More Coin Grabbing...

You have in front of you 4 boxes. You are told one contains 4 gold coins, another 3 gold and 1 silver, another 2 gold 2 silver, and finally 1 gold and 3 silver. You are then blindfolded, asked to pick a box and draw a coin. Its gold! You are asked to draw another coin from the same box. Given that you have drawn a gold coin, what is the probability you will draw a silver coin on this draw?

@joe-shmo said
You have in front of you 4 boxes. You are told one contains 4 gold coins, another 3 gold and 1 silver, another 2 gold 2 silver, and finally 1 gold and 3 silver. You are then blindfolded, asked to pick a box and draw a coin. Its gold! You are asked to draw another coin from the same box. Given that you have drawn a gold coin, what is the probability you will draw a silver coin on this draw?

(1/4)(1/3 + 2/3 + 1) = 0.5

@ashiitaka said
(1/4)(1/3 + 2/3 + 1) = 0.5

Sorry, feel free to try again!

@joe-shmo said
Sorry, feel free to try again!

Okay, time to use the conditional probability formula!

Let event A = getting a gold coin on the first draw
Let event B = getting a silver coin on the second draw

Pr(A) = 0.25(1+0.75+0.5+0.25) = 5/8
Pr(A and B) = 0.25((3/4) x (1/3) + (2/4) x (2/3) + (1/4) x 1) = 5/24

P(B|A) = (5/24)/(5/8) = 1/3

Edit: Removed the duplicate line. Conditional probability is trippy, but that's why I love it.

@ashiitaka said
Okay, time to use the conditional probability formula!

Let event A = getting a gold coin on the first draw
Let event B = getting a silver coin on the second draw

Pr(A) = 0.25(1+0.75+0.5+0.25) = 5/8
Pr(A and B) = 0.25((3/4) x (1/3) + (2/4) x (2/3) + (1/4) x 1) = 5/24

P(B|A) = (5/24)/(5/8) = 1/3





P(B|A) = P(B and A)/P(A)

There ya go.

For anyone looking for further explanation:

This is a conditional probability question, meaning we must update the probabilities of having chosen a certain box , given that we have already chosen a gold coin.

Note that the gold coins are not evenly distributed among all the boxes. For an example that should intuitively feel correct lets look at the extremes here. It should feel correct that if we drew a gold coin it is more likely we are drawing out of the box with ALL gold coins than the box containing only 1 gold coin. Upon knowing this new information ( drawing a gold coin ) our probabilities shift from the evenly distributed chance of picking a certain box of 1/4.

For a really obvious case imagine simply 2 boxes each with a million coins: one contains 1 gold coin the rest sliver, the other is exactly the opposite. Now if you have done this game again and you have a drawn a gold coin it should be apparent that you are way more likely to have chosen the box with 999,999 gold coins, as opposed to drawing the single gold out of the box containing 999,999 silver coins., and that is why we update the probalites from 1/2 for this case.

So we are adding up the probabilities of being in a certain box ( given a first coin gold ) and then drawing a silver coin from that box.

All Gold
P( Box 1 ) = 4/10 ( box 1 contains 4 out of 10 total gold coins )
P( S | Box 1) = 0 ( there are no silver coins in Box 1 )

3G,1S
P( Box 2 ) = 3/10 ( box 2 contains 3 out of 10 total gold coins )
P( S | Box 2) = 1/3 ( if we have chosen box 2 there will be 1 silver coin and 2 gold coins remaining in the box )

2G,2S
P( Box 3 ) = 2/10 ( box 3 contains 2 out of 10 total gold coins )
P( S | Box 3) = 2/3 ( if we have chosen box 3 there will be 2 silver coin and 1 gold coin remaining in the box )

1G,3S
P( Box 4 ) = 1/10 ( box 4 contains 1 out of 10 total gold coins )
P( S | Box 3) = 1 ( if we have chosen box 4 there will be 3 silver coins remaining in the box - as we have removed the only gold )

Thus;

P(S | first draw G ) = 4/10 ‧ 0 + 3/10 ‧ 1/3 + 2/10 ‧ 2/3 + 1/10 ‧ 1 = 1/3

If anyone wants clarification on any part please let me know.

Hint:
Reveal Hidden Content

Follow up with the same boxes:

This time you are blindfolded, you choose a box and draw a coin. You are then instructed to draw another coin from the same box. What is the probability the coin is silver?

@joe-shmo said
Follow up with the same boxes:

This time you are blindfolded, you choose a box and draw a coin. You are then instructed to draw another coin from the same box. What is the probability the coin is silver?

That's much easier, and much less counter-intuitive.

First, let's observe that it doesn't matter what colour your first coin was. You didn't see it anyway. Once your box is chosen, the chance of your second coin being silver is the same as that of your first coin being silver, or your third, or your last.

So, let's just tote them up.

¼ you drew from the all-gold box, and your odds are 0.
¼ you drew from box 2, with silver at ¼.
¼ you drew from box 3, with silver at ½; and
¼ you drew from box 4, with silver at ¾.

So your total chance of your second coin being silver is (¼+½+¾😉×¼, or 1½×¼, which is 3/8.


And frankly, we could have avoided all of this palaver just by noticing that in fact, what you've just done is toss a couple of coins about and end up drawing one coin at random from the lot, so the odds of it being silver were all going to be 6 silver out of 16 coins in total - making, again, 3/8.

Demonstrating rather nicely the difference between pre- and post-knowledge probabilities.

@shallow-blue said
That's much easier, and much less counter-intuitive.

First, let's observe that it doesn't matter what colour your first coin was. You didn't see it anyway. Once your box is chosen, the chance of your second coin being silver is the same as that of your first coin being silver, or your third, or your last.

So, let's just tote them up.

¼ you drew from the all-gold ...[text shortened]... n, 3/8.

Demonstrating rather nicely the difference between pre- and post-knowledge probabilities.

"And frankly, we could have avoided all of this palaver just by noticing that in fact, what you've just done is toss a couple of coins about and end up drawing one coin at random from the lot, so the odds of it being silver were all going to be 6 silver out of 16 coins in total - making, again, 3/8."

yep! good catch! I completely missed it.

I went about it using conditional probabilities.

P( Second Draw Silver ) = P( S ) ∙ P( S | S ) + P( G ) ∙ P( S | G ) = 3/8

It works, but its embarrassingly inefficient!