Expert: Scott A Wilson - 11/13/2011

Question
I was having a discussion with a friend, who like me, is an engineer, so we have pretty good knowledge of math, but apparently not enough. It comes from a game we were theoretically "creating" and would like to know which option yields the higher expected value. Like I said, we both have master's degrees in engineering #me IE and him ME# so we would probably understand your explanation but just can't figure out the details.

Suppose you had a deck of cards and your goal is to get the highest total of two "scoring" cards. The cards are valued 2-13 for 2 through king, and aces are worth 15, except for the "bonus" Ace of spade which is 25. So a king and a 7 would be 20.

We came up with two scenarios:
1. Pick three cards and discard the lowest card. The two highest are your scoring cards.
2. Pick 4 cards and your scoring cards are the highest of either the middle two or the highest and lowest card. #So, if you picked an Ace of Clubs, 10, 6 and 2, you would take either 15+2 or 10+6, so obviously your total would be 17.#

Which option #best two out of three, of the highest of the 4 card option# would yield a higher expected value? Off the top of my head, I am pretty sure the standard deviation would be significantly higher for option 1 but I am not sure how to do the math.

Answer
The way to do this would be to count the number of possibilities.
On the first one, if three cards are drawn, the total number of choices is C(52,3).
That is, 52!/(3!(52-3)!).  That would be 52*51*50/(3*2*1).  This is the same as (52/2)(51/3)50.
That is 26*17*50 = 442*50 = 22,100.

There are C(4,3) ways to get all 2's, and that is 4.

There are C(8,3) ways to get all 2's and 3's,
which is 8*7*6/(3*2*1) = 56, but we have to throw out the 4 that only had 2's,
so there is 52 ways to have a three.

There are C(12,3) ways to get all 2's, 3's, or 4's, and that is C(12,3).
That is 12*11*10/6 = 220, but 56 of these choices have no 4 in them,
so it is 220 - 56 = 164.

There are C(16,3) wways to get from 2's to 4's, and that is 16*15*14/6 = 560.
Now there were 220 ways to not get only 1's, 2's, and 3's in that set,
so 560 - 220 = 340.

It comes down to the chance of getting each card as a high number:
2     4
3    52
4   164
5   340
6   580
7   884
8  1252
9  1684
10  2180
J  2740
Q  3364
K  4052
A  4804.

To make it more complicated, now we have to look at the 2nd highest number and add them,
and that is a lot more complicated.  To get 3-2's, there are 4 ways of doing that.
To get a 3 and 2-2's, there are 24 ways of doing that.  To get 2-3's and a 2,
there are 24 ways of doing that.   Again, to get 2-2's, there are 4 ways of doing that.
The total is 4+24+24+4 = 56, with only 4 ways of getting straight 2's, and that is what
the above table tells us.   The sum is the highest two cards, so for 3-3's or 2-3's and
a 2, the sum is 6, and there are 4+25=28 ways.  For 1-s and 2-2's, there are 24 ways,
and for 3-2's, there are 4 ways, but that was counted already in 2's alone.

For each of the higher numbers, it gets even more complicated, but can be done in the same style.
It might be easier to use a spreadsheet to put all the numbers in for three cards, and then find the sum of the bigger two.

What I got for the score and the number of ways was
4    1
5    3
6    7
7   12
8   19
9   27
10  37
11  48
12  61
13  75
14  91
15 108
16 124
17 138
18 151
19 159
20 163
21 162
22 157
23 147
24 133
25 114
26  91
27  63
28  69
29   0
30  37.

Note that there is no way to get a 29 since that requires a 15 and a 14,
and 14 is not a possibility.



When dealing with four cards, the concept seems even harder.
However, it can be done.  In just a little while longer, in fact, I did it.
Now to me, that took awhile, but to you, it hardly even passed.
One moment while I paste it in ...

4     1
5    10
6    39
7    92
8   185
9   318
10   511
11   760
12  1089
13  1490
14  1991
15  2580
16  2355
17  3144
18  2839
19  2498
20  2161
21  1804
22  1471
23  1134
24   841
25   560
26   343
27   154
28   142
28     0
30    49.


Looking at the average, I got 19.5 for the used 2 out of 3 dice and 17.5 for the one
that used 2 out of 4 dice.  As far as the variance, I got 25.7 for the 1st one and
16.2 for the 2nd one.

What this says is that the 1st one is better (with 3 dice used),
but it also has a highter variance, yet at the mean minus a standard deviation,
the 1st one is still higher.

When the number of standard deviations looked at gets out to around 1.88,
both have the same minumum value.