Expected value of highest card in 4 card draw from standard deck

Question

Suppose you draw 4 cards from a standard 52 cards deck, what is the expected value of the highest card among the 4 cards? (Ace = 1, ..., King = 13) An accurate result is not necessary, I am more interested in a fast method to approximate it without using a calculator (if the accurate result is not simple enough).

I thought about (falsely) assuming that, by denoting the values of the cards as $X_1,\dots,X_4$, $$ P(\max(X_1,\dots,X_4)\leq x) = P(X_1\leq x)^4 $$

Then, for simplicity calculating the maximum of 4 i.i.d. uniforms on $[0,1]$ gives $$ P(\max(U_1,\dots,U_4)\leq x) = \int_0^1x4x^3dx = \frac{4}{5} $$ and finally, $$ 12\cdot\frac{4}{5}+1 = 10.6 $$ was my approximation for the max of 4 i.i.d. uniforms on $[1,13]$.

How bad is that?

As the 4 draws are in fact not independent because, loosely speaking, drawing a small card reduces the probability of drawing another small card, we have $$ P(\max(X_1,\dots,X_4)\leq x) < P(X_1\leq x)^4 $$ and thus the expected value should be a little higher, and of course the draws are not uniform from $[1,13]$, does that make a big difference?

In general, how would you approach these kind of questions, for example finding the expected value of the square of the number of red cards in that 4 card draw? I somehow always assume that the draws are independent, e.g. approximate the probability for a red card by $1/2$ to make the computation of the expected value as simple as possible and then try to correct the answer in the right direction, but I am sure there are better approaches.

Answer 1

$$P[\max\le k]=\frac{{4 k}\choose{4}}{52\choose4}=\frac{4k(4k-1)(4k-2)(4k-3)}{52(51)(50)(49)}$$ $$E[\max]=\sum_{k=1}^{13}kP[\max=k]=P[max\le 1]+\sum_{k=2}^{13}k(P[max\le k]-P[max\le k-1])$$ $$=\frac{32577}{2975}\approx 10.95$$

For an approximation, I would use the fact that the maximum of 4 independent uniform random variables has a Beta distribution with parameters 4 and 1 see here. The mean of that Beta distribution is 0.8. These are not Uniform(0,1) and not independent, but they are Unif(0.5,13.5) rounded to the nearest integer. Thus, if I took Unif(0,1) random variables, multiplied them by 13 and added 0.5, I would have Unif(0.5, 13.5) random variables. The maximum of those would have mean approximately 0.8*13+0.5=10.9. This approximation works very well in this case.

Answer 2

You want a rough estimate, so I will calculate in my head and check against the calculator sometimes to see how far off I am getting. P(no kings) = $$\frac {48 \choose 4}{52 \choose 4} = \frac{48\cdot47\cdot46 \cdot45}{52\cdot51\cdot50\cdot49}\approx .92^4 \approx .68 $$ so there is very roughly a 1/3 chance that the max is 13. P(no kings or queens) = $$\frac {44 \choose 4}{52 \choose 4} = \frac{44\cdot43\cdot42 \cdot41}{52\cdot51\cdot50\cdot49}\approx .85^4 \approx .5 $$ (I am using rough fractions and tricks like $$(1-1/n)^4 \approx 1-4/n$$ So at this point a rough lower bound of the expected value is roughly $1/3*13 + 1/6*12 + 1/2*(1) \approx 13/2 = 6.5$ At this point taking a wild stab at it, I would guesstimate that P(13) = 1/3, P(12) = 1/6, P(10) = 1/12, P(9) = 1/24, which gives roughly 4.3 + 2 + .9 + .4 or about 7.6 and that is probably fairly low. After all, the probability of the max being 1 is $$\frac{1}{52 \choose 4} \approx 1/12500$$ (okay, it's exactly 1/270725, but the point is, ignorably small). So the distribution is probably still weighted more heavily to the high end, so let's try $P(13) = 1/3, P(12) = 1/6, P(11) = 1/6, P(10) = 1/6$, and $P(9) = 1/6$, giving about 4.3 + 2 + 1.83 + 1.67 + 1.5 + 1.33 to finish, or ...well, 12.5 is a bit high. So this is a bit too rough, and taking a while. Still, I wouldn't be surprised if the true answer is more than 11. Let's try trimming the low values: $P$(all cards ≤ 5) = $$\frac{20 \choose 4}{52 \choose 4} \approx \frac{20\cdot 19 \cdot 18 \cdot 17}{52\cdot51\cdot50\cdot49} \approx .4^4 = .0256$$. The true value is .01789 so not bad. The same trick for all cards less than 8 gives roughly $.6^4 = .1296$ and we won't go too far wrong giving say 7 that probability and only other values over 8 get values. So all the values up to 8 contribute about 1 to the expected value. That just leaves 9 10 and 11 to estimate. $$P(no cards above 10) = \frac {40 \choose 4}{52 \choose 4} = \frac{40\cdot 39 \cdot 38 \cdot 37}{52\cdot51\cdot50\cdot49} \approx .75^4 = 81/256 \approx 1/3$$, so I wasn't too far wrong with my guess of 1/6 for P(max = 11). No cards above 9 would involve 36/52 and the like, or about ${\frac{2}{3}}^4 = 16/81 \approx .2$, meaning $P(10) \approx .13$. No cards above 8 uses 32/52 so about $.6^4 = .13$, making $P(9) \approx .07$ So my new estimate becomes $$13/3 + 12/6 + 11/6 + 10\cdot .13 + 9 \cdot .07 + 1$$ or $$4.33 + 2 + 1.833 + 1.3 + .63 + 1 = 11.1$$ or so.