Solve this for reps

06-28-2022, 05:28 AM

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#1

Schnitzl
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Schnitzl
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Solve this for reps

I've got two boxes, both have 2 bills inside.

One box has $1 and $100
Second box has $100 and $100

Now I choose a box at random and pull out a $100. What are the chances I pull another $100 from that box?

06-28-2022, 05:42 AM

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#2

xxAchillesxx

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50/50. But if you were to switch boxes your chances would go up to 66%.

No idea if I'm right or not but it seems like a similar to the Monty Hall Problem.

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06-28-2022, 05:44 AM

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#3

piramparam
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2/3

1096 ng/dl

06-28-2022, 05:55 AM

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#4

Pterodactyl314
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First instinct was to say 50/50 but I see 3 scenarios

1. Pull 100 then pull 1
2. Pull 100 then pull 100
3. Pull 100 then pull 100

But the last 2 might be irrelevant. Ignoring the initial odds the 100 your holding is from box A or B and you have 2 options now. Pull 100 or pull 1

I’ll go with 50%

See you tomorrow

06-28-2022, 05:58 AM

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#5

Bodhy
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Yep, it's 50%. The same box criterion is the key. If you could change boxes it would make the problem more difficult and a different probability, but you either pull out a $100 or a $1 on the next pull.

Back off, Warchild.

Seriously.

06-28-2022, 06:07 AM

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#6

Gainzzz
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There's about a 350% chance OP is a ****g0t

06-28-2022, 06:08 AM

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#7

Johnez
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50/50
You are given 2 choices still and have yet to figure out which of the 2 choices you've picked.

Virtue is its own reward.

06-28-2022, 06:11 AM

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#8

BraddlesMcGee
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Originally Posted By Schnitzl⏩

I've got two boxes, both have 2 bills inside.

One box has $1 and $100
Second box has $100 and $100

Now I choose a box at random and pull out a $100. What are the chances I pull another $100 from that box?

I'm redoing my answer.

I will only pull 100 if I have box B, therefore the question can be phrased as, given that I pulled 100 from the randomly selected box, what are the chances that that is box B.

There's a 0.25 (0.5*0.5) chance of choosing box A and pulling 100 from it.

1 - 0.25 = prob I'm holding box B.

Answer = 0.75

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06-28-2022, 06:45 AM

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#9

Schnitzl
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Still no correct answer in sight

06-28-2022, 06:54 AM

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#10

Redfish225
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Originally Posted By Anachron⏩

Bout 1 in 350.

This^^^

I loe e.

06-28-2022, 07:05 AM

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#11

SUPERH0T
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2 in 3

You are more likely to be in the universe where you picked a bill from the one with 2 x 100 dollar bills.

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06-28-2022, 07:22 AM

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#12

numberguy12
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Originally Posted By xxAchillesxx⏩

50/50. But if you were to switch boxes your chances would go up to 66%.

No idea if I'm right or not but it seems like a similar to the Monty Hall Problem.

Originally Posted By Pterodactyl314⏩

First instinct was to say 50/50 but I see 3 scenarios

1. Pull 100 then pull 1
2. Pull 100 then pull 100
3. Pull 100 then pull 100

But the last 2 might be irrelevant. Ignoring the initial odds the 100 your holding is from box A or B and you have 2 options now. Pull 100 or pull 1

I’ll go with 50%

Originally Posted By Bodhy⏩

Yep, it's 50%. The same box criterion is the key. If you could change boxes it would make the problem more difficult and a different probability, but you either pull out a $100 or a $1 on the next pull.

Originally Posted By metroins⏩

This is correct

Originally Posted By Johnez⏩

50/50
You are given 2 choices still and have yet to figure out which of the 2 choices you've picked.

Originally Posted By BraddlesMcGee⏩

I'm redoing my answer.

I will only pull 100 if I have box B, therefore the question can be phrased as, given that I pulled 100 from the randomly selected box, what are the chances that that is box B.

There's a 0.25 (0.5*0.5) chance of choosing box A and pulling 100 from it.

1 - 0.25 = prob I'm holding box B.

