Odds of guessing suit from a deck of cards, with perfect memory

Solution 1:

The approximate odds of guessing the suit properly is 35.4%, quite a bit higher than 25%. This is easy to calculate with Monte Carlo analysis, though that does not easily allow a more precise answer. A combination of elementary probability and MC would be best though I have not attempted this.

I can't immediately think of a good way to find the exact rational probability since there are many ways for the number of each suit to change throughout the draws. Perhaps a combinatorialist could help here?

Solution 2:

Here's a possibly helpful formula (that seems very hard to put in closed form). If you have already drawn $n$ cards, then $P(\mathrm{success})$ is

$$ \sum_{\spadesuit+\heartsuit+\diamondsuit+\clubsuit=n;\ 0\leq \spadesuit,\heartsuit,\diamondsuit,\clubsuit\leq13}\frac{13-\min(\spadesuit,\heartsuit,\diamondsuit,\clubsuit)}{52-n}\cdot\frac{\binom{n}{\spadesuit, \heartsuit, \diamondsuit, \clubsuit}\binom{52-n}{13-\spadesuit, 13-\heartsuit, 13-\diamondsuit, 13-\clubsuit}(13!)^4}{52!} $$

(This formula uses the multinomial coefficient, which in this case counts the number of ways to choose $\spadesuit$ positions to be spades, $\heartsuit$ of the remaining positions to be hearts, etc. out of the $n$ first cards to be drawn.)

The first factor is the probability of success given a partition $\spadesuit+\heartsuit+\diamondsuit+\clubsuit=n$ (the partition represents how the first $n$ cards were distributed, disregarding the order they came in). The rest is the probability of the first $n$ cards providing that partition (counting all the possible orderings, divided by the total possible orderings of a deck).

Added later:

The sum above conditions over the partition that represents the cards which have already been drawn. We could also sum over partitions that represent the cards yet to be drawn. That makes computing more efficient for the second half of the deck. If $r$ cards remain, then $P(\mathrm{success})$ is

$$ \sum_{\spadesuit+\heartsuit+\diamondsuit+\clubsuit=r;\ 0\leq \spadesuit,\heartsuit,\diamondsuit,\clubsuit\leq13}\frac{\max(\spadesuit,\heartsuit,\diamondsuit,\clubsuit)}{r}\cdot\frac{\binom{r}{\spadesuit, \heartsuit, \diamondsuit, \clubsuit}\binom{52-r}{13-\spadesuit, 13-\heartsuit, 13-\diamondsuit, 13-\clubsuit}(13!)^4}{52!} $$

Evaluating these sums requires enumerating partitions. This could be done with some programming, but I did it for $r,n\leq10$ in Excel and found the following $P_n(\mathrm{success})$:

$$ \begin{align} P_0&=0.25 & P_1&\approx0.2549 & P_2&=0.26 & P_3&\approx0.2653 \\ P_4&\approx0.2686 & P_5&\approx0.2710 & P_6&\approx0.2733 & P_7&\approx0.2762 \\ P_8&\approx0.2788 & P_9&\approx0.2809 & P_{10}&\approx0.2830 & &\cdots\\ &\cdots & &\cdots & P_{42}&\approx0.4005 & P_{43}&\approx0.4122 \\ P_{44}&\approx0.4258 & P_{45}&\approx0.4392 & P_{46}&\approx0.4548 & P_{47}&\approx0.4836 \\ P_{48}&\approx0.5201 & P_{49}&\approx0.5514 & P_{50}&\approx0.6176 & P_{51}&=1 \end{align} $$

Solution 3:

overall odds for any given draw over the course of 52 picks

If I rephrased your question as "how much should you be willing to pay to play the game where I will show you $n$ cards out of 52 and if you guess the next remaining card then I give you a dollar" would an answer to this question be suitable? Just to be clear, in this game you would have to pay up front before you see any cards (although you know the number $n$ beforehand), and you should be willing to pay up to a quarter when $n=0$ and up to a dollar when $n=52-1$. This is not really an answer, I just wanted to understand the question.

Alternatively, would you allow the player to see the $n$ cards (rather than just the number $n$) before deciding how much to pay? Or on the other hand would you not even allow $n$ to be known, but you would require the player to decide how much to pay and then $n$ is chosen uniformly at random between 0 and 51 at the beginning of the game?

