Carnival Game Problem

[Sonyvaio]

1. You are at a carnival and there is a game being played.

2. The game consistents of 40 tiles on a table numbers 1 through to 40.

3. Prior to the game starting, you have to pick 2 numbers, between 1-40. Completely random numbers.

4. Once the game starts, the host will turn over 10 of the 40 tiles.

In order to win the game, ONLY one of your 2 chosen tiles has to be revealed in the 10 tiles she has turned over.

If both do that is fine, but in order to win, you only need ONE of them turned over.

If I run this game 100 times what's my chances of winning ?

My solution

Event 1. Pick 2 balls out of 40 with only 1 needing to be correct 2/40 =1/20

Event 2. Pick 2 balls out of 10
2/10 =1/5

Probability of both events =1/20 * 1/5

=1/100

Out of 100 games I will win one time


Just need to check if this is correct

Thanks

 

[Deleted member 4993]

1. You are at a carnival and there is a game being played.

2. The game consistents of 40 tiles on a table numbers 1 through to 40.

3. Prior to the game starting, you have to pick 2 numbers, between 1-40. Completely random numbers.

4. Once the game starts, the host will turn over 10 of the 40 tiles.

In order to win the game, ONLY one of your 2 chosen tiles has to be revealed in the 10 tiles she has turned over.

If both do that is fine, but in order to win, you only need ONE of them turned over.

If I run this game 100 times what's my chances of winning ?

My solution

Event 1. Pick 2 balls out of 40 with only 1 needing to be correct 2/40 =1/20

Event 2. Pick 2 balls out of 10
2/10 =1/5

Probability of both events =1/20 * 1/5

=1/100

Out of 100 games I will win one time


Just need to check if this is correct

Thanks

I will first calculate the chance of NOT picking either of the two balls in 10 picks

Chance of NOT picking winning ball in the first draw = 38/40

Chance of NOT picking winning ball in the second draw = 37/39

Chance of NOT picking winning ball in the third draw = 36/38
.
.
Chance of NOT picking winning ball in the tenth draw = 29/31..........................edited

Chance of winning = 1 - (38*37*36...*29)/(40*39*38...*31)

 

[Dr.Peterson]

1. You are at a carnival and there is a game being played.

2. The game consists of 40 tiles on a table numbers 1 through to 40.

3. Prior to the game starting, you have to pick 2 numbers, between 1-40. Completely random numbers.

4. Once the game starts, the host will turn over 10 of the 40 tiles.

In order to win the game, ONLY one of your 2 chosen tiles has to be revealed in the 10 tiles she has turned over.

If both do that is fine, but in order to win, you only need ONE of them turned over.

If I run this game 100 times what's my chances of winning ?

My solution

Event 1. Pick 2 balls out of 40 with only 1 needing to be correct 2/40 =1/20

Event 2. Pick 2 balls out of 10
2/10 =1/5

Probability of both events =1/20 * 1/5

=1/100

Out of 100 games I will win one time


Just need to check if this is correct

Thanks

No, it isn't correct.

In fact, "I will win one time" isn't even a probability ("chances of winning"), but a claim that you are certain to win exactly once, which is very wrong.

If you want additional help in addition to this (which is what you asked for), try explaining WHY you thought what you did might be correct. Why those divisions, what do the results mean, why multiply the them?

Then back up and think about, in one game, what is the probability that you will NOT win?

 

[Sonyvaio]

No, it isn't correct.

In fact, "I will win one time" isn't even a probability ("chances of winning"), but a claim that you are certain to win exactly once, which is very wrong.

If you want additional help in addition to this (which is what you asked for), try explaining WHY you thought what you did might be correct. Why those divisions, what do the results mean, why multiply the them?

Then back up and think about, in one game, what is the probability that you will NOT win?

Sorry english is not my first language after rustling around forums and going thru examples online ...I know I am wrong

After digging this is what I came up with
The probability of winning in a single round is (2 choose 1)(38 choose 9)/(40 choose 10) + (2 choose 2)(38 choose 8)/(40 choose 10) = 0.44

So
I can most probably won 4.4 games if I played 10

 

[Dr.Peterson]

Sorry english is not my first language after rustling around forums and going thru examples online ...I know I am wrong

After digging this is what I came up with
The probability of winning in a single round is (2 choose 1)(38 choose 9)/(40 choose 10) + (2 choose 2)(38 choose 8)/(40 choose 10) = 0.44

So
I can most probably won 4.4 games if I played 10

Your probability for one game is correct (0.4423), though I'm not sure of the reasons for your calculation, since you didn't explain it. I used a much simpler method, based on first finding the probability that neither of my cards matches (that is, I don't win).

But the question, as you stated it, was, "If I run this game 100 times what's my chances of winning?" That is asking for a probability, not for the expected number of wins out of 10 games. So unless you mistranslated the question, you have not answered it.

On the other hand, the answer to the question as stated is 1: you are virtually certain to win at least once out of 100 games, because the probability of losing all 100 games is essentially zero (3.7*10^-36). So there is a good chance that wasn't the question you were really asked.