BrainDen.com - Brain Teasers

## Question

There are two bowls that you and a challenger must eat from.  After flipping a coin you were selected to pick the bowl that each would eat from.  In the first bowl there are three out of five poisonous apples.  In the second bowl, there are two out of five poisonous apples.  Whoever eats from the first bowl must eat two apples at random from the bowl.  Whoever eats from the second bowl must eat three random apples from the second bowl.  Which bowl should you pick to eat?

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• 1

assuming five apples in each bowl,

Spoiler

you have a 2/5*1/4 chance to survive in the first bowl, or 10%. You have a 3/5*2/4*1/3 chance to survive with the second bowl, or a 10% chance. Its the same chance either way. I say it doesn't matter. (I assume that after eating an apple you take it out of the bowl.)

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• 1

Probabilities of something occuring in sequence multiply together, they don't add.

We could calculate the probability where we don't eat 2 poisonous or where we do.  I chose to calculate where we do.

For the first bowl.  We only care if you eat 2 poisonous apples. The probability the first apple eaten is poisonous is 3/5.   Assuming you ate a poisonous apple because that is the only scenario we care about, the probability the second apple is poisonous is 2/4.  This is the only scenario eating from this bowl that we die.  Multiply this together (3/5 * 2/4) and the chance of dying is 30%.   70% chance you survive.

For the second bowl, we do the same thing, but there are multiple possibility branches for this one.

The probability the first apple eaten is poisonous is 2/5.  The probability the second apple is poisonous is 1/4 if you already ate a poisonous one.
So a 10% chance you die after eating the second apple (assuming it is fast poison).  If you are still alive then the 2nd apple was not poisonous so that had a probability of 3/4.  That makes he probability for the third equal to 1/3. So 2/5 * 3/4 * 1/3 is 10% chance of dying.
The chance the first apple is not poisonous is 3/5.  So for the second the chance is 2/4, leaving the last at 1/3.    So multiply these out and you get 10%%.
Due to having 3 scenarios that result in death we have to add those probabilities.  10% + 10% + 10% = 30%.  70% chance we survive.

So both bowls have the same odds.  Others arrived at this answer, but with the wrong math.  Cheers.

Edited by Taciav

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I choose the second bowl. The other person has to go first. I never have to eat an apple. :)

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the odds of eating a poison apple  for the 1st bowl  = 3/5 + 3/4 =  12/20 + 15/20 = 27/20 = !.35 chances of eating the P/A

2nd bowl =2/5 + 2/4= 2/3 = 24 /60 +30/60 + 40/60 = 84/60 = 1.4 CHANCES

Therefore you would eat from the 1st bowl with a 0.05 better chance of avoiding the P?A

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I agree with Flamebirde's answer. But I would add a bit more to it.

Spoiler

His answer would be correct if you assume that biting into a poisoned apple brings immediate death.

Pick from the first bowl and get
P(survival) = 2/5 x 1/4 = (2x1)/(5x4)
Pick from the second bowl and get
P(survival) = 3/5 x 2/4 x 1/3, but cancel out the 3 in the numerator and 3 in the denominator and you have (2x1)/(5x4) so the same thing.

However, if you don’t just keel over dead right after biting into a poison apple and just get a horrible stomachache or something, then it might be better if you only ate one poison apple instead of two, or two instead of three. If that’s the case (or in more general terms, if it’s better to eat fewer poisoned apples even if the probability of eating any non-zero number of poisoned apples is unchanged) then I’d rather eat from the bowl with only two poisoned apples. Since the question doesn’t specify whether apples are immediately lethal or not, that seems to me like the best course of action with the information that’s given.

Edit: Scratch that. If someone poisoned three of five apples and you had to choose two to eat, your probability of choosing one poisoned apple would be the same as if you chose three apples and then someone came by and randomly poisoned two of the five in play and you ask how many of those two apples that got poisoned are in your hands. It ends up being symmetrical, so it doesn't matter which bowl you pick.

Edited by plasmid
posted before thinking :p

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On 5/25/2018 at 6:25 AM, bonanova said:

The OP does not say how many apples there are. It says the proportion that are poisonous.

Question: was that the intent?

I meant them to be fractions so five in each

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Spoiler

Revised survival probabilities are

Bowl 1: (2/5)(1/4) = 0.1.
Bowl 2: (3/5)(2/4)(1/3) = 0.1.

Horrible odds. I would totally decline the challenge.

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OP forgot to state that eating 2 poison apples results in death, while eating only 1 will not cause harm.

Therefore Taciav has the correct solution.

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For the bowl with 2 poisonous apples out of 5, and you have to eat 3, there are 3 possible ways to die:

The first way to die is by eating 2 poisonous apples in a row. The first time there's a 2/5 chance to eat a poisonous apple, which reduces to a 1/4 chance to eat a second one. Multiply these two together:

(2/5)*(1/4) = 10%

Second way to die is if you eat one good apple and then two bad ones:

(2/5) * (2/4) * (1/3) = 6.6%

Third way to die is if you eat one bad one, then a good one, and then a bad one:

(2/5) * (1/4) * (1/3) = 3.33%

10% + 6.6% + 3.33% = (Basically) 20%

This means that the chance of dying is 20%, so a 80% chance of survival.

