May 25th, 2021 at 10:59:02 AM
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If I were to deal 13 cards from a shuffled (presumed random) deck of cards, how many different ranks should I expect to see?

Suited89

Suited89

some people need to reimagine their thinking

May 25th, 2021 at 4:13:15 PM
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After some serious number crunching, I get 188,474 / 20,825, which is about 9.05.

May 26th, 2021 at 6:59:44 AM
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Quote:ThatDonGuyAfter some serious number crunching, I get 188,474 / 20,825, which is about 9.05.

I agree. Using a Markov Chain, here is the average number after 1 to 52 cards.

Cards | Exp Ranks |
---|---|

1 | 1.00000000000000 |

2 | 1.94117647058824 |

3 | 2.82588235294118 |

4 | 3.65642256902761 |

5 | 4.43505402160864 |

6 | 5.16398559423769 |

7 | 5.84537815126050 |

8 | 6.48134453781513 |

9 | 7.07394957983193 |

10 | 7.62521008403361 |

11 | 8.13709483793517 |

12 | 8.61152460984394 |

13 | 9.05037214885954 |

14 | 9.45546218487395 |

15 | 9.82857142857143 |

16 | 10.17142857142860 |

17 | 10.48571428571430 |

18 | 10.77306122448980 |

19 | 11.03505402160860 |

20 | 11.27322929171670 |

21 | 11.48907563025210 |

22 | 11.68403361344540 |

23 | 11.85949579831930 |

24 | 12.01680672268910 |

25 | 12.15726290516210 |

26 | 12.28211284513810 |

27 | 12.39255702280910 |

28 | 12.48974789915970 |

29 | 12.57478991596640 |

30 | 12.64873949579830 |

31 | 12.71260504201680 |

32 | 12.76734693877550 |

33 | 12.81387755102040 |

34 | 12.85306122448980 |

35 | 12.88571428571430 |

36 | 12.91260504201680 |

37 | 12.93445378151260 |

38 | 12.95193277310920 |

39 | 12.96566626650660 |

40 | 12.97623049219690 |

41 | 12.98415366146460 |

42 | 12.98991596638660 |

43 | 12.99394957983190 |

44 | 12.99663865546220 |

45 | 12.99831932773110 |

46 | 12.99927971188480 |

47 | 12.99975990396160 |

48 | 12.99995198079230 |

49 | 13.00000000000000 |

50 | 13.00000000000000 |

51 | 13.00000000000000 |

52 | 13.00000000000000 |

It's not whether you win or lose; it's whether or not you had a good bet.

May 26th, 2021 at 10:43:01 AM
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9.05 is about 69.6% of the 13 ranks appearing.

If you take this same problem but assume infinite deck, I believe the answer is 1 - 1/e =~ 63.2% as the number of ranks approaches infinity. For instance, if you assume 1,000 ranks and draw 1,000 cards, the average number of ranks appearing is very close to 632.

I got this via simulation. Assuming it’s correct, can any of our calculus gurus get to the answer directly? I imagine the solution being an infinite series using inclusion-exclusion that turns out being an expression of 1/e...sort of like the derangement formula

If you take this same problem but assume infinite deck, I believe the answer is 1 - 1/e =~ 63.2% as the number of ranks approaches infinity. For instance, if you assume 1,000 ranks and draw 1,000 cards, the average number of ranks appearing is very close to 632.

I got this via simulation. Assuming it’s correct, can any of our calculus gurus get to the answer directly? I imagine the solution being an infinite series using inclusion-exclusion that turns out being an expression of 1/e...sort of like the derangement formula

Last edited by: Ace2 on May 26, 2021

It’s all about making that GTA

May 26th, 2021 at 10:59:37 AM
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Quote:WizardI agree. Using a Markov Chain...

My method:

First, determine all of the partitions of 13 such that no partition has more than 4 in it. This represents how many cards of particular ranks are in the hand.

For example, (4, 3, 3, 3) is a four of a kind and three threes of a kind, and (2, 2, 2, 1, 1, 1, 1, 1, 1, 1) are three pairs and seven singles.

Take (4, 3, 2, 2, 1, 1) as an example.

There are 13 possible choices for the four of a kind rank.

For each one, there are 12 possible choices for the three of a kind rank, and for each one, there are four possible threes (for example, three Aces could be Ace of spades, hearts, and clubs, Ace of spades, hearts, and diamonds, Ace of spades, clubs, and diamonds, or Ace of hearts, clubs, and diamonds).

