#11
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Re: Another Boredom Filler
Since people seem to be somewhat interested, these are the results of making it through any number of cards without a match.
Card 1 in X 52 61.5612 51 56.9152 50 52.5458 49 48.5413 48 44.8914 47 41.5559 46 38.348 45 35.4083 44 32.6861 43 30.1386 42 27.8102 41 25.6838 40 23.7293 39 21.9168 38 20.2873 37 18.7315 36 17.3382 35 16.0514 34 14.839 33 13.7105 32 12.6682 31 11.7049 30 10.8071 29 9.9822 28 9.2113 27 8.5042 26 7.8502 25 7.2597 24 6.7077 23 6.2028 22 5.7351 21 5.3004 20 4.8953 19 4.5186 18 4.1738 17 3.8565 16 3.5594 15 3.2864 14 3.0338 13 2.7993 12 2.5886 11 2.3924 10 2.2122 9 2.0442 8 1.8893 7 1.7457 6 1.6136 5 1.4908 4 1.3765 3 1.2717 2 1.1741 1 1.0833 0 1 It gets bunched up when I post it, but you get the idea. |
#12
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Re: Another Boredom Filler
The thought I've been having lately is that if you were to translate the question into counting from 1-52 and matching the exact card instead of rank only, the problem simply becomes a Fixed Point calculation based on the permutations...and should tend to 1/e for no fixed points, I believe. Although, even that isn't a closed form solution. From what I've been able to read about this, as the sample size goes up, the probability of no matches tends toward 1/e^4.
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#13
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Re: A Similar Problem
[ QUOTE ]
Shuffle 2 decks of cards Then simultaneously deal from the top 1 card per deck at a time. What is the probability that you will get through the whole deck without ever dealing the same 2 cards simultaneously? [/ QUOTE ] This is much easier than the original problem. Answer in white: <font color="white">From inclusion-exclusion: P(no matches) = 1 - P(1 or more match) = 1 - [52*51! - C(52,2)*50! + C(52,3)*49! - ... - C(52,52)*0!] / 52! = 1 - (1 - 1/2! + 1/3! - ... - 1/52!) = 1 - 1 + 1/2! - 1/3! + ... + 1/52! Note that this is approximately the Tayor series for e^-1 =~36.8%. [ QUOTE ] Then for an encore what is the probability for n Decks (1 < n < 52) [/ QUOTE ] Answer in white: <font color="white"> EDIT: This solution is approximate because when there are k matches, some of these can be the same card, so there will not always be n^k ways to match k cards. The approximation should be good since the cases where the same card is matched more than once should make a small contribution. P(no matches) = 1 - P(1 or more match) = 1 - [n*52n*(52n-1)! - n^2*C(52n,2)*(52n-2)! + n^3*C(52n,3)*(52n-3)! - ... - n^(52n)*C(52n,52n)*0!] / (52n)! = 1 - (n - n^2/2! + n^3/3! - ... - n^(52n)/(52n)! ) = 1 - n + n^2/2! - n^3/3! + ... + n^(52n)/(52n)! This is approximately the Tayor series for e^-n.</font> |
#14
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Re: Another Boredom Filler
[ QUOTE ]
Could you sketch out how the I/E method would be applied to this problem? [/ QUOTE ] See my solution in this thread for similar problems. It is not e^-4 for this problem since there are not always 4^n ways to match n cards, since some of the C(52,n) would choose the same rank. This is what makes inclusion-exclusion tedious for this problem, but not for the problem where we match the exact card with 1 deck. Note that on average there will be 52*1/13 = 4 matches, so we would need more than 4 terms. The other error due to the truncation of the Taylor series is insignificant. |
#15
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Re: Another Boredom Filler
I don't have a good method to solve this problem. However I am wondering the following:
Why is it that the approximate answer (12/13^52 = about 1 in 64.2) yields a probability significantly lower than the "actual" answer of about 1 in 61.5? My first thought of approaching this problem was doing the "tree" analysis suggested by LetYouDown. It gets way too complicated, but take the first 2 cards. THere are 3 scenarios: 1. First card is an ace, a match, game over, you lose. Probability 1/13. 2. First card is a 2. Probability 1/13. Probability of a match next card is now 1/17. Probability of a 2 the first card, and no match the 2nd card is 1/13*16/17 = 16/221. 3. First card is a 3-K. Probability 11/13. probability of a match next card is now 4/51. Probability of a 3-K 1st card and no match the 2nd card is 11/13*47/51 = 517/663. Total probability of no match the 1st 2 cards is 16/221 + 517/663 = 565/663 = 0.852187029 12/13 ^ 2 = 0.852071006. Thus, for 2 cards at least, the probability of no matches is HIGHER with the approximate method compared to the actual answer. It's not clear to me why this changes when you continue to go through more cards. Edit: I am stupid, and can't count. |
#16
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My 2nd answer is approximate
Donkeyradish's second problem has the same issue as the original, in that when we choose C(52n,k) matches, some of these can be the same card, so there won't always be n^k ways to match the k cards. In fact, the original problem can be viewed as a special case of this problem with 4 decks of 13 cards each.
The approximation should be better in this case because for a low number of matches (low order terms), the cases where we match the same card make a small contribution. The cases where it is likely that we match the same card are represented by the high order terms, and these make a small contribution because these terms are small, the series having already converged for all practical purposes. |
#17
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Re: 2nd answer is approximate
I am assuming by the title of this post that you assumed I could compare 2 numbers and determine which one was greater than the other. Since that assumption was false, your reply makes no sense to me, my post had no purpose, I am stupid, or perhaps my assumption of your assumption was incorrect. Or something.
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#18
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Re: 2nd answer is approximate
[ QUOTE ]
I am assuming by the title of this post that you assumed I could compare 2 numbers and determine which one was greater than the other. Since that assumption was false, your reply makes no sense to me, my post had no purpose, I am stupid, or perhaps my assumption of your assumption was incorrect. Or something. [/ QUOTE ] I don't understand what you're saying. I responded to my own post and pointed out that my solution to the second problem that donkeyradish asked was an approximation, though it should be a good one. What 2 numbers are you talking about? |
#19
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Re: 2nd answer is approximate
nevermind, I saw the reply and for some reason I thought it was to my post, which is why it confused me.
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#20
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Re: 2nd answer is approximate
Bruce, do you think that the only real method for getting an exact answer here would be to go through the inclusion/exclusion principle for all of the terms? I realize that as it deviates from the expected number of matches, the fluctuation will be negligible at some point. However, my thoughts at this point lean toward there not being a more elegant, exact solution as opposed to an approximation.
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