Solution 4:

Here's code that calculates the exact probabilities for all cards by iterating over all possible tuples of counts $(n_\spadesuit,n_{\heartsuit},n_\diamondsuit,n_\clubsuit)$ of suits that could be left at each stage and summing over the product of the probability of guessing correctly and the probability that the tuple of counts occurs,

$$ \frac{\max(n_\spadesuit,n_{\heartsuit},n_\diamondsuit,n_\clubsuit)}{n_\spadesuit+n_\heartsuit+n_\diamondsuit+n_\clubsuit}\cdot\frac{\binom{13}{n_\spadesuit}\binom{13}{n_\heartsuit}\binom{13}{n_\diamondsuit}\binom{13}{n_\clubsuit}}{\binom{52}{n_\spadesuit+n_\heartsuit+n_\diamondsuit+n_\clubsuit}} \;. $$

The result is (with the first number specifying the number of cards already seen):

0 : 1/4 (0.25)
1 : 13/51 (0.2549019607843137)
2 : 13/50 (0.26)
3 : 13/49 (0.2653061224489796)
4 : 16783/62475 (0.2686354541816727)
5 : 106093/391510 (0.27098413833618556)
6 : 2461329/9004730 (0.27333734603924825)
7 : 463489/1677900 (0.27623159902258776)
8 : 10757773/38591700 (0.2787587227305353)
9 : 1553669/5531477 (0.2808777836371732)
10 : 657371689/2323220340 (0.28295709954054554)
11 : 453097047/1587533899 (0.2854093681309164)
12 : 9135476103/31750677980 (0.2877253868013309)
13 : 19717365683/68037167100 (0.2898028610453418)
14 : 3309516509/11339527850 (0.2918566410152606)
15 : 93820396735/318867523142 (0.29423001693784706)
16 : 567356569949/1913205138852 (0.29654769288850896)
17 : 1470709561568/4923689695575 (0.29870070059243387)
18 : 8393986192969/27900908274925 (0.30084992611200445)
19 : 1336231110539/4405406569725 (0.3033161841910114)
20 : 359214427949/1174775085260 (0.3057729368422033)
21 : 1602940622179/5202575377580 (0.3081052182514681)
22 : 48455050826299/156077261327400 (0.3104555424294373)
23 : 4108466315759/13119537908680 (0.31315632794054454)
24 : 357267426009/1130994647300 (0.3158878133171515)
25 : 26326745113747/82653088824684 (0.3185210073587066)
26 : 6637039822613/20663272206171 (0.3211998446514621)
27 : 37228812888137/114795956700950 (0.32430421730897874)
28 : 2685312701058/8199711192925 (0.32748869294018335)
29 : 3439900369607/10405150755160 (0.33059591836294394)
30 : 86828291680/260128768879 (0.3337896536941231)
31 : 18444074519/54640701640 (0.3375519340970161)
32 : 2005767881871/5873875426300 (0.34147266264624354)
33 : 9635253619927/27900908274925 (0.3453383497406197)
34 : 573381038243/1641229898525 (0.34936058547209436)
35 : 282384301238/797168807855 (0.3542340072209197)
36 : 573033822439/1594337615710 (0.3594181162086006)
37 : 235645681349/646353087450 (0.3645773276626127)
38 : 29372048121/79376694950 (0.37003364954287504)
39 : 59847262801/158753389900 (0.3769825818440681)
40 : 5233528283/13607433420 (0.3846080389640444)
41 : 4339393/11062954 (0.39224541654968464)
42 : 1376123/3435700 (0.4005364263468871)
43 : 265142/643195 (0.41222646320322764)
44 : 16430899/38591700 (0.4257625085186711)
45 : 7909123/18009460 (0.4391649166604662)
46 : 356087/783020 (0.4547610533575132)
47 : 40283/83300 (0.48358943577430974)
48 : 1733/3332 (0.5201080432172869)
49 : 703/1275 (0.5513725490196079)
50 : 21/34 (0.6176470588235294)
51 : 1/1 (1.0)

The results agree with those of alex.jordan. The average over all $52$ cards is

$$ \frac{4581073495766609}{12946021439565200}\approx35.4\%\;, $$

in agreement with Charles's simulation.