For each one of these, there are C(11,2) = 55 possible choices for the two pair ranks, and for each one, there are six possible pairs.

For each one of these, there are C(9,2) = 36 possible choices for the two single ranks, and for each one, there are four possible cards.

The total = 13 x 12 x 4 x 55 x 6 x 36 x 4 = 29,652,480 13-card hands.

The number of different ranks in each of these hands is 6.

For every possible partition (IIRC, there are 39 of them), multiply the number of hands by the number of different ranks, then sum these 39 numbers and divide by the total possible number of hands, which is C(52,13).

June 16th, 2021 at 1:47:19 PM
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If you assume infinite deck, then the limit is indeed 1-1/e. I have a proof, but it is 3 pages long and too mathematical to include in a reply. If there is a way to attach a pdf, I am happy to share it with anyone who is interested.

June 16th, 2021 at 3:30:46 PM
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I'm interestedQuote:GMIf you assume infinite deck, then the limit is indeed 1-1/e. I have a proof, but it is 3 pages long and too mathematical to include in a reply. If there is a way to attach a pdf, I am happy to share it with anyone who is interested.

It’s all about making that GTA

June 17th, 2021 at 8:03:34 AM
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Here is the link to the paper on my OneDrive. Delete the space after //

https:// 1drv.ms/b/s!AhDS5R4jVo9yk0SAPjJWg1Wf2ZxP?e=eZjpir

https:// 1drv.ms/b/s!AhDS5R4jVo9yk0SAPjJWg1Wf2ZxP?e=eZjpir

June 17th, 2021 at 8:06:20 AM
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Quote:GMI have tried to include a link to it on my OneDrive, but the system won't let me, not even as plain text. :(

New posters can’t post links until they have 20 posts. If you post it as a broken link it will appear.

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

June 18th, 2021 at 8:52:00 AM
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I am quite confident that I have the answer for the general question when there are n ranks, k cards of each rank (so n=13, k=4 for a standard deck of playing cards) and we draw N cards, but I don't have a proof yet.

The expected number of different ranks is n(1-C(kn,N)/C(k(n-1),N)),

where C(r,s) denotes the binomial coefficient r choose s, the number of ways of choosing s different objects from a set of r objects if don't care about their order.

If n=N=13, k=4, then C(52,13)=635013559600, C(48,13)=192928249296, 1-192928249296/635013559600=14498/20825 and the expected number of different ranks is 13*14498/20825=188474/20825 in agreement with everybody else's calculations.

If we fix n=13, k=4 and let N vary, then my formula gives the numbers which are in the Wizard's table.

If we fix n and N, and let k tend to infinity, then the number of expected ranks tends to n(1-(1-1/n)^N), which is the answer in the infinite deck case.

If we fix k, and let N and n tend to infinity in such a way that N/n has a limit c, then expected number of ranks divided by n tends to 1-(1-(c/k))^k.

If k=4 and N=n, then this expression gives 1-(3/4)^4=175/256=0.68359375.

If k tends to infinity, then 1-(1-(c/k))^k tends to 1-1/e^c, which is the limit in the infinite deck case.

The expected number of different ranks is n(1-C(kn,N)/C(k(n-1),N)),

where C(r,s) denotes the binomial coefficient r choose s, the number of ways of choosing s different objects from a set of r objects if don't care about their order.

If n=N=13, k=4, then C(52,13)=635013559600, C(48,13)=192928249296, 1-192928249296/635013559600=14498/20825 and the expected number of different ranks is 13*14498/20825=188474/20825 in agreement with everybody else's calculations.

If we fix n=13, k=4 and let N vary, then my formula gives the numbers which are in the Wizard's table.

If we fix n and N, and let k tend to infinity, then the number of expected ranks tends to n(1-(1-1/n)^N), which is the answer in the infinite deck case.

If we fix k, and let N and n tend to infinity in such a way that N/n has a limit c, then expected number of ranks divided by n tends to 1-(1-(c/k))^k.

If k=4 and N=n, then this expression gives 1-(3/4)^4=175/256=0.68359375.

If k tends to infinity, then 1-(1-(c/k))^k tends to 1-1/e^c, which is the limit in the infinite deck case.

Last edited by: GM on Jun 18, 2021