Answer = 0.75

Originally Posted By jreacher⏩

25%

No. It's 2/3. Do trials with playing cards and find the probability by long term behavior if you must

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06-28-2022, 07:26 AM

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#13

Drew23
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1:2 that you pull the 100/100 box and 1:1 that the first is 100 and 1:1 that the second is 100
1:2 that you pull the 1/100 box and 1:2 that you pull the 100 first.

75% that the first bill you pull is 100.
50% that the second bill is 100.

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06-28-2022, 07:32 AM

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#14

numberguy12
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Originally Posted By Drew23⏩

1:2 that you pull the 100/100 box and 1:1 that the first is 100 and 1:1 that the second is 100
1:2 that you pull the 1/100 box and 1:2 that you pull the 100 first.

75% that the first bill you pull is 100.
50% that the second bill is 100.

Nope, you're more likely to be in the 2nd box.

Here's another way to look at it, without getting into conditional probability notation:

By drawing the initial $100, I either picked the $100 from Box 1, picked the first $100 from box 2, or picked the second $100 from box 2. Each of these is equally likely. 3 equally likely possibilities, 2 of which mean Im in box 2 and the next bill will be $100. Thus probability is 2/3.

Edit: here's how to test with playing cards

Spoiler!

Make two piles of 2 cards: first has 2 red, other has one red, one black. In this sense a red card represents a $100, the black represents $1. For each trial, with the cards turned face down, move the piles around until you dont know which is which, pick one pile at random. Then shuffle the two cards in that pile. Pick the top card. If it's a red, record what the other card is (red/black). If it's a black,this does not count as a trial and record nothing, as that doesnt fit the given info, we didnt select a $1 bill first.

You should have a list with Red and Black at the top, with tick marks under each, for each trial there should be 1 tick mark recorded under Red or Black (whatever the 2nd card was). I argue the red tick marks should be roughly double that of the black at the end of a very long series of trials. This is because as follows: instead of even moving the piles around, you could substitute a fair coin flip to see which pile you are choosing: put the pile of two reds on the left, put the pile of red/black on the right. Heads = you pick the pile on the left, Tails = you pick the pile on the right. So 1/2 chance of either pile being selected: equal chance. Now flip the coin again to see which card in the pile you are picking. Heads = top card, Tails = bottom card. If its the pile with 2 reds, it doesnt matter, Red is getting a tick mark regardless. But if the other pile is selected, only half of the time will a tick mark get generated for Black, depending on if the 2nd coin flip flipped the red or black card.. Thus at the end of a very long series of trials, you should have roughly X tick marks under Red, and 1/2X tick marks under Black. This means you should expect to be in the 2-red pile X/ (X+1/2X) = X/(3/2X) = 2/3 of the time.

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06-28-2022, 07:54 AM

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#15

Schnitzl
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Numberguy and SUPERH0T came in clutch! 2/3rd is correct

06-28-2022, 07:58 AM

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#16

Cleveland33
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Originally Posted By Bodhy⏩

Yep, it's 50%. The same box criterion is the key. If you could change boxes it would make the problem more difficult and a different probability, but you either pull out a $100 or a $1 on the next pull.

This must be very embarrassing for you

Originally Posted By Schnitzl⏩

Numberguy and SUPERH0T came in clutch! 2/3rd is correct

06-28-2022, 04:30 PM

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#17

numberguy12
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To follow up on the response that mentioned 2 choices. 2 choices does not mean it's a 50/50.

Here are some typical examples:

-I will win the powerball lottery with the ticket I just bought
-I will not win the powerball lottery with the ticket I just bought

-The Cardinals will win the World Series
-The Cardinals will not win the World Series

-I will roll a 1 on this fair 6-sided die
-I will not roll a 1 on this fair 6-sided die

-An asteroid will obliterate the earth tomorrow
-An asteroid will not obliterate the earth tomorrow

In all the above cases, there are 2 possibilities, but in none of them are the chances 50/50. Nor does the fact there are 2 possible boxes you could be in, mean its 50/50 in this scenario.

