Probability of getting the smallest value of cards

[songoku]

Homework Statement

Martin plays game with 4 cards numbered 1, 2, 3 and 4. In each round, a random sample of n cards is taken without replacement (1≤ n ≤ 4). Find the probability of getting 2 as the smallest values

Relevant Equations

Probability

There are 3 cases of getting 2 as smallest values:
(1) Taking one card (n = 1) → Probability = 1/4

(2) Taking two cards (n = 2) → Probability = 1/4 x 2/3 x 2! = 1/3

(3) Taking three cards (n = 3) → Probability = 1/4 x 1/3 x 1/2 x 3! = 1/4

Total probability = 1/4 + 1/3 + 1/4 = 5/6

But the solution is 5/24.

I don't know where my mistake is but maybe I misinterpret the question. I also don't understand how many rounds the game will take (the question says "in each round")

Thanks

 

[Delta2]

You should multiply each probability by 1/4 because each probability of the sum is actually $P(A_kB)=P(A_k)P(B/A_k)$ where
$A_k$={ the event to draw k cards},
$B$={the event to get 2 as the smallest value}
and $P(A_k)=1/4$.
and the probability of what is asked is $$P(B)=\sum_kP(A_k)P(B/A_k)$$
Also I am a bit in doubt on how you calculate $P(B/A_k)$ (though their final values seem correct to me but I calculated them in another way) if you can expand on that please.

 

[PeroK]

Homework Statement:: Martin plays game with 4 cards numbered 1, 2, 3 and 4. In each round, a random sample of n cards is taken without replacement (1≤ n ≤ 4). Find the probability of getting 2 as the smallest values
Relevant Equations:: Probability

There are 3 cases of getting 2 as smallest values:
(1) Taking one card (n = 1) → Probability = 1/4

(2) Taking two cards (n = 2) → Probability = 1/4 x 2/3 x 2! = 1/3

(3) Taking three cards (n = 3) → Probability = 1/4 x 1/3 x 1/2 x 3! = 1/4

Total probability = 1/4 + 1/3 + 1/4 = 5/6

But the solution is 5/24.

I don't know where my mistake is but maybe I misinterpret the question. I also don't understand how many rounds the game will take (the question says "in each round")

Thanks

I suspect the assumption is that you choose $n$ uniformly from the set $\{1,2,3,4 \}$. Then the probability is the average of the ones you calculated (including the case where $n = 4$).

This is what @Delta2 has calculated.

 

[Delta2]

BTW I think the problem asks for P(B) in one round of the game.

 

[songoku]

You should multiply each probability by 1/4 because each probability of the sum is actually $P(A_kB)=P(A_k)P(B/A_k)$ where
$A_k$={ the event to draw k cards},
$B$={the event to get 2 as the smallest value}
and $P(A_k)=1/4$.
and the probability of what is asked is $$P(B)=\sum_kP(A_k)P(B/A_k)$$
Also I am a bit in doubt on how you calculate $P(B/A_k)$ (though their final values seem correct to me but I calculated them in another way) if you can expand on that please.

BTW I think the problem asks for P(B) in one round of the game.

I think I understand.

I suspect the assumption is that you choose $n$ uniformly from the set $\{1,2,3,4 \}$. Then the probability is the average of the ones you calculated (including the case where $n = 4$).

This is what @Delta2 has calculated.

I can understand why I need to multiply my answer by 1/4 (since I need to consider the possibility of drawing 1, 2, 3 or 4 cards in one round) but sorry I don't understand the logic related to average probability. From calculation perspective, the result will be the same but when doing other practice questions, I never think of finding average probability, always probability only.

Is there any way to explain about average probability intuitively? And also is there any way to know whether we need to calculate average probability or not? Maybe my last question can be rephrased as: how can I know I need to calculate average probability in this case

Thanks

 

[PeroK]

The average probability is just when you get when you have several options all with different probabilities. I would have thought the concept is self explanatory.

 

[songoku]

Thank you very much for the help and explanation Delta2 and PeroK

 

[Steve4Physics]

Hi Guys. I can’t get the official answer (5/24). As far as I can tell, what I’ve done is equivalent to @Delta’s Post #2 method/formula. It’s frustrating – a consequence of old age presumably.

