Yesterday I posted a brain teaser on twitter that boggled almost everyone who read it. Cudos to BusTechno1 and stringsn88keys for getting it right. Here is the text of the twitter:
Teaser (3/5 diff) 3 coins in hat: Double sided head, dbl tail & 1 head/tail. Blind draw one, you see head. What’s % that other side is head?
The overwhelming majority of people said the answer was 50%. Obviously this is wrong (wouldn’t be much of a teaser would it!). The right answer is 2/3, because you could be looking at 3 different heads and 2 of them have another head on the opposite side.
And despite my best attempts at explanation on twitter, email and text, many people just could not understand how it wasn’t 50%. After all, you’re either holding the double head coin or the regular one, right?
And yes it is true there are only two coins you could be holding, but that doesn’t mean the odds are equal that you’re holding one versus the other. After all, when you hold a lottery ticket in your hand, it can either be a winner or a loser. Certainly that’s not a 50-50 shot!
I was going to simply post a probability chart, but Andy beat me to that last night. FYI he also refernces a 6sided die, one with all 6s on it and one normal. He does this because in one of my tweets trying to explain it I used this example. Look at that chart, it helps explain the solution.
My math mentor taught me long ago that one great way to find the correct answer in a tough question is to exaggerate it, find the answer then un-exaggerate it back to normal and apply the same principles. So let me do that with this problem, but exaggerate it even further than a 6 sided die:
For the coin problem, nearly everyone could agree that the double tailed coin played no role in the solution. It can be ignored, since we see a head on the mystery coin, we know the double tail coin is still in the hat and not in play. So I will use 2 items for my exaggeration:
Imagine you have two decks of cards. They are shuffled but remain seperate from each other. They are rubberbanded to remain intact. So you have two decks of cards, each rubberbanded so their cards cannot mix. One is a normal 52 card deck. The other is a trick deck with all Ace of Spades. 52 aces of spades. Place them in a hat.
You blindly draw one deck out face down (unable to see any of the cards). You randomly pick one card out of the deck and see that it is the ace of spades. What are the chances that you have the trick deck?
The correct answer is 52/53. Since there are 53 AoS in the hat and 52 of them belong to the trick deck. Its an almost lock that you have the trick deck. IF you can agree on this, then you WILL agree on the coin problem. If you don’t agree, I likely cannot explain it to you. However, you’re welcome anytime for a betting game. We’ll do this deck trick and when we don’t get the A0S we’ll start over. When we do, I’ll give you 2 to 1 odds on $100 that you have the normal deck. We can play all night.
Assuming you agree that the deck is more likely the trick deck, let’s slowly unexaggerate it. Take out one card from each deck (be sure to leave the AoS in the normal deck). Now you have 51 in each. Again, if you pull out a deck blindly and see an AoS, surely you can still agree the odds are heavily in favorite of that deck being the trick deck. (51/52).
The KEY is that when you pull that random card out and it is the AoS your odds change drastically from 50-50 on the trick deck to nearly 99%. After all, if you pulled the 4 of hearts, it would be 0%, right?
- Now take one more out. 50 cards in each deck. You draw the AoS. Odds are 50/51 its trick deck.
- Now take one more out. 49 cards in each deck. You draw the AoS. Odds are 49/50 its trick deck.
- Now take one more out. 48 cards in each deck. You draw the AoS. Odds are 48/49 its trick deck.
- Now take one more out. 47 cards in each deck. You draw the AoS. Odds are 47/48 its trick deck.
- Now take one more out. 46 cards in each deck. You draw the AoS. Odds are 46/47 its trick deck.
- Now take one more out. 45 cards in each deck. You draw the AoS. Odds are 45/46 its trick deck.
- Now take one more out. 44 cards in each deck. You draw the AoS. Odds are 44/45 its trick deck.
- ………
- Now take one more out. 5 cards in each deck. You draw the AoS. Odds are 5/6 its trick deck.
- Now take one more out. 4 cards in each deck. You draw the AoS. Odds are 4/5 its trick deck.
- Now take one more out. 3 cards in each deck. You draw the AoS. Odds are 3/4 its trick deck.
- Now take one more out. 2 cards in each deck. You draw the AoS. Odds are 2/3 its trick deck.
Now that two card deck is no different (odds-wise) than our two coins. There are just 2 coins with 2 sides each, instead of 2 decks with 2 cards each. In one deck, the two cards were different and one was AoS (same in one coin, one was heads, other was tails) and in another deck, the two cards were same and both the AoS (same in the other coin, the double head).
Still not convinced?
Imagine in our deck problem that you glued the two cards together so their backs were touching and you could see both cards when you flipped it over. There’s your two coins.
The fact that the coins are physical objects does not stop them from possessing two sides each. And the solution still is valid, you could be looking at any of three heads, two of which have a head on the other side.
Finally, if somehow you still don’t believe me. I challenge you to this:
Get 3 sheets of paper.
On paper one write “side A: tails” on one side and “side B: tails” on the other.
On paper two write “side C: tails” on one side and “side D: heads” on the other.
On paper three write “side E: heads” on one side and “side F: heads” on the other
Blindly draw one and randomly pick a side to look at. Write down which letter and value you got as well as what value (heads/tails) was on the other side. Do that about 2000 times.
At the end each letter will show up about 1/6 of the time. For this puzzle, throw out all A, B and Cs, since they are tails and the problem said we saw a head.
Of the remaining 1000 or so, you will find that you have about 1/3 Ds, 1/3 Es and 1/3Fs. And you will see that on all Ds the other side is a tail, on all Es and Fs, the other side is a head. ie 1 tail for every 2 heads, or 2/3.
You will discover that it is in fact 2/3. And if you really did that 2000 times you’d have about 667 successes of the 1000 trials that counted (initial head).
I hope you understand it now, and don’t feel bad if it fooled you. This was a very tricky problem. Had I given the trick deck problem instead, you would have seen it was nearly 99%. The coins fool you into thinking its either coin1 or coin2 and both have equal chances. But remember, that new information that one side is a head changes the odds completely.
It is similar to a lottery drawing. Before the numbers are drawn you have a 1 in 200 million chance of winning. After the first number is drawn, if it matches your card, your odds go up dramatically. They must, because if the number didn’t match, they plummit dramatically (to zero).
Hope that helps, and pass it along. Its a fun one that can trick even the best math people. 
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thakilla
