$Using probability to calculate expected time a task would take with "bad" luck$

Question

$Using probability to calculate expected time a task would take with "bad" luck$

Base assumptions:

- There is a video game where you can bring one item with you into an "expedition".
- There are a total of 20 items you can choose from.
- 1 Expedition lasts a maximum of 4 minutes.
- There are 75 maps each player can be sent to for any given expedition, with a 1/75 random and equal chance.
- Let's assume a person is extremely unlucky and inefficient, and needs to see a map 6 times on average to explore the entirety of it.

The task:

- There is a secret easter egg that requires you to bring a [specific item] to a [specific spot] of a [specific map] to trigger a flag, but none of the information in square brackets is known.
- Each time, the item, the map, and the spot are different.
- A player needs to trigger 7 flags in this manner to complete the easter egg.
- In addition, the order of flags triggered is determined, and it doesn't work if you trigger them out of order. As such, once a flag is triggered, you may need to search all previous maps all over again.

The question:

If there are 20 players working collectively (they all have different items, and they go on separate expeditions simultaneously, meaning they each have their own individual random chance to encounter any given map), but they cooperate and corroborate information about the correct item, map, and spot (the clear conditions are the same for everyone), how long is it expected to brute force all 7 flags?

Keep in mind that all 19 other players have to essentially wait for 1 player to get the correct item on the correct spot of the correct map before they can explore the next item/map combination.

EDIT: I'll give the bounty even for an approximation or less-detailed and simplified answer, so here is the new simplified question below:

All trials are essentially independent from one another, so instead of brute forcing all 7 flags, let's just approximate how much time it'd take on average to trigger 1 flag and multiply that by 7.
In addition, 20 items are split between 20 people, but let's just say that 1 person can test all 20 items all at the same time on a given map (which means the correct item is irrelevant, only the correct map).
So all I'm asking is the estimated time for 1 person to get a successful trial with "bad" luck (needs to see a map 6 times on average to trigger the flag, with a 1/75 chance to see each map). And then multiply that value by 7.

If anyone has a better easy method of approximating the answer, feel free to do that instead.

Probability

Hayitzchance

8

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Erdos

+2

This is a project-like question that may take several hours to answer thoroughly and carefully. You should consider offering a significantly higher bounty to provide enough incentive for users to invest their time, especially if you are serious about getting a detailed response.
- Hayitzchance
  
  0
  
  Hello, Thank you for the clarification. I'm not a math expert (which is why I'm here), so maybe it would help if I simplified the situation a bit as I'd be satisfied with just an approximation
- Erdos
  
  +1
  
  Any reasonable answer to this question would take around two hours to complete. I’m not aware of any profession in North America where someone would work two hours for just $10. I live in the USA, where the minimum wage is over $15 an hour!
- Hayitzchance
  
  -2
  
  I don't really have a understanding of why it would take 2 hours to complete (not doubting you, just looking for clarification).
- Hayitzchance
  
  -2
  
  Is there no formula for just calculating estimated time it would take for an event of X% probability to occur? Or is it not that simple? And just finding the average probability it'd take for 1 person to get 1/75 chance 6 times (doesn't have to be consecutive). Again, I only need approximations. Thanks

Answer

The answer is accepted.

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Erdos

+2

This is a project-like question that may take several hours to answer thoroughly and carefully. You should consider offering a significantly higher bounty to provide enough incentive for users to invest their time, especially if you are serious about getting a detailed response.

Hayitzchance

0

Hello, Thank you for the clarification. I'm not a math expert (which is why I'm here), so maybe it would help if I simplified the situation a bit as I'd be satisfied with just an approximation

Erdos

+1

Any reasonable answer to this question would take around two hours to complete. I’m not aware of any profession in North America where someone would work two hours for just $10. I live in the USA, where the minimum wage is over $15 an hour!

Hayitzchance

-2

I don't really have a understanding of why it would take 2 hours to complete (not doubting you, just looking for clarification).

Hayitzchance

-2

Is there no formula for just calculating estimated time it would take for an event of X% probability to occur? Or is it not that simple? And just finding the average probability it'd take for 1 person to get 1/75 chance 6 times (doesn't have to be consecutive). Again, I only need approximations. Thanks
Hayitzchance

0

Hello, Thank you for the clarification. I'm not a math expert (which is why I'm here), so maybe it would help if I simplified the situation a bit as I'd be satisfied with just an approximation
Erdos

+1

Any reasonable answer to this question would take around two hours to complete. I’m not aware of any profession in North America where someone would work two hours for just $10. I live in the USA, where the minimum wage is over $15 an hour!
Hayitzchance

-2

I don't really have a understanding of why it would take 2 hours to complete (not doubting you, just looking for clarification).
Hayitzchance

-2

Is there no formula for just calculating estimated time it would take for an event of X% probability to occur? Or is it not that simple? And just finding the average probability it'd take for 1 person to get 1/75 chance 6 times (doesn't have to be consecutive). Again, I only need approximations. Thanks

Answer 1

Answers can only be viewed under the following conditions:

The questioner was satisfied with and accepted the answer, or
The answer was evaluated as being 100% correct by the judge.

View the answer

This is a basic expected value question. The number of minutes for finding a single flag is of the form:

$$\sum_{k=1}^4 p(k)*4*k$$

where k is the round number, and $p(k)$ is the probability that it's found on that round. There are a maximum of four rounds, since each of the 20 members of the party can search a map and there's only 75 maps.

Since each person can check each item at a given spot, and finding each spot is part of the expedition, really we're trying to figure out the probability that the person is on the right map for that round.

What's the probability they'd find it in one round? Assuming independence, $p(1) = \frac{20}{75}$. The next one is $p(2)= \frac{55}{75}\frac{20}{55} = \frac{20}{75}$, because while there's 55 places remaining, we're also assuming it was not in the first 20. All together, the expected amount of time to find one spot is:

$$\frac{20}{75}*4 + \frac{20}{75}*8 + \frac{20}{75}*12 + \frac{15}{75}*16 = \frac{720}{75}$$

Which is 9 minutes and 36 seconds per round. Multiplied by 7 rounds is 67 minutes and 12 seconds. This does not account for "unlucky person" who requires multiple expeditions per map

Larrybo

49

$Using probability to calculate expected time a task would take with "bad" luck$

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