Can anyone point out/hint where I’ve gone wrong? Thanks.
______

Probability that n=1 is ¼ (since there are 4 equal-possibility outcomes for n (1,2,3 or 4)).
n=1. One card chosen.
Probability of the card being the ‘2’ = ¼.
Overall probability that n=1 and the ‘2’ is chosen = ¼ x ¼ = 1/16.
______

Probability that n=2 is ¼.
n=2. Two cards chosen.
Probability of one of the cards being the ‘2’ = ½ (because the six possible pairs are 12, 13, 14, 23, 24 and 34, and the ‘2’ is present in three of the six possibilities).
Probability that n=2 and the ‘2’ is chosen = ¼ x ½ = 1/8.
______

Probability that n=3 is ¼.
n=3. Three cards chosen.
Probability of one of the cards being the ‘2’ = 3/4 (because the four possible triplets are 123, 124, 134 and 234, and the ‘2’ is present in three of the four possibilities).
Probability that n=3 and the ‘2’ is chosen = ¼ x 3/4 = 3/16.
______

Probability that n=4 is ¼.
n =4. Four cards chosen.
Probability of one of the cards being the ‘2’ = 1.
Probability that n=4 and the ‘2’ is chosen = ¼ x 1 = ¼.
______

Total probability that ‘2’ is chosen = 1/16 + 1/8 + 3/16 + ¼ = 10/16 = 5/8. But his is three times too big.
AAGH. (Sound of distress. Not an acronym.)

 

[PeroK]

Hi Guys. I can’t get the official answer (5/24). As far as I can tell, what I’ve done is equivalent to @Delta’s Post #2 method/formula. It’s frustrating – a consequence of old age presumably.

Can anyone point out/hint where I’ve gone wrong? Thanks.
______

Probability that n=1 is ¼ (since there are 4 equal-possibility outcomes for n (1,2,3 or 4)).
n=1. One card chosen.
Probability of the card being the ‘2’ = ¼.
Overall probability that n=1 and the ‘2’ is chosen = ¼ x ¼ = 1/16.

Okay

______

Probability that n=2 is ¼.
n=2. Two cards chosen.
Probability of one of the cards being the ‘2’ = ½ (because the six possible pairs are 12, 13, 14, 23, 24 and 34, and the ‘2’ is present in three of the six possibilities).
Probability that n=2 and the ‘2’ is chosen = ¼ x ½ = 1/8.
______

Out of the six possibilities, 2 is the lowest in two cases. That gives $\frac 1 4 \times \frac 1 3 = \frac 1 {12}$.

Probability that n=3 is ¼.
n=3. Three cards chosen.
Probability of one of the cards being the ‘2’ = 3/4 (because the four possible triplets are 123, 124, 134 and 234, and the ‘2’ is present in three of the four possibilities).
Probability that n=3 and the ‘2’ is chosen = ¼ x 3/4 = 3/16.

Again, out of the four possibilities, 2 is the lowest only once. So $\frac 1 4 \times \frac 1 4 = \frac 1 {16}$

______

Probability that n=4 is ¼.
n =4. Four cards chosen.
Probability of one of the cards being the ‘2’ = 1.
Probability that n=4 and the ‘2’ is chosen = ¼ x 1 = ¼.

2 can never be the lowest in this case.

______

Total probability that ‘2’ is chosen = 1/16 + 1/8 + 3/16 + ¼ = 10/16 = 5/8. But his is three times too big.
AAGH. (Sound of distress. Not an acronym.)

 

[Steve4Physics]

Okay

Out of the six possibilities, 2 is the lowest in two cases. That gives $\frac 1 4 \times \frac 1 3 = \frac 1 {12}$.

Again, out of the four possibilities, 2 is the lowest only once. So $\frac 1 4 \times \frac 1 4 = \frac 1 {16}$

2 can never be the lowest in this case.

Thankyou @PeroK! I had a mental convolution.. My brain completely changed the problem into calculating the probability that a '2' was drawn - not that '2' needed to be the smallest in the draw. I need to go and lie down. Thanks again.