I’m honestly stuck. I’ve seen people say 1/2, others 2/3. Something Monty Hall Problem, Bayes Theorem but I’m still confused so here I am.
Edit: I believe you are not allowed to change your box choice on the 2nd “turn” as that would make having two boxes pointless, wouldn’t it?
In: 10
If you reach into a random box and take out a random bill, there are four possibilities, all equally likely. You will either grab the $1 bill, the $100 bill in the mixed box, the first $100 bill in the all-hundreds box, or the second $100 bill in the all-hundreds box. You grabbed a $100 bill, so we know the first possibility didn’t happen, but the other three are all still equally likely. So, there was a 1/3 chance you grabbed the $100 from the mixed box, a 1/3 chance you grabbed the first $100 bill in the all-hundreds box, and a 1/3 chance you grabbed the second $100 bill in the all-hundreds box. You will grab a $100 bill next if either of the second possibilities happened, so there is a 2/3 chance of getting another $100.
We can exaggerate the situation to make it more obvious. Let’s say the first box has a million bills in it, all of them $1 bills except for one $100 bill. The other box has a million $100 bills. If you reach into a random box and grab a $100 bill, you know that you almost certainly reached into the all-hundreds box, so the next bill you grab from that box will almost certainly be another $100, with a much better than 1/2 chance.
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This is tricky, because the specific wording of the question is super important here, and it’s probably where a lot of people are getting caught up on.
Let’s look at it this way: let’s say the first box has 1 $100 bill and 999 $1 bills (in other words, it would take forever to find the one $100 bill in it) and the second box is all $100 bills (in other words, it’s only possible to draw $100 bills from it).
You and your friend decide to take out bills for fun, to see if you can pull out a $100 bill. Your friend picks a box and draws a bill from it.
Nope, $1.
They put the bill back and the box back, mix around the boxes, pick a box again and draw a bill from it.
Nope, another $1.
They put the bill back and the box back, mix around the boxes, pick a box again and draw a bill from it.
Nope, another $1.
They put the bill back and the box back, mix around the boxes, pick a box again and draw a bill from it.
Oh hey, a $100 bill.
Then your friend turns to you and says, “I bet you dinner that the next bill I draw from this box is going to be a $100 bill.”
Would you take this bet?
I hope not, because it seems *reaaaaally* likely to me that he has the box with all $100 bills in it.
So this bet is in the form of “If I pick a random box **and** take out a $100 bill, what is the chance that the next bill I take from that box is another $100 bill”?
And the odds would be very very likely that you would draw another $100 bill. Much higher than 50%, even though there are only two boxes and only one has more than one $100 bill in it.
**But** if the bet is “If I pick a random box, what is the chance that I will draw one $100 bill, then draw another $100 bill?” then the odds are 50/50, because this will only happen if you pick the box with all the $100 bills in it, which is a 50% chance.
So back to your original question, with the very specific way it is worded, the probability is 2/3, because, since you drew the first $100 bill, you’ve demonstrated some evidence that the box you picked was a little more likely to be the one with two $100 bills in it. Like in the case where there were 1,000 bills in each box and one has 999 $1 bills, drawing a $100 bill from the box first gives you a little information about which box you might have.
Of the box you pulled the $100 or either box?
The box you already chose, you have a 50% chance.(you’ve already either selected box a or box b, a 1:2 choice)
If you can pick either box for second draw, you have a 66% chance of getting another $100 bill. (There are 3 bills remaining, $100, $100, and $1. You will get the $100 bill 2:3 times)
E: it’s an interesting problem because you have a statistically better chance to get another $100 if you have the option of switching boxes, even if you don’t utilize the option.
However it would always make sense to select the other box for your second draw, if possible.
Let’s stamp the bills.
The first box has one $100 bill and one $1 bill. Let’s put a red stamp on the back of the $100 bill and a black stamp on the $1 bill.
The second box has two $100 bills. Let’s put a green stamp on the back of one, and a blue stamp on the back of the other.
You pick a box at random, then draw a bill at random from the box.
What are the chances that the bill has a red stamp? 25%.
What are the chances that the bill has a black stamp? 25%.
What are the chances that the bill has a green stamp? 25%.
What are the chances that the bill has a blue stamp? 25%.
Oh wait, by the way, the bill doesn’t have a black stamp (it’s a $100 bill).
What are the chances that the bill has a red stamp? 33%.
What are the chances that the bill has a green stamp? 33%.
What are the chances that the bill has a blue stamp? 33%.
Note that you can only draw a second $100 bill from the box if the first $100 bill you drew had a green or blue stamp. So the chances of that are 66%.
Are you always picking the second bill from the same box as you pulled out the initial $100, or shuffling them up and picking at random again? The math is very different in either case.
When talking probabilities the exact wording of the question matters as well.
“What are the chances I pick $100 from a random box then $100 again in the next pick”
vs
“If I pick $100 from a random box, what are the chances I get $100 again in the next pick”
Have very different meanings.
Eli5 answer:
There are 3 $100 bills. You picked a bill randomly, meaning each bill had the same chance. But we know you didn’t pick the $1 bill. So, there’s a 1/3 chance you picked each one of the $100 bills.
There’s a 1/3 chance you picked the one in the box with the $1 bill.
