Two diamonds drawn on two consecutive draws from a deck of cards you win $37. Otherwise pay me $5. If you play 713 times how much do you expect to win or lose

Question

I need help with this question, it is very confusing

Answer 1

The probability of drawing 2 diamonds is 1/16 because there are 4 suits in the pack. So there is a 1/16 chance of winning $37. The probability of losing is 15/16, which is a forfeit of $5.

Assuming the cards are each returned to the pack after drawing, the expectation over 713 games is 713/16*37=$1648.81 in winnings and 713*15/16*5=$3342.19 in losses.

The expectation is a loss of $1693.38 approximately.

There are other variations. For example if the cards are not returned to the pack until after drawing the second card, the odds for drawing 2 diamonds is 1/4*12/51=1/17 and the odds of failure are 16/17, making the gains=$1551.82 and the losses=$3355.29, with a total loss of $1803.47.

Another variation is to start by drawing two cards (game 1); return both cards to the pack; draw one card from the pack and compare it with the second card from game 1—this is game 2. After 713 games, 714 cards in total would have been drawn, because 2 cards are drawn for game 1 and thereafter only 1 card is drawn. This changes the results, but the odds are 1/17 for success and 16/17 for failure as before. In this case the gains are 714*37/17=42*37=$1554 and the losses are 16*714*5/17=42*80=$3360, with an overall loss of $1806. [Because this solution has amounts in exact dollars, it looks like the answer that was expected.]

ALTERNATIVE APPROACH

Consider the following sequence of 16 draws:

The following sequence of 17 cards (where S=spades, C=clubs, D=diamonds, H=hearts):

SCCDCHCSDDHDSHHSS

consists of the following 16 draws:

1. SC
2. CC
3. CD
4. DC
5. CH
6. HC
7. CS
8. SD
9. DD
10. DH
11. HD
12. DS
13. SH
14. HH
15. HS
16. SS

All 16 possible permutations of two card suits are included in these draws, and the permutation DD appears as number 9. It requires 17 cards to be drawn to create this list. So DD occurs once in 16 draws.

Now we turn to the probabilities. There are 13 cards of each suit. So when the first card is drawn, there is a probability of 13/52=1/4 of picking a particular suit. That leaves 51 cards in the pack and we have 12 cards of that suit remaining, giving us a probability of 12/51=4/17. Combine the probabilities and we get 1/4×4/17=1/17 of picking two cards of the same suit, and 1-1/17=16/17 of failing to do so. It means that we win $37 only once and we lose 16×5=$80, with a net loss of $43. After drawing the second card we return both cards to the pack, remembering the suit of the second card. Now we draw a third card with a probability of 1/4 of picking the required suit. But we are comparing the suit of the third card with that of the second card, for which the probabity 4/17 applied. So again the probability is still 1/17 (=4/17×1/4) for two cards of the same suit. The probability of 1/17 persists all along the sequence as we draw just one card to make up a pair.

The second S in draw 16 starts the next sequence, a further 16 draws. So we can now continue the sequence. We start from draw 1, because it began with S, but we could start anywhere where the first card is S, so, for example, we could start at draw 13, continue to draw 16 then follow up with draws 1-12. Note that after drawing the second card for each draw both cards are returned to the pack.

It happens that 713=714-1=42×17-1. This means we used 714 cards for 713 draws. So after 42 sequences we lose 42×43=$1806.