> There are three boxes:
> – a box containing two gold coins,
> – a box containing two silver coins,
> – a box containing one gold coin and one silver coin.
>
> Choose a box at random. From this box, withdraw one coin at random. If that happens to be a gold coin, then what is the probability that the next coin drawn from the same box is also a gold coin?
My thinking is this… Taking a box at random would be 33% for each box. Because you got one gold coin it cannot be the box with TWO silver coins, therefore the box must be either the gold and silver coin or the box with two gold coins. Each of which is equally likely so the chance of a second gold coin is 50%
I understand that this is a veridical paradox and that the answer is counter intuitive. But apparently the real answer is 66% !! I’m having a terrible time understanding how or why. Can anyone explain this like I was 5?
In: Mathematics
The key to the problem is that the solution treats the two coins in each box as separate scenarios (Choosing the coin on the right side of box 1 is not the same as choosing the coin on the left side of the same box).
Let’s look at the three possibilities:
1. We have chosen the only gold coin in box 3 (SILVER, GOLD*).
2. We have chosen the left drawer’s coin in box 1, which contains (GOLD*, GOLD).
3. We have taken the right drawer’s coin in box 1 (GOLD, GOLD*).
In case (A), the remaining coin in the box is silver.
In case (B), the other coin is gold.
In case (C), the remaining coin is also gold.
As you can see, out of the three possible cases, knowing that the first coin is gold, two are favorable. The probability of the other coin being gold is then twice the probability of it being silver. Therefore, the requested probability is 2/3, which is 66.67%, greater than the intuitively calculated probability of 1/2 or 50%.
Label the possibilities: box 1 has coin 1 (gold) and coin 2 (gold), box 2 has coin 3 (gold) and coin 4 (silver), box 3 has coin 5 (silver) and coin 6 (silver). There are now six equally likely draws:
* Box 1, coin 1: gold
* Box 1, coin 2: gold
* Box 2, coin 3: gold
* Box 2, coin 4: silver
* Box 3, coin 5: silver
* Box 3, coin 6: silver
There are three ways you can get a gold coin, and two of those three ways are from box 1.
The problem with considering that it’s just box 1 or box 2 when drawing a gold coin is that these two boxes are not equally likely. This is because all of the draws from box 1 will be considered, while half of the draws from box 2 are discarded because this is conditional on drawing a gold coin.
Another way to look at it: let’s say you have two boxes. One has 99 gold coins and 1 silver coin. The other has 99 silver coins and 1 gold coin. If you draw a gold coin, is it equally likely that it came from either of the boxes?
If you pull out a gold coin, there’s only two possible boxes, but there were two random events that got us here.
| GG | x | SS | x | GS | x
— | — | — | — | — | —
1/3 | x | 1/3 | x | 1/3 | x
G | G | S | S | G | S
1/2 | 1/2 | 1/2 | 1/2 | 1/2 | 1/2
1/6 | 1/6 | 1/6 | 1/6 | 1/6 | 1/6
This is where we come in. We can see there are 6 ways we could have gotten here, but only 3 of those have had us pull a gold coin. If we just look at those 3, 2 of those came from the gold box, and one came from the mixed box. Therefore, we have a 2/3 chance for the gold coin to be in the box, and a 1/3 chance for there to be a silver one.
because there are 2 gold coins in one of the boxes, the odds that you are holding that box after pulling 1 gold coin is actually 66% since you are twice as likely to have pulled the gold coin from that box as the other.
The GG box has a G1 and a G2 coin
The SS box has an S1 coin and an S2 coin
The GS box has a G3 coin and a S3 coin.
There are six scenarios (before we even choose a box):
1) We choose GG and then choose G1 followed by G2
2) We choose GG and then choose G2 followed by G1
3) We choose SS and then choose S1 followed by S2
4) We choose SS and then choose S2 followed by S1
5) We choose GS and then choose G3 followed by S3
6) We choose GS and then choose S3 followed by G3
We are in 1, 2, or 5. We don’t know which one. But 2 out of those 3 choices will yield a gold coin on our second pull.
Perhaps it would be easier to understand if you were to put a number on each coin.
Put the numbers G1 and G2 on the two gold coins that will be in the same box, S1 and S2 on the two silver coins in the same box, and G3 and S3 for the two remaining coins in the last box.
You’re equally likely to pick each box, and equally likely to pick each coin in the box. So, each coin has a 1/6 probability of being picked. Since we know that the coin you picked was golden, each of the golden coin still has the same probability, but as there’s only 3 more possibilities now, you only get 1/3. So, G3, the coin in the same box as the other silver coin, also only has 1/3 odds of being picked. And that’s the only possibility in which you draw a gold coin in the first one and a silver coin in the second one. So, the probability of drawing a gold coin twice is 1 – 1/3, or 2/3.
There are three possible outcomes of the experiment: reset, success, and fail. If you pick a silver coin first we “reset” and discard the outcome. If you pick a gold coin first, then a silver, it’s a fail. If you pick a gold coin first, then another gold, it’s a success.
The question asks for P(success) / (P(success) + P(fail)). We can calculate this as follows:
33% – picked the silver-only box and picked a silver coin, reset
33% – picked the gold-only box, next coin is gold, success
16.5% – picked the silver/gold box and picked the gold coin first, next coin is silver, fail
16.5% – picked the silver/gold box and picked the silver coin first, reset
P(success) = 33%
P(fail) = 16.5%
P(success) / (P(success) + P(fail)) = 33% / (33% + 16.5%) = 66%.
Label all the coins.
Box one contains gold coins G1 and G2. Box two contains silver coins S1 and S2. Box three contains coins G3 and S3 (which are gold and silver, respectively).
Pick a box at random, then randomly draw a coin. The possible options for the coins you get are:
– G1, and then the next coin must be G2.
– G2, and the next coin must be G1.
– S1, and the next coin must be S2.
– S2, and the next coin must be S1.
– G3, and the next coin must be S3.
– S3, and the next coin must be G3.
So if you draw a gold coin, you know your second coin must be G1, G2, or S3. Two of these three outcomes mean you have the double coin box. 66%.
A way to visualise it that I haven’t seen anyone mention is to ignore the individual boxes.
Box 3 might as well not exist – we didn’t draw a silver coin so its existence is entirely irrelevant because we can’t have drawn from it.
So take the remaining two boxes and *pretend they’re just one big box*
It contained 3 gold, one silver coin, but one gold has been drawn.
So the superbox now contains two gold, one silver coin – ergo, 66% gold 33% silver
Here’s one way to see why it’s not 50-50
Suppose you have three boxes:
* a box containing 1,000,000 gold coins,
* a box containing 1,000,000 silver coins,
* a box containing 999,999 silver coins and 1 gold coin.
Choose a box at random. From this box, withdraw one coin from the 1,000,000 at random. If that happens to be a gold coin, then what is the probability that the next coin drawn from the same box is also a gold coin?
If we follow your logic from earlier, you would say it can’t be the silver coin only box so it must be one of the other two boxes and your argument gives you 50%.
But it’s not 50% because the chance of you picking the gold coin in the box not all gold was literally 1 in a million: if you pick that box and randomly pick a coin it is almost certain to be silver. So you are overwhelmingly likely to have picked the gold only box if you see a gold coin. So the correct answer is 999,999 in a million not 50%.
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