✭✭
    I got 7.7%

    P(A) on the first card is .08 (4/52), P(A) on the second card given !A on the first card is (3/51*48/52), each successive probability of first Ace is 4/remaining * sum(P(A)) on any earlier card.

    P(second A) is 3/remaining cards * P(first A | card number)

    Total prob = sum(P(A, 1_52)) * P(second A | P(A, 1_52))
  • NinjahNinjah Red Chipper Posts: 1,138 ✭✭✭✭
    I missed the part where it said another ace. Either way, it's a branch of mathematics in which I haven't traveled very far.
  • TheGameKatTheGameKat Posts: 2,633 -
    Is there a general solution in which the number of cards dealt before the first A peels is unknown? It's not obvious to me. If we can include that number in the solution, then it's trivial.
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  • EazzyEazzy Red Chipper Posts: 983 ✭✭✭✭
    TheGameKat wrote: »
    Is there a general solution in which the number of cards dealt before the first A peels is unknown? It's not obvious to me. If we can include that number in the solution, then it's trivial.

    no the question is before you deal any cards......whats the odds that the next card after the first ace is dealt will be another ace....
  • TheGameKatTheGameKat Posts: 2,633 -
    Hmmm. I can express it as a series but I'm guessing there's a more intelligent way of doing it with factorials.
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  • CrodigglyCrodiggly SF Bay AreaRed Chipper Posts: 34 ✭✭
    TheGameKat wrote: »
    Hmmm. I can express it as a series but I'm guessing there's a more intelligent way of doing it with factorials.

    It's a finite series (the first A has to come in 49 cards) so a spreadsheet solves.

    I'm on the edge of my seat to see if my answer was right @Eazzy
  • TheGameKatTheGameKat Posts: 2,633 -
    Crodiggly wrote: »
    TheGameKat wrote: »
    Hmmm. I can express it as a series but I'm guessing there's a more intelligent way of doing it with factorials.

    It's a finite series (the first A has to come in 49 cards) so a spreadsheet solves.

    I'm on the edge of my seat to see if my answer was right @Eazzy

    Right. Ironically it's easier to do it analytically if the series is infinite, but that would require an unusual deck of cards.

    Still wondering if there's some sort of argument from symmetry to do this quicker tho.
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  • EazzyEazzy Red Chipper Posts: 983 ✭✭✭✭
    The math s incredibly simple..

    I you shuffle a deck of cards.....
    what the odds that the bottom card is an ace....

    Same deck now deal out the cards till you turn over an ace.....
    whats the odds the bottom card is an ace....

    There is no difference between the bottom card or the card after an ace once an ace is turned over (in terms of this question).

    its still 13 to 1.....


  • TheGameKatTheGameKat Posts: 2,633 -
    edited November 9
    Hmmm. So the first A is really just acting as a marker as a spot in the deck, and every deck slot has a 1/13 probability of being occupied by an A. Counter-intuitive but correct. Might have @Doug Hull simulate it to demonstrate numerically.
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  • TheGameKatTheGameKat Posts: 2,633 -
    edited November 9
    You can actually demonstrate this for a reduced problem explicitly. Apologies to anyone who glanced at this and thought we were discussing PLO.

    Suppose you restrict yourself to a 4-card deck:

    :As :Ah :Ks :Kh

    We can again ask, if we deal out the cards, what is the probability that an A will immediately follow the first A? The trap is to assume we are now choosing one A from 3 cards giving a probability of 1/3, whereas the correct probability is actually 1/2 - the equivalent of 1/13 in the full-deck example.

    You can also ask the question, what is the probability that the card dealt after the :As is the :Ah ? Is the answer now 1/3? 1/2? Or something else?

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  • EazzyEazzy Red Chipper Posts: 983 ✭✭✭✭
    TheGameKat wrote: »
    You can actually demonstrate this for a reduced problem explicitly. Apologies to anyone who glanced at this and thought we were discussing PLO.

    Suppose you restrict yourself to a 4-card deck:

    :As :Ah :Ks :Kh

    We can again ask, if we deal out the cards, what is the probability that an A will immediately follow the first A? The trap is to assume we are now choosing one A from 3 cards giving a probability of 1/3, whereas the correct probability is actually 1/2 - the equivalent of 1/13 in the full-deck example.

    You can also ask the question, what is the probability that the card dealt after the :As is the :Ah ? Is the answer now 1/3? 1/2? Or something else?

    If you look for a particular Ace that changes the problem because the AH could be dealt out before you get to the As

    Another way to look at the solution is to realize...that the cards dealt before the the first ace on average, will offset the fact that the first Ace takes an ace out of the deck...

    In the 4 card example
    if the first card was and ace its 2 to 1 the next card is an ace
    If its dealt 2nd then it its even odds...
    and if the ace is dealt 3rd its 100 %.....

