Brain Teasers
Under Which Cup?
Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.Probability
You decide to play a game with your friend where your friend places a coin under one of three cups. Your friend would then switch the positions of two of the cups several times so that the coin under one of the cups moves with the cup it is under. You would then select the cup that you think the coin is under. If you won, you would receive the coin, but if you lost, you would have to pay.
As the game starts, you realise that you are really tired, and you don't focus very well on the moving of the cups. When your friend stops moving the cups and asks you where the coin is, you only remember a few things:
He put the coin in the rightmost cup at the start.
He switched two of the cups 3 times.
The first time he switched two of the cups, the rightmost one was switched with another.
The second time he switched two of the cups, the rightmost one was not touched.
The third and last time he switched two of the cups, the rightmost one was switched with another.
You don't want to end up paying your friend, so, using your head, you try to work out which cup is most likely to hold the coin, using the information you remember.
Which cup is most likely to hold the coin?
As the game starts, you realise that you are really tired, and you don't focus very well on the moving of the cups. When your friend stops moving the cups and asks you where the coin is, you only remember a few things:
He put the coin in the rightmost cup at the start.
He switched two of the cups 3 times.
The first time he switched two of the cups, the rightmost one was switched with another.
The second time he switched two of the cups, the rightmost one was not touched.
The third and last time he switched two of the cups, the rightmost one was switched with another.
You don't want to end up paying your friend, so, using your head, you try to work out which cup is most likely to hold the coin, using the information you remember.
Which cup is most likely to hold the coin?
Hint
Write down the possibilities. Remember that there are only three cups, so if the rightmost cup wasn't touched...Answer
The rightmost cup.The rightmost cup has a half chance of holding the coin, and the other cups have a quarter chance.
Pretend that Os represent cups, and Q represents the cup with the coin.
The game starts like this:
OOQ
Then your friend switches the rightmost cup with another, giving two possibilities, with equal chance:
OQO
QOO
Your friend then moves the cups again, but doesn't touch the rightmost cup. The only switch possible is with the leftmost cup and the middle cup. This gives two possibilities with equal chance:
QOO
OQO
Lastly, your friend switches the rightmost cup with another cup. If the first possibility shown above was true, there would be two possibilities, with equal chance:
OOQ
QOO
If the second possibility shown above (In the second switch) was true, there would be two possibilities with equal chance:
OOQ
OQO
This means there are four possibilities altogether, with equal chance:
OOQ
QOO
OOQ
OQO
This means each possibility equals to a quarter chance, and because there are two possibilities with the rightmost cup having the coin, there is a half chance that the coin is there.
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Comments
This was one of the funnest teasers I have attempted. I enjoyed thinking out the possibilities and figuring out where it might be. I'm glad I came up with the right answer.
Got it wrong; but loved the explanation that made me see the error of my ways. Good teaser!
Got it correct! I loved it! Is there a similar one in the future?
By the way, the question should say "Which CUP holds the coin?"
Great job, it's going in my favorites!
By the way, the question should say "Which CUP holds the coin?"
Great job, it's going in my favorites!
Great job my friend, I loved the explaination!
If I want to stay online, my iPad won't let me get a piece if paper and a pen to write down possibilities. That's why I got this wrong and why does my iPad keep auto correcting things so I can't do things? But anyway, that was cool. I wish that were mine!
Apr 20, 2012
Great job, it's going in my favorites!
You (whever posted this teaser) said "He switched two of of the cups 3 times." which seems to say "Two of the cups were swapped 3 times." when your solution (and logic validity) leads me to believe you meant "On 3 occasions, he swapped two cups.". If the rightmost cup was swapped and then not swapped, how can he have swapped the same cups three times in only 3 moves?
I hope my point is clear and that that line gets clarified.
I hope my point is clear and that that line gets clarified.
Loved this teaser
I nailed it and it felt good
I used ABC though to represent the cups
I ended up with 4 possibilities which are
CBA
BAC
ACB
BAC
the probability then of the coin being under the first cup is 1/4 which is also the same with the probability of it being under the second cup. There are two possibilities however for the coin being under the rightmost cup which are the two BAC above. Hence it has the highest probability which is 1/2
I nailed it and it felt good
I used ABC though to represent the cups
I ended up with 4 possibilities which are
CBA
BAC
ACB
BAC
the probability then of the coin being under the first cup is 1/4 which is also the same with the probability of it being under the second cup. There are two possibilities however for the coin being under the rightmost cup which are the two BAC above. Hence it has the highest probability which is 1/2
When I first read this i groaned and almost avoided it...then I read a few comments that p-eop-le actually managed got solve it. When I started, I found it OK - and FUN to do! Into the Favourites.
You can actually generalize this to any arbitrary initial condition by modeling this as a Markov process! The initial state is a vector S0 denoting the probability of being in the initial state (S0 = {0,0,1} in this case). Then each one of the swaps can be modeled with a transition probability matrix (which are symmetric in this case because swapping cup A with cup B is equivalent to swapping cup B with cup A). The first one is
T1 = {{0.5, 0, 0.5}, {0, 0.5, 0.5}, {0.5, 0.5, 0}} (i.e. 50% chance to swap either cup 1 or 2 with 3)
and the second is
T2 = {{0, 1, 0}, {1, 0, 0}, {0, 0, 1}} (i.e. 100% chance to swap cups 1 and 2, and do nothing with cup 3)
The third swap is identical to T1; therefore the final state is given by T1@T2@T1@S0 (where @ denotes matrix multiplication). Using this formalism, you can get the final probability distribution starting from any initial state! For the case of S0 = {0,0,1}, the final answer is {0.25, 0.25, 0.5} (i.e. 25% chance to be in either cup 1 or 2, and 50% chance to be in cup 3.
T1 = {{0.5, 0, 0.5}, {0, 0.5, 0.5}, {0.5, 0.5, 0}} (i.e. 50% chance to swap either cup 1 or 2 with 3)
and the second is
T2 = {{0, 1, 0}, {1, 0, 0}, {0, 0, 1}} (i.e. 100% chance to swap cups 1 and 2, and do nothing with cup 3)
The third swap is identical to T1; therefore the final state is given by T1@T2@T1@S0 (where @ denotes matrix multiplication). Using this formalism, you can get the final probability distribution starting from any initial state! For the case of S0 = {0,0,1}, the final answer is {0.25, 0.25, 0.5} (i.e. 25% chance to be in either cup 1 or 2, and 50% chance to be in cup 3.
HEY BOYS REPLY
You are playing the shell game. Therefore, the probability that it is under any of them is zero.
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