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Probability of the sum of three random cards from a standard deck of playing cards

Ok, I was watching TV and saw hold'em poker playing. In that game three cards are flipped over in the center of the table. If one was to give the cards valuations (Ace=1, 2=2,....and so forth Jack, Queen, King are all 10), what would be the average point total of the three cards. A proof on how this is calculated would be greatly appreciated. Thank you in advance.
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3 Answers

There are 22,100 ways to select 3 cards from a deck of 52 cards.  This spreadsheet image shows the total number of combinations of cards that will yield each sum from 3 to 30, and then computes the actual expected value of the total of the three cards: http://www.wyzant.com/resources/files/259851/avgcardtotal_png
[ The actual spreadsheet is available at http://www.wyzant.com/resources/files/259846/avgcardtotal ]
 
Warning:  This is a very large spreadsheet, so when you bring it up you may need to reduce the size of your view to avoid scrolling.
 
The actual expected total value of the three cards is 19.6153846153846 according to this calculation.  The standard deviation is approximately 5.352847465.
 
The way the spreadsheet works is there is a number in each column of a row indicating how many ways three numbers having the specified relationship indicated by the column heading can be chosen to add up to the value in column A.   
 
For example, the number 27 has a 1 in the column under COMBIN(16,2)*COMBIN(4,1), also a 1 under the column COMBIN(16,1)*(COMBIN(4,1))2, and a 1 under the column COMBIN(4,3).  This corresponds to the fact that there are three ways for the three card values to add to 27.  First 27 can come in 1 way from 2 cards of value 10 and one of value less than 10; second there is 1 way in which 27 can result from one card of value 10 and two cards of value less than 10 of unequal values; third there is one way that 27 can result from three cards of equal value less than 10.
 
Each way of expressing the number in column A, indicated in columns C-I, can be chosen from the cards in the number of ways indicated in row 4 of the column. So for each row the total number of combinations of cards that can add up to the number in column A  is the sum of the products of the number in each of the columns C-I and the number of combinations formula in row 4 of that corresponding column.  The value of that sum appears in column B. 
 
You can get a rough estimate by finding first the average point total of any one card and multiplying it by 3: The total number of points in the deck is
4(1+2+3+....+9+10+10+10+10)=340, so the average per card is 340/52=6.54.
For three cards drawn from three complete decks, you would have on average 3(6.54)=19.6 points.
 
However, your three cards come from the same deck, and this where it gets tricky. Drawing the first card will effect the average points of the remaining 51 cards: for example, if you drew a king (of value 10) first, then the average point number for the second card will be lower than 6.54 (it would be (340-10)/51=6.47).
To find the exact answer, you would have to go through all possibilities ( (king, 7, 8), (ace,5,10), etc.) and calculate the value, then find the average. I suspect it would still be close to 19.6.
There are 52C3 possible 3 card combinations = 52*51*50/(3*2*1)
= 52*17*25 = 22,100

Expected values for 1st card:
1 (4/52)
2 (4/52)
3 (4/52)
4 (4/52)
5 (4/52)
6 (4/52)
7 (4/52)
8 (4/52)
9 (4/52)
10 (16/52) = 40(4/52)

Total expected value for 1st card:
(4/52)(40 + 45) = (1/13)(85) = 85/13 = 6.53846153846154

Expected values for 2nd card given 1st had value 10:
1 (4/51)
2 (4/51)
3 (4/51)
4 (4/51)
5 (4/51)
6 (4/51)
7 (4/51)
8 (4/51)
9 (4/51)
10 (15/51) = 150/51

Total expected value for 2nd card given 1st had value 10:
(4/51)(45)+150/51 = (180+150)/51 = 330/51 = 110/17 = 6.47058823529412

At this point I might notice how monumental this approach will be, so I might estimate the expected value for 3 cards as 3 * 6.5 = 19.5.

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