For the second bowl there is only one way to die, and that's if you eat 2 bad ones in a row:

(3/5) * (2/4) = 30%

So that means there's a 30% chance of death and a 70% chance of survival, respectively.

I can't figure out how people get to it being 70% for both.

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On 10/5/2019 at 5:39 PM, Silas said:

For the bowl with 2 poisonous apples out of 5, and you have to eat 3, there are 3 possible ways to die:

The first way to die is by eating 2 poisonous apples in a row. The first time there's a 2/5 chance to eat a poisonous apple, which reduces to a 1/4 chance to eat a second one. Multiply these two together:

(2/5)*(1/4) = 10%

Second way to die is if you eat one good apple and then two bad ones:

(2/5) * (2/4) * (1/3) = 6.6%

Third way to die is if you eat one bad one, then a good one, and then a bad one:

(2/5) * (1/4) * (1/3) = 3.33%

10% + 6.6% + 3.33% = (Basically) 20%

This means that the chance of dying is 20%, so a 80% chance of survival.

For the second bowl there is only one way to die, and that's if you eat 2 bad ones in a row:

(3/5) * (2/4) = 30%

So that means there's a 30% chance of death and a 70% chance of survival, respectively.

I can't figure out how people get to it being 70% for both.

You wouldn’t get 6.66% because you’re choosing that you didn’t eat a poisoned one so you would have to do 3/5*2/4*1/3 for that one because the 3/5 is choosing that you ate the alright one and you wouldn’t get 3.33% because you did the same thing wrong as you would have to choose the odds of 2/5*1/4*2/3 because the 2/3 is that you eat the healthy one at the end which you would already be dead. Your answer would be if you had eaten all three poisoned ones in a row.

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Riddles are often a play on words so here's my guess -

"In the first bowl there are 3 out of 5 Poisonous Apples." That means that there are a total of 5 Poisonous Apples. It does NOT mean that there are 3 Poisonous Apples and 2 Regular Apples in Bowl 1. It means that there are 3 Poisonous Apples in Bowl 1 and 2 Poisonous Apples in Bowl 2.

I believe Bowl 1 has 3 Poisonous apples and nothing else. Bowl 2 has 2 Poisonous apples and nothing else, meaning you can't even eat 3 apples because there are only 2 apples in the bowl.

Eat from Bowl 2 because you won't have to eat any apples, especially after Bowl 1 eater is poisoned.

Edited by James
update

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On 5/24/2018 at 1:14 AM, flamebirde said:

assuming five apples in each bowl,

Reveal hidden contents

you have a 2/5*1/4 chance to survive in the first bowl, or 10%. You have a 3/5*2/4*1/3 chance to survive with the second bowl, or a 10% chance. Its the same chance either way. I say it doesn't matter. (I assume that after eating an apple you take it out of the bowl.)

What happens if you need two apples to die from poison overdose and one poisoned apple is fine for you. would it be :

4/5 * 3/4 * 2/3 = 40% Survival for the Second Bowl   and

3/5 * 2/4 = 30% Survival for the First one ?

You don't die instantly in this extra riddle from this video:

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It doesn't matter which bowl because it says there are five poisonous apples, they are just separated differently. Three poisonous apples in one bowl and two in the next one. So for those people trying to calculate the odds stop, because which ever bowl you pick all the apples are poisoned. It's called a riddle, not a math equation.

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There's a way to understand that the odds are equal for both bowls without doing any math (for both the one poison apple = death and the two poison apples = death variants).

Label the apples in each bowl as either common (C) or rare (R), so each bowl has CCCRR.  For bowl 1, the common apples are poisonous and the rare apples are not.  For bowl 2 the opposite.

When you choose the apples that you are going to eat (from whichever bowl) you are going to divide the apples into two groups - those you will eat, and those you won't.

In the one apple = death variant, the only way to survive is to divide the apples into two specific groups: CCC in one group and RR in the other.  This is true for bowl 1 and bowl 2 - which means that the probability is the same for both.    For bowl 1, you will then eat the two apples (both rare/nonpoisonous).  For bowl 2, you will then eat the three apples (all common/nonpoisonous apples).

In the two apples = death variant, you will only die in one way: if you divide the apples such that the group of two apples contains two common apples.  That is, CRR in one group and CC in the other.  Again, this is true for bowl 1 and bowl 2, so the probabilities are the same.  (The order of apples doesn't matter within the groups because we will either eat all or none of the apples in a given group.)    If we choose bowl 1, we eat the smaller group, and only die if that smaller group contains two common/poisonous apples.  If we choose bowl 2, we eat the larger group and will only die if the smaller group (the one we didn't eat) contains two common/nonpoisonous apples (as this means that both of the poisonous apples were in the group we ate).

So, without doing any math, you can see that the probabilities are equal, even if you can't tell what the probabilities are.

This understanding of the problem also makes calculating the probability of death easier, if you want to calculate it.

For one apple = death, you only survive if the small group has two rare apples.  2/5 * 1/4 = 1/10 (10% survival rate/90% survival rate)

For two apples = death, you only die if the small group has two common apples. 3/5 * 2/4 = 3/10 (30% death rate/70% survival rate)

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