    On average thats 0% of the time....
  • TheGameKatTheGameKat Posts: 2,633 -
    Nuh uh. Same concept as before. One card, 4 slots. Probability =1/4. You can explicitly write out all 24 possible permutations and find 6 satisfy the condition.
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  • jeffncjeffnc Red Chipper Posts: 4,814 ✭✭✭✭✭
    That's not really how the question was worded.

    Assume we have turned over X cards until we found the ace. The probability of the next card being an ace is 3/(52-X).

    While you seem to be looking for some sort of average number, it will rarely be the case that the odds of turning over an ace after you've found the first one will be 1/13. The only time that will be true is if you just happened to find the first ace on the 13th card.
  • jeffncjeffnc Red Chipper Posts: 4,814 ✭✭✭✭✭
    Eazzy wrote: »
    its still 13 to 1.....

    1 out of 13 and 13 to 1 are not the same thing.

  • jeffncjeffnc Red Chipper Posts: 4,814 ✭✭✭✭✭
    You seem to be asking this question now: what is the probability that the first 2 aces that come off the deck are back to back.
  • TheGameKatTheGameKat Posts: 2,633 -
    edited November 9
    jeffnc wrote: »
    That's not really how the question was worded.

    Assume we have turned over X cards until we found the ace. The probability of the next card being an ace is 3/(52-X).

    While you seem to be looking for some sort of average number, it will rarely be the case that the odds of turning over an ace after you've found the first one will be 1/13. The only time that will be true is if you just happened to find the first ace on the 13th card.

    No, it's always 1/13. Or if you prefer, the average is 1/13 and a probability is in some sense an average.
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  • jeffncjeffnc Red Chipper Posts: 4,814 ✭✭✭✭✭
    TheGameKat wrote: »
    No, it's always 1/13.

    Well obviously this isn't true. "You turn over cards until an ace comes. whats the odds that the next card will be another Ace"

    The first card in the deck is an ace. The probably the next card is an ace is 3/51.

    The 23rd card in the deck is the first ace. The probability the next card is an ace is 3/29.

    Probably just poorly worded first attempt. Reading through the thread I think he is actually asking what is the chance the first 2 aces are back to back.

  • persuadeopersuadeo Red Chipper, Table Captain Posts: 4,167 ✭✭✭✭✭
  • TheGameKatTheGameKat Posts: 2,633 -
    The issue with these things is frequently making sure we're trying to solve the same problem.

    The point is, a probability for this general situation cannot then say "let's suppose the first card is an A." It's like an EV equation. We don't claim the probability of hitting a flush draw is 1 when we hit it and 0 when we miss. Similarly, in this general problem, on any given realization you can claim the probability is not 1/13 once the first A has been revealed - it then follows the formula you gave.

    If you do this with my 4-card simplification and simply write down every permutation, you can see this explicitly simply by averaging all the 0s and 1s.
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  • TheGameKatTheGameKat Posts: 2,633 -
    So yes, the question is better worded, what is the probability the first two aces are back to back.
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  • jeffncjeffnc Red Chipper Posts: 4,814 ✭✭✭✭✭
    I mean, @Eazzy said the math is "incredibly simple", while at the same time avoiding showing any math at all :) @Crodiggly is obviously on the right track, although it's quite brute force. There might be a more elegant solution.
  • TheGameKatTheGameKat Posts: 2,633 -
    The elegant solution is to recognize that before any cards are revealed, the probability of a card of any rank being in any slot is 1/13. The problem is asked in this way because the intuitive assumption is that the fact an A is revealed acts as a conditional prior to the probability the next card is an A. It doesn't.

    For a related problem: What's the probability that the :As is dealt immediately after:

    1) Another Ace
    2) The :Jd
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  • TheGameKatTheGameKat Posts: 2,633 -
    persuadeo wrote: »
    This is just so much fun.

    Compared to what I'd be doing otherwise, this is definitely fun.
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  • jeffncjeffnc Red Chipper Posts: 4,814 ✭✭✭✭✭
    TheGameKat wrote: »
    The elegant solution is to recognize that before any cards are revealed, the probability of a card of any rank being in any slot is 1/13.

    Yes, but I meant that still has to jibe with a real math answer that adds up all the probabilities. I'm sure there's an equation.

  • TheGameKatTheGameKat Posts: 2,633 -
    jeffnc wrote: »
    TheGameKat wrote: »
    The elegant solution is to recognize that before any cards are revealed, the probability of a card of any rank being in any slot is 1/13.

    Yes, but I meant that still has to jibe with a real math answer that adds up all the probabilities. I'm sure there's an equation.

    Sure just using brute force would get there.
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