Current Events > There are three bottles of water. One of them is poisoned. You must drink one.

(+) Yeah! (+)

There are three bottles of water. One of them is poisoned. You must drink one.

Your rival is there. Your rival must also drink one.

You get to choose whether youd like to go first and third, or second.

If you drink the poisoned water, you die.

You can choose to go first, and if you dont drink the poisoned water, your rival must drink. If they still do not drink the poisoned water, you must drink the last water and die.

Or you can choose to go second, let your rival drink first, and hope they die on the first drink, or that youre able to not drink the poisoned water on your turn, leading them to have to drink it on the third turn.

What do you choose?

Pour all three on the floor and go home.
No rivalry is worth death.

Question: Can I drink all three?

Asking for a buddy.

Questionmarktarius posted...

Pour all three on the floor and go home.
No rivalry is worth death.

No you cant do that it ruins it

not this shit again.

I've seen a debacle over this and it's not nice.

MarbyIsBack posted...

Question: Can I drink all three?

Asking for a buddy.

No if you drink all three then you get in trouble

Ricemills posted...

not this shit again.

I've seen a debacle over this and it's not nice.

Wait what

Medussa posted...

this is one of those things that looks kinda complicated, but really isn't. everytime you drink, you have a 1/3 chance of dying, so you should only drink once.

the math:

if you drink first, you have a 1/3 (2/6) chance of picking the death bottle. and a 4/6 chance of not. half of those times (2/6), your opponent will drink the death bottle, and half the time you will in round three. so, you die 2/6 + 2.6 times, or 2/3 times. your opponent dies the other 1/3 of the times.

Wow!

I didnt know the answer. I honestly would have picked the other option.

Medussa posted...

this is one of those things that looks kinda complicated, but really isn't. everytime you drink, you have a 1/3 chance of dying, so you should only drink once.

the math:

if you drink first, you have a 1/3 (2/6) chance of picking the death bottle. and a 4/6 chance of not. half of those times (2/6), your opponent will drink the death bottle, and half the time you will in round three. so, you die 2/6 + 2.6 times, or 2/3 times. your opponent dies the other 1/3 of the times.

What's the math on if you chug all 3 at the same time?

MarbyIsBack posted...

What's the math on if you chug all 3 at the same time?

You 100% get in trouble

The first bottle is a red herring. At the end of the day, the final odds are 50/50, with the only difference that whoever goes first has an additional 1/3 chance to die in round one.

So going second is best.

https://gamefaqs.gamespot.com/a/user_image/8/7/1/AAfMcrAAD2y3.jpg

Kim_Seong-a posted...

The first bottle is a red herring. At the end of the day, the final odds are 50/50, with the only difference that whoever goes first has an additional 1/3 chance to die in round one.

So going second is best.

I..

think youre using the term red herring wrong

Smunchie posted...

https://gamefaqs.gamespot.com/a/user_image/8/7/1/AAfMcrAAD2y3.jpg

Aww

i drink all of them?

MICHALECOLE posted...

You 100% get in trouble

sounds like none were actually poisoned then

If my rival goes first and dies, do I still have to go? I'm not really thirsty.