The other box has two $100 bills, So there’s a 2/3 chance you picked one of those two.
Omgggg I get to finally use my actuarial test skills for something other than work!!!
Ok so first we need to find the probability of getting a 100 dollar bill. This is Pr(N1=100) = (1/2)*1+(1/2)*(1/2) = 3/4
Now we simply need to calculate the Pr of the second bill being $100 given the first bill was $100.
Pr(N2=100|N1=100) = ((1)*(1/2)+(0)*1/2)/(3/4) = 2/3
This is Bayes Theorem.
To help with other problems like this, think about probabilities like a shape, where 100% of all possible probabilities are within the shape. When you are given information that has already happened, you can take away areas of the shape because you have more information, meaning you can be more sure of your answer. You must readjust the total scope of the total probabilities, so you divide by the amount of remaining area of the shape.
In the first step, we found what the probability of the first outcome was. The second outcome is dependent on the first outcome, so we have to adjust the scope of the probabilities to account for this.
To make this problem more simple, we knew a $100 bill was taken from the randomly chosen box. We know that there is 3 bills left, 2 of which are $100. Since the PR of each box is equal, we can jump straight to 2 $100 bills / 3 total bills
[Edit: I was wrong.]
Two boxes.
Box 1 ($100, $1)
Box 2 ($100, $100)
From the perspective of having grabbed a $100 bill, the probability of getting another in each box looks like this:
Box 1 (0% probability of drawing $100)
Box 2 (100% probability of drawing $100)
Combining these probabilities results in a 50% chance of drawing another $100 bill.
It’s 2/3. There’s plenty of explanations already in the topic, so I coded up a small simulation to test your scenario. You can run the linked code by clicking on the triangle button at the top.
[First simulation.](https://tio.run/##lZPBbqMwEIbPyVPMRpUWSkuTXtleVt3DHiqtlGOUgwMmcQI2soeGKOLZs@MBGlp1pW0kwjCe/xvPj9mLV3G/zw6XiyorYxH2lIgLobfxbTKdVvWmUCmkhXAOXoTScJ5O@qRDgXR7NSqD0i8FS7RKb1drEHbrQsCdNUc3Iv5qUlmhMkyZKI2QWXFcwBPMk3FiPh9SubEQ@LziDN1@AC3Tj@IoChnkhas1td2YRjoqPJ@p5s4XtnfQx22bwMMDLCXWFRXirqvu5ZAr6/CnaUjt@4UQvAjcxVbozJRBCLfwGDLgj0oPNJn041ukcT3nPUUVBWEYvxq469U/sWvmPtPgnR42HpBbU37eJ4fg2ufbkx@z92GSGk2ltWTi75z1HdP7Csrp7wg3bEhmfMxwLpNNJa0qpcZhLTW1N/7tmdbpdZSccQdVARpWatkQwp78Dlr/521m7pUJR0WOd1vx/UEUVorsxPvSg31O0gDZO/8WcA9f87CDfDCR3re0YwfHrcjCwUA@flGUvM3ySe3V7v6wjur9tTw5lGVsaowr@h6w0MHs2cojzCDqD3xE8c2C9@jiWZj8l4pc63QUjJTttL1c/gI) This is the scenario if you draw the first bill from one box, and the second bill from the other box. The result I’m getting is:
> Drew 24991 $1 bills.
> Drew 50331 $100 bills.
[Second simulation.](https://tio.run/##jVPBbqMwED0nXzEbVVooLU16ZXupuoceKq2UY5SDA07iBGxkDw1RxLdnZwwhVN1qiwQMM@89D2/snXgX97tsfz6rojQWYUeJOBd6E98m43FZrXKVQpoL5@BNKA2n8ahLOhRIr3ejMii4FMzRKr1ZLEHYjQsBt9Yc3EDxd53KEpXxKiOlETIrDjN4gmkyTEynl9TaWAg4r3yGXr@AynRRHEWhF2LiYknLrkwtHQFPJ8LcMbC5gy5umgQeHmAusSoJiNsW3dMh3Ron9bOpScCXFrxwCMGbwG1shc5MEYRwC4/h0kv9UekeBGM7EVgr6/BZ5fmrziTrfKmQwCcOwfsWFh@V2vVeyJoWDivGr60pyGPZ0fpG1hBcRX88sQ2dT6PUaFS6kl7vde3ZrSL7Dsrpnwg33rDMcEwzpj3BMFmX0qpCarzUUlPxYPpvqtO4Cp9xe1UCGs/UsiYJe@QOGn7wGLzuVRMOiibStsLrg8itFNnR96UvXjlJP5B9MmsG9/ClYS3nv44Npcmyi2F@O0ZR0vf@D@zV3m7zDvB8z48OZRGbCuOSzgfmOpi8WHmACUTdAYgovpn5Jl08CZNvscillkfBgNmMm/P5Lw) This is the scenario is you draw both bills from the same box. The result I’m getting is:
> Drew 25056 $1 bills.
> Drew 50022 $100 bills.
Of course there’s randomness involved so the results vary a bit, but they’re always around this 25000 and 50000. So in both scenarios you have a 2/3 chance to draw a second $100 bill after having already drawn one in the first place.
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