Question

Suppose I ask you to pick any four cards at random from a deck of 52, without replacement, and bet you one dollar that at least one of the four is a face card (i.e., Jack, Queen, or King). Should you take the bet? Why? (Hint: See how the probability of this event compares to 50%. If this is too hard, try it with replacement first.) (b) What if the bet involves picking three cards at random instead of four? Should you take the bet then? Why? (c) Refer to the posted Rcode folder for this part. Please answer all questions.

Answer 1

Answer:

a) No, because you have only 33.8% of chances of winning the bet.

b) No, because you have only 44.7% of chances of winning the bet.

Step-by-step explanation:

a) Of the total amount of cards (n=52 cards) there are 12 face cards (3 face cards: Jack, Queen, or King for everyone of the 4 suits: clubs, diamonds, hearts and spades).

The probabiility of losing this bet is the sum of:

- The probability of having a face card in the first turn

- The probability of having a face card in the second turn, having a non-face card in the first turn.

- The probability of having a face card in the third turn, having a non-face card in the previous turns.

- The probability of having a face card in the fourth turn, having a non-face card in the previous turns.

1) The probability of having a face card in the first turn

In this case, the chances are 12 in 52:

$P_1=P(face\, card)=12/52=0.231$

2) The probability of having a face card in the second turn, having a non-face card in the first turn.

In this case, first we have to get a non-face card (there are 40 in the dech of 52), and then, with the rest of the cards (there are 51 left now), getting a face card:

$P_2=P(non\,face\,card)*P(face\,card)=(40/52)*(12/51)=0.769*0.235=0.181$

3) The probability of having a face card in the third turn, having a non-face card in the first and second turn.

In this case, first we have to get two consecutive non-face card, and then, with the rest of the cards, getting a face card:

$P_3=(40/52)*(39/51)*(12/50)\\\\P_3=0.769*0.765*0.240=0.141$

4) The probability of having a face card in the fourth turn, having a non-face card in the previous turns.

In this case, first we have to get three consecutive non-face card, and then, with the rest of the cards, getting a face card:

$P_4=(40/52)*(39/51)*(38/50)*(12/49)\\\\P_4=0.769*0.765*0.76*0.245=0.109$

With these four probabilities we can calculate the probability of losing this bet:

$P=P_1+P_2+P_3+P_4=0.231+0.181+0.141+0.109=0.662$

The probability of losing is 66.2%, which is the same as saying you have (1-0.662)=0.338 or 33.8% of winning chances. Losing is more probable than winning, so you should not take the bet.

b) If the bet involves 3 cards, the only difference with a) is that there is no probability of getting the face card in the fourth turn.

We can calculate the probability of losing as the sum of the first probabilities already calculated:

$P=P_1+P_2+P_3=0.231+0.181+0.141=0.553$

There is 55.3% of losing (or 44.7% of winning), so it is still not convenient to bet.