This could be generalized to any discrete data set, but I'll just refer to poker hands here.

After seeing n random two card hands, what is the probability that I have seen every hand in an unknown range?

For an example:

Someone is dealt two cards. Sometimes he mucks his hand and sometimes he exposes them. He always takes the same action with any specific strategically unique hand. What is the probability that I know his action for each of the 169 possible two card combinations after n dealt hands?

This is probably not hard to answer with a simulation. Do we need to assume a prior distribution? I'd be most interested in a mathematical approach since I could probably program a simulation myself pretty easily. I'm not sure how to approach the problem mathematically.

See the Coupon Collector Problem. There is a formula to solve this for any size set.

If one takes all 169 starting hands as equally likely (which, of course, they aren't..but which should be enough to give us a ballpark figure) then the expected value comes out to roughly 960 hands.
(169* 169th Harmonic number)

Of course, if his strategy is "always expose" you don't need an expected value

You can easily weight them by their expected frequency to calculate an accurate answer.

Pairs 78/1326
Suited non-pairs 312/1326
Offsuit non-pairs 936/1326

Sorry I haven't popped back in. I looked up the coupon collector problem and found that 960 hands estimate also. However, the problem is a bit more nuanced since we don't know the size of the set we're "collecting," isn't it?

You'll have to pick one and refine your guess based on the hands you've seen. You will never know for sure a villain's exact range.

Right, but we're never guaranteed to have seen a complete range over a finite number of hands even if we know what it is. I'm interested in knowing how many hands are needed to have 50%, 90%, 95% etc. confidence that the range is complete.

We shouldn't have to pick a range. If we go 10000 hands without seeing any new hands, we can be pretty confident the range is complete without having a prior assumption of what the range is.

Actually, we do need to assume a prior distribution. In this case I would just assume that all villain ranges are equiprobable.

There are ways to do this but they are difficult calculations.
https://www.google.com/search?q=coup...dence+interval

People have written their PhD thesis on this exact problem.

Well, the 960 value is your E(x)...so this gives you a 50% certainty.

That's not true and it is not what he asked.

Hmm, I think it is true assuming equal frequencies (just as a math exercise). If the average number of trials to obtain all coupons is 960, then after 960 trials you have a 50% chance to have completed the set. Is that not correct?

I was thinking of mean vs. median.

It's been awhile since I worked with problems related to the coupon-collector problem. I believe the distribution of collection times is a combination of various geometric distributions and itself is not from a common distribution family. I think the distribution values (as a formula) may be available.

Since this is fairly easy to do, and in the absence of such distribution formulas, I ran a simulation of 1 million trials of the equal-probability coupon collector case with 10 items. I chose 10 so that the simulation would not take very long. Maybe I'll do the same thing for 169 but the time it will take will be much longer and presenting all the output will be more arduous.

