I don't think you have enough information to answer this question. You are selecting or choosing 4 cards - 1 from each suit - at 2 different times. When the problem talks about "the selected card" it is not clear which card is being referred to and which cards it is being matched to. Sorry. Check to see if you have copied the problem correctly.
If the problem is actually saying that if your second card in a suit matches the first card in that suit for one suit, you win all the money you staked; if you match in 2 suits, you get 2x the money you staked, etc., then you could answer the question.
Let s = the total money staked (s/4 is the money staked on each suit).
Let X = the number of matches you get when you select the cards the second time
Let S = the amount of money win.
The probability that you match in any one suit is 1/13. The probability that you do not match any one suit is 12/13.
The probability that you match all 4 suits is (1/13)^4. There is only 1 way this occurs. 0.000035
The probability that you do not match any of the 4 suits is (12/13)^4. There is only 1 way this occurs. 0.7260
The probability that you match 1 suit and do not match any of the other 3 is [(1/13) (12/13) (12/13) (12/13)]. There are 4 ways this occurs, so the total probability of matching only 1 suit is 4[(1/13) (12/13)^3]. 0.2420
The probability that you match 2 out of the 4 suits is [(1/13) (1/13) (12/13) (12/13)]. There are 6 ways this can occur, so the total probability of matching 2 suits is 6[(1/13)^2 * (12/13)^2]. 0.0302
The probability that you match 3 out of the 4 suits is [(1/13)^3 * (12/13)]. There are 4 ways this can occur, so the total probability of matching 3 out of 4 suits is 4[(1/13)^3 * (12/13)]. 0.000420
The only possible values the number of matching suits is 0, 1, 2, 3, or 4, This corresponding amounts won are 0, s, 2s, 3s, or 4s. Combining the probabilities with the possible amounts won, we get the following probability model:
S | 0 s 2s 3s 4s
P(S) | 0.7260 0.2420 0.0302 0.000420 0.000035
Note: the probabilities do not add up to exactly 1 because of rounding errors.
The expected amount the gambler will get is the sum of S times P(S) for all 5 possibilities:
0(.7260) + s(0.2420) + 2s(0.0302) + 3s(0.00042) + 4s(0.000035) = 0.3042s
On average, if the gambler plays the game many, many times, he will win back 0.3042 times his original stake each time he plays.
Let me know if you have any questions.
