Question 1161243
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If you draw a red Ace, you get $1.  There are 2 red aces in the deck, so the probability of drawing a red Ace is 1/26.  The expected value of drawing a red ace is then (1/26)(1) which is approximately $0.04.


The club Ace gets you $2, 1 club ace, 1/52 times $2 is the same $0.04.


The spade Ace is worth $3, 1/52 times $3 gives E(AS) = $0.06.


For the 2s, you have 2 times 1/26, 4 times 1/52, and 6 times 1/52.


The rest are summarized in the following table:


<pre>


 A♥♦	 1	  1/26	 $0.04 
 A ♣	 2	  1/52	 $0.04 
 A ♠	 3	  1/52	 $0.06 
 2♥♦	 2	  1/26	 $0.08 
 2 ♣	 4	  1/52	 $0.08 
 2 ♠	 6	  1/52	 $0.12 
 3♥♦	 3	  1/26	 $0.12 
 3 ♣	 6	  1/52	 $0.12 
 3 ♠	 9	  1/52	 $0.17 
 4♥♦	 4	  1/26	 $0.15 
 4 ♣	 8	  1/52	 $0.15 
 4 ♠	12	  1/52	 $0.23 
 5♥♦	 5	  1/26	 $0.19 
 5 ♣	10	  1/52	 $0.19 
 5 ♠	15	  1/52	 $0.29 
 6♥♦	 6	  1/26	 $0.23 
 6 ♣	12	  1/52	 $0.23 
 6 ♠	18	  1/52	 $0.35 
 7♥♦	 7	  1/26	 $0.27 
 7 ♣	14	  1/52	 $0.27 
 7 ♠	21	  1/52	 $0.40 
 8♥♦	 8	  1/26	 $0.31 
 8 ♣	16	  1/52	 $0.31 
 8 ♠	24	  1/52	 $0.46 
 9♥♦	 9	  1/26	 $0.35 
 9 ♣	18	  1/52	 $0.35 
 9 ♠	27	  1/52	 $0.52 
10♥♦	10	  1/26	 $0.38 
10 ♣	20	  1/52	 $0.38 
10 ♠	30	  1/52	 $0.58 
 F♥♦	20	  3/26	 $2.31 
 F ♣	40	  3/52	 $2.31 
 F ♠	60	  3/52	 $3.46 
			$15.48
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The expected value of the game is the sum of the individual expected values.  By the way, don't add the numbers in the expected value column -- they are all rounded and you will get an incorrect sum.  The correct sum of the unrounded values which is then rounded to two digits is shown.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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