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# Math problem: When should you choose open chest?

You win a race with 3 people in the race, there are 5 chests before you, one contains $1000, the other 4 nothing, everyone will open a chest until the $1000 has been won, for winning the race you get to decide the order, when should you open the chest?

### 8 Answers

- Anonymous4 weeks ago
First i make a bargain that i will let the others choose the chest they want but they have to split it 20% with me. Then I bet 20% of of all my money on each person that they will find the right chest (including myself.) (and that nobody will find it..) then I roll a 5 sided dice to see what chest i should get I pick it and if its the right one i get $1000 and whatever i bet on me + the money i bet on others because it's one of those betting things where everyone puts money in and winner takes all. (split between winners of course.) This should mean i get more money but my logic might be flawed, (I saw it in a betting anime.) then if i don't get it and one of the others get's the money i get $200 + what i bet on them + what others bet on others + what i bet on others. Which should mean i still gain money. and for the last one where nobody get's it atleast I'll likely get 43% of my money i bet back or maybe even more depending on who also picked nobody and who even betted in the first place and how much they betted.

I don't know i don't try gambling or betting i just saw this in a anime and thought it might apply here? If I'm wrong please press thumbs down so I can know.\

Edit: i might have misunderstood the question. Just ignore me please.

- RichardLv 61 month ago
Since I have no idea what game this is in, I can't tell, not knowing the RNG stats

- JohnLv 71 month ago
it doesn't actually matter. no matter your position, it's 20% that you'll get it.

the winner in the race goes first and has a 1 in 5 chance or 20%

the second person has a 1 in 4 chance but you also have to take into account the chance that the first person loses, which is 4 in 5. 1/4 x 4/5 = 4/20=1/5=20%

the third person has a 1 in 3 chance and you factor in the other people's chances of losing so that's 1/3 x 3/4 x 4/5 = 12/60=1/5=20%

the last person has a 1 in 2 shot of getting the chest but it's still 20% that they'll even get to make a choice and select the correct chest 1/2 x 2/3 x 3/4 x 4/5= 24/120=1/5=20%

the remaining 20% is that no one selects the correct chest

i actually learned this from a mario party video believe it or not. the game bowser's big blast has a similar concept where there are 5 levers to press with one loser each round. it doesn't matter what order the players press the levers, everyone has an equal chance of losing

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- 1 month ago
One can easily say going LAST would yield the best chances, but you gotta take into consideration if it's found before then...at the same time it's needed to be known if the winners can only choose ONCE, if not the second chooser can win flawlessly if the other two don't find it at all.

If not, maybe going 2nd or 3rd may be best...but it's all probability 🤔

- 1 month ago
theres a few points here 3 people in the race and not by your description 3 OTHER people in the race., but 5 chests, that means dependant on whether somebody or not chooses correctly, it could be that at least 1 player may get to open 2 chests, otherwise some chests are not opened, and obviously having multiple goes increases your probability of finding the loot

first person 1/5 20%

1/4 25%

1/3 33%

1/2 50%

i would say to go 2nd 25% because if nobody finds it thereafter , it would be your turn again when there is only 1 chest and 100% probablity left..

anyway not much to do with video games this question lol

- jimanddottaylorLv 71 month ago
Go last. 3 chests will be open, so there is a good chance that the prize will be in one of the other 2. The 1st person to open one has a 1 in 5 chance of choosing correctly. The 2nd person has a 1 in 4 chance of choosing correctly. the 3rd (or last) person to choose has a 1 in 3 chance of choosing correctly.

If there had been 5 people in the race then when the 4th person chooses a chest, you know that you have either won or lost, before the last chest is opened.

- Anonymous1 month ago
This sounds like a probability problem.