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♫1:2:3:4 Pythagoras crew ♫ ♫ 🧮

Nullius in verba

06-28-2022, 04:53 PM

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#18

iwant2beswole
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Originally Posted By numberguy12⏩

To follow up on the response that mentioned 2 choices. 2 choices does not mean it's a 50/50.

Here are some typical examples:

-I will win the powerball lottery with the ticket I just bought
-I will not win the powerball lottery with the ticket I just bought

-The Cardinals will win the World Series
-The Cardinals will not win the World Series

-I will roll a 1 on this fair 6-sided die
-I will not roll a 1 on this fair 6-sided die

-An asteroid will obliterate the earth tomorrow
-An asteroid will not obliterate the earth tomorrow

In all the above cases, there are 2 possibilities, but in none of them are the chances 50/50. Nor does the fact there are 2 possible boxes you could be in, mean its 50/50 in this scenario.

Strong post to sig ratio you always deliver in these threads

watchout your comments boyo ↓

06-28-2022, 04:55 PM

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#19

johnnydeep1
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Originally Posted By Schnitzl⏩

I've got two boxes, both have 2 bills inside.

One box has $1 and $100
Second box has $100 and $100

Now I choose a box at random and pull out a $100. What are the chances I pull another $100 from that box?

33%

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06-28-2022, 08:07 PM

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#20

Darkius
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I like this problem.

06-28-2022, 08:52 PM

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#21

Pterodactyl314
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Ah shouldn’t have tripled guessed myself.

Just looking at the possibilities

1. Box A: 1
2. Box A: 100
3. Box B: 100
4. Box B: 100

You said you have a 100 so option 1 is out

2. Box A: 100 you then grab 1
3. Box B: 100 you then grab 100
4. Box B: 100 you then grab 100

Which gives us 2/3

So it’s not 50/50 because half of the time you choose Box A you get a 1 and op’s scenario doesn’t exist.

What I still don’t fully understand is calculating odds part way through a process…

75% of the time this situation is possible. You discard 25% of your attempts but with the 75% successful arrivals to scenario there’s then a 66% of choosing a 100 next.

So what you have to take into account is we’ve discarded the $1 result. This gives us 50% chance of having Box B and 25% chance of having box A.

To break it down further
25% you then grab 100
25% you then grab 100
25% you then grab 1

50/75 (2/3) you get 100
25/75 (1/3) you get 1

Ok I think I see how it all relates now. I was wondering how the original 50% odds disappeared but it was just converted into new odds.

Maybe a simpler way to look at this is
A. 25% of the time you’re tossing the results
B. 50% of the time you’re grabbing a 100 next
C. 25% of the time you’re grabbing a 1 next

Then just ignore A and our total working percentage is 75%. It’s weird comprehending an undefined element.

See you tomorrow

06-28-2022, 09:05 PM

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#22

DuracellBunny
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Originally Posted By Pterodactyl314⏩

Ah shouldn’t have tripled guessed myself.

Just looking at the possibilities

1. Box A: 1
2. Box A: 100
3. Box B: 100
4. Box B: 100

You said you have a 100 so option 1 is out

2. Box A: 100 you then grab 1
3. Box B: 100 you then grab 100
4. Box B: 100 you then grab 100

Which gives us 2/3

So it’s not 50/50 because half of the time you choose Box A you get a 1 and op’s scenario doesn’t exist.

What I still don’t fully understand is calculating odds part way through a process…

75% of the time this situation is possible. You discard 25% of your attempts but with the 75% successful arrivals to scenario there’s then a 66% of choosing a 100 next.

So what you have to take into account is we’ve discarded the $1 result. This gives us 50% chance of having Box B and 25% chance of having box A.

To break it down further
25% you then grab 100
25% you then grab 100
25% you then grab 1

50/75 (2/3) you get 100
25/75 (1/3) you get 1

Ok I think I see how it all relates now. I was wondering how the original 50% odds disappeared but it was just converted into new odds.

There are 3 $100 bills, so once you state that a $100 bill was drawn, the chance of it being any given box is 1/3 for each $100 bill in that box.

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