Note EV(number of selections to select all 10) = 10*H(10) = 29.28968

Number of Selections Required	Tally	Cumulative Tally
1	0	0
2	0	0
3	0	0
4	0	0
5	0	0
6	0	0
7	0	0
8	0	0
9	0	0
10	328	328
11	1699	2027
12	4267	6294
13	8146	14440
14	13003	27443
15	18629	46072
16	24322	70394
17	29475	99869
18	34734	134603
19	38587	173190
20	41420	214610
21	43791	258401
22	44498	302889
23	45066	347965
24	44679	392644
25	43444	436088
26	42569	478657
27	40356	519013
28	39047	558060
29	36782	594842
30	34464	629306
31	32324	661630
32	29856	691486
33	27476	718962
34	25584	744546
35	23211	767757
36	21372	789129
37	19857	808986
38	17921	826907
39	16266	843173
40	14865	858038
41	13732	871770
42	12396	884166
43	11031	895197
44	10294	905491
45	9123	914614
46	8346	922960
47	7525	930485
48	6900	937385
49	6017	943402
50	5569	948971
51	5082	954053
52	4563	958616
53	4073	962689
54	3743	966432
55	3369	969801
56	3011	972812
57	2732	975544
58	2417	977961
59	2202	980163
60	2039	982202
61	1798	984000
62	1569	985569
63	1389	986958
64	1220	988178
65	1209	989387
66	1110	990497
67	959	991456
68	825	992281
69	801	993082
70	696	993778
71	601	994379
72	558	994937
73	492	995429
74	451	995880
75	370	996250
76	378	996628
77	352	996980
78	288	997268
79	248	997516
80	260	997776
81	225	998001
82	198	998199
83	204	998403
84	151	998554
85	120	998674
86	130	998804
87	136	998940
88	118	999058
89	93	999151
90	82	999233
91	67	999300
92	67	999367
93	68	999435
94	62	999497
95	51	999548
96	52	999600
97	31	999631
98	34	999665
99	38	999703
100	29	999732
101	23	999755
102	25	999780
103	15	999795
104	24	999819
105	17	999836
106	21	999857
107	14	999871
108	18	999889
109	13	999902
110	10	999912
111	9	999921
112	5	999926
113	4	999930
114	5	999935
115	8	999943
116	6	999949
117	3	999952
118	9	999961
119	4	999965
120	4	999969
121	4	999973
122	1	999974
123	2	999976
124	0	999976
125	3	999979
126	5	999984
127	3	999987
128	1	999988
129	1	999989
130	2	999991
131	1	999992
132	0	999992
133	1	999993
134	1	999994
135	0	999994
136	0	999994
137	1	999995
138	1	999996
139	1	999997
140	0	999997
141	0	999997
142	1	999998
143	0	999998
144	1	999999
145	0	999999
146	0	999999
147	0	999999
148	0	999999
149	0	999999
150	0	999999
151	0	999999
152	0	999999
153	0	999999
154	0	999999
155	1	1000000

Obviously, any pertinent cumulative percentiles can be easily read off the table.

I am dusting off some cobwebs in that part of my brain devoted to information on the coupon-collector problem. As we know, the problem is typically presented and fairly easily solved assuming all the items have identical probabilities of being selected.

The unequal-probability case is more challenging. The formula for the Expected Number of Required Selections, given below, is a trifle daunting. And, I believe, it is beyond even high-speed computer's capability to solve for large numbers of items.

Let there by M items with probability of selection being Pi (for i=1 to M). Then:

EV = {Sum[Q=1 to M-1] (-1)^(M-1-Q) * Sum[|J|=Q] 1/(1 - Sum[j in J] Pj)} + (-1)^(M-1)

If I remember correctly, if there are a small number of "classes" in which items may belong, I think there is a way to calculate the expected value via the formula using partitions and such.

There may be some value in running a simulation of the 169 2-card poker hands using their correct probabilities (given above by NOG) and comparing the results to the case assuming equal probabilities. I have given this virtually zero thought but my first guess would be that the unequal-probability EV would be larger than the equal-probabiity EV.

May I ask what language you are using? I programmed the same sim in C++ and it takes two seconds for 10 coupons. For 169 coupons it takes about 39 seconds (with part of that time just being output of the table).

He was collecting the full set of all 10 coupons 1 million times. How many times did you collect them?

1 million... I said the same sim. I could have made a mistake, but I get roughly the same output as him.

In the past I've always been suspicious of whosnext's reports of how long his simulations take. Since this one was very easy for me to implement myself I thought I'd double-check his results.

I am not a tech guy so I don't know everything that contributes to the speed of a program, but I use C++ on a laptop which cost about $300 about 3-4 years ago.

EV of the unequal (true) probability case of the 169 2-card poker hands is around 1646, much higher than the EV of the case assuming equal probabilities of 965.

I guess it is obvious that the unequal probability case will typically require more selections as unequal means that there will be many low probability items (e.g., suited hands).