expected value question

Question

Phil is told there are 3 aces and 3 kings in a pile of cards. In each turn, a card is drawn without replacement, and Phil earns $1 if he guesses the drawn card correctly. Phil plays 6 turns following the optimal strategy. How much money should Phil expect to earn?

Answer 1

Assuming a shuffled deck of 3 Aces and 3 Kings for a total of 6 cards to start.

Given the optimum strategy of

1. guess the card type that has the highest remaining count. If the counts are equal, guess either one. (keep track of cards drawn)

NOTE. draws are NOT independent as knowledge of prior draws changes the guess strategy.

First draw: prior knowledge = 3 of one card, 3 of second card. P success = .5 Random guess (Expected value of winnings for first draw is .5 * $1 or $.5)

Second draw: prior knowledge = 2 of one card , 3 of another [2:3]. Guess the other. P success = 3/5 or .6. Calculation of expected winnings follows as above = $.60

Third draw: We do not ignore the prior draws and use that knowledge to make next guess. Either we have 1 of one card and 3 of another. OR 2 of one card and 2 of another (odds are 2/5 of the first combination [ 1:3 ] and 3/5 of the second combination [ 2:2 ]. Since we know which of these combinations we have we will pick the best choice for each combination. Expected value = .4 * 3/4 + .6 *2/4 or 8/12 ($.60)

Fourth draw: Following above logic our expected number of wins = $.65

here we can simplify calculations by using reverse logic:

Sixth draw: We know what last 5 draws were so we know what the remaining card is.

Probability of success is 1 and expected value of win is $1

Fifth draw : Following reverse logic from 6th draw we know that we will have 2 cards, either an A,K or 2K or 2A with proportion of occurrence of .5, .25, .25. For either 2K or 2A we will know what the next draw must be. Only in the case of the A + K pair will the chance be .5 of success. Expected value for successes will be E(guess=True) = .5 * .5 + .25 *1 + .25*1 or .75 Expected win = $.75

Sum of all wins for 6 guesses and draws = $4.10

Answer 2

The optimal strategy is to pick one specific card value (either ace or king) if the unrevealed count is equal, and whichever unrevealed count is larger if those counts are not equal. Before the first pick, the unrevealed count is equal (3 aces, 3 kings), so you could pick either one since the probability is 50/50.

I chose to pick aces whenever the unrevealed count was equal. After all cards of a certain type have been revealed, then clearly you should pick the other card type. Note also that If you reveal an ace on the first try, then the kings have a higher unrevealed count (3 kings, 2 aces unrevealed); thus, the optimal strategy would be to bet on kings on the next pick.

There are 20 possible sequences:6!/3!/3! since there are 6 cards, 3 aces, 3 kings.(I presume there are no other cards being used.)

Here are the 20 sequences and the pay-offs associated with each sequence and using the above strategy:

AAAKKK $4

AAKAKK $4

AAKKAK $5

AAKKKA $4

AKAKKA $5

AKKAAK $5

AKAKAK $6

AKKAKA $4

AKKKAA $4

AKAAKK $5

KAAAKK $4

KAAKAK $5

KAAKKA $4

KAKAAK $4

KAKKAA $3

KAKAKA $3

KKAAAK $4

KKAAKA $3

KKAKAA $3

KKKAAA $3

The total of the pay-offs is $82.00, and since there are 20 equally likely outcomes (as listed above), the expected value (pay-off) is $82/20 = $4.10

Note that the odds aren't always 50/50. For instance if you draw a king on the first pick, then the odds are 3/2 (60%) that you will pick an ace on the next attempt.

If anyone finds an error in the above calculations, please post a comment. I haven't exhaustively examined my calculations due to time constraints, and I would welcome anyone who would like to check my work.

Answer 3

To solve this problem, we need to use the concept of expected value and probability, considering Phil is following an optimal strategy.

Problem Breakdown

1. Card Composition: There are 6 cards in total: 3 aces and 3 kings.
2. Optimal Strategy: Since there are equal numbers of aces and kings, the optimal strategy for guessing the card is to guess the card type that has the highest remaining count. If the counts are equal, Phil can guess either one, as it makes no difference in probability.

Calculating Expected Earnings

Let's calculate the expected earnings for each turn and sum them up:

First Turn:

1. Probability of guessing correctly: 3/6=1/2 (since there are 3 aces and 3 kings).
2. Expected earnings: 1/2×1=0.5

Second Turn:

1. Regardless of the outcome of the first turn, there will still be 5 cards left, with either a 2:3 or 3:2 ratio of aces to kings. Thus, the probability of guessing correctly remains 1/2
2. Expected earnings: 1/2×1=0.5.

Third Turn:

1. Similar to the second turn, the probability of guessing correctly remains 1/2 due to the remaining cards.
2. Expected earnings: 1/2×1=0.5.

Fourth Turn:

1. Again, the probability of guessing correctly remains 1/2.
2. Expected earnings: 1/2×1=0.5.

Fifth Turn:

1. The probability remains 1/2.
2. Expected earnings: 1/2×1=0.5

Sixth Turn:

1. The probability remains 1/2
2. Expected earnings: 1/2×1=0.5.

Total Expected Earnings

Adding up the expected earnings from all 6 turns:

0.5+0.5+0.5+0.5+0.5+0.5=3

So, Phil should expect to earn $3 by playing 6 turns using the optimal strategy.