Don’t Flip Out
You are blindfolded, and are told if you can correctly solve the following, the blindfold will be removed. You are given 99 coins that are heads up, and an unknown number of coins that are tails up. You never remove the blindfold, you do not peek, …. You can count the coins, put them in arbitrary many piles, flip whichever coins you want, but remember, when you flip and when you sort, you DO NOT know which ones are heads up, which are tails up. In the end, you must end up with just two piles, each containing an equal number of heads. How do you do this?
Communicated by Eric Adelizzi.
you don’t know which coins are heads or tails
you are blindfolded
you don’t know which one is a tail to flip…
The problem does not say anything about the final two piles containing an equal number of tails, just an equal number of heads. I first was thinking of averaging but once I realized the denominator (ratio) of heads to tails does not have to be the same, the problem is simplified. So why wouldn’t I just flip one of the tail coins and make the heads an even 100, then from there, I can just create any combination of piles where I could have 1/x = 1/x all the way to 50/x = 50/x. This would just increase the tail pile but would not affect the relationship of heads between the two piles.
No. You have some number of coins. You know that 99 of the coins are heads, but you do not know which ones. You want to have two piles where the two piles have the same number of heads. By this, I mean heads face up.
everything has heads. two piles of equal numbers would have equal numbers of heads. Not heads up.
you don’t have a tails pile — you don’t know which is which, all mixed together
You take one from the “tails” pile, flip it, and add it to the 99 heads. Then sort that pile of 100 into two piles of 50.
I don’t see how that solves the problem. You need to say how many are in each pile and what you flip and when. Email to you bounced, you can email me at [email protected]
100?
Email to you bounced; please email me at [email protected]
Email to you bounced; please email me at [email protected]
Answer!
I need the answer!
no — want same number of heads face up.
for the die hard: is it the water jug problem? A variant of that might be on the site in additional problems. it’s a great one, and available on YouTube: https://www.youtube.com/watch?v=5_MoNu9Mkm4
Technically, to every tail, there is a head (each coin has a head and a tail). To split them into even piles would simply require knowing the number of coins (99 + n) and finding a number that divides evenly into it. When the coins are split into the piles, each pile will be “…containing an equal number of heads…”
I have a fun riddle I use with my students in HS. I got it from Die Hard. Let me know if you’d like me to share!
email to you bounced — please email me at [email protected]
99 is odd so how can it be divided in two and if you flip a coin then you are not certain.
email me sjm1 AT williams.edu
I have no idea how to solve this can you plz tell me the answer or a hint
yes, have to use all
do I have to use all the coins for the final two piles? pls send me a hint also.
well done! not posting as it’s the soln; email me at sjm1 AT williams.edu
you can do it with 100% certainty — will email a hint
Do you get them all together and flip each one? This way you have a 50% chance of them landing on heads and a 50% chance of tails. This will be the quickest I’ve got one if this is right, if not can I have a hint please. Brilliant site!
Not sure what you mean — sjm1 AT williams.edu
444444
your email didn’t work — email me at sjm1 AT williams.edu — wrong answer here
I would put all the coins together. Flip each coin and separate it into two piles without looking, one being heads and one being tails. The probability of getting a head and tail after a flip is 50/50.
sure /s/
maybe I’m getting dumber, but can’t think of how. Hate to do this, but could you please send me a hint.
but how do you know the two piles have the same number of heads now? //s
Can’t you just flip all the coins, then split the coins into two piles?
The coins are all in one big pile initially sjm1 AT williams.edu
1. take 99 coins which are in one pile already
2. flip all 99 coins so you get tails up
3. merge two groups of coins
4. now you have n number of coins tails up (hope we have have an equal number :))
5. next we take one coin turn it head up put on one side then next coin head up put on other side until you end with no more coins.
Anonymous: “doesnt he have 2 piles now?….” I might have misread what you posted. could you email me at sjm1 AT williams.edu
but you need two piles. email me at sjm1 AT williams.edu for a hint. ..s
1. put 99 coins in one pile
2. flip ALL of them
3.You now have equal heads
will do
Please send a hint for “Don’t Flip Out”.
done
Please send a hint to my email.
but that’s not the soln as they aren’t necessarily the same number up //s
You should just split the group in half and they will both have the same amount of heads, but they may not all be “heads up”.
but you don’t know which ones are heads up….
Just put aside the 99 u know for sure are heads up and leave the others
email me at sjm1 AT williams.edu and include the text of the riddle and I’ll send a hint ..s
send hint plz!
but that might not solve it, depending on the configuration
Can I just turn over one coin??
I’ll send a more detailed hint. //s
Steve,
I read the hint you gave me also above. Still there is no answer.
First let me get something clear, when you say “…equal number of heads” you mean heads up, correct?
If that is the case, whatever piles i make, i will have a mix of heads up and tails up no matter how many times i flip then, so i will never know which ones are heads up.
For example, if i make a pile of three coins, i will have all tails up, all heads up or a mix, if i flip them, i will just have the opposite, still an unkown number of heads up.
I’ll send a hint first. //s
I couldnt figure this one out. Could you send me the answer, please?
Thank you.
It’s hard to read the write-up. I don’t think it’s correct. First of all, n can be any number and need not be even….
first count the number of coins(n) and get the number of tails by ( n – 99 )(since n must be even number n-9 will be an odd number). if the number of tails is less than that of number of heads flip all the coins. now we will have number of tails greater than heads or equal number heads and tails. now divide the coins into two piles with equal number of coins. now since the number of head coins is a odd number two piles cannot have equal number of heads which means number of head coins in one group will be less than that of number of heads of other group. and the number of heads will also be less than number of tails in the group with less number of heads. now take one group and start to flip the coin of that group one by one until the number of heads will be equal. if the number of heads will not be equal even after you flipped all the coins of that group then flip the whole group at once.(you took the group with greater number of heads). now take other group and flip all the coins one by one and eventually a time will come when you will have equal number of heads in both the group.
I HOPE I MADE IT CLEAR.(*hey man i know the solution of GEOMETRY but dont know how to send it to you)
you can flip any coin, head or tail, but you don’t know if it’s a head or tail as you’re blindfolded.
I suppose you can not flip any heads, but sort the tails… i ‘m not sure if it’ll work, it seems to work
not necessarily — if you flip all the coins, every 2 flips returns you to where you started; email me at sjm1 AT williams.edu for hints / solns.
This may sound a tad off, but what if you sorted all of your coins into two equal piles. Then, you flip all the coins a large number of times. As the amount of flips approaches infinity, the odds would have it that you have an equal amount of heads and tails in each pile!
You would divide both piles in half and put those halves together??
I like your originality,but there is a less destructive soln too. //s
Cut all of the coins in half and put one half of each coin into each of the two piles, being careful not to flip any of them. This is cheating though.
sure == have two piles, one with 99 and one with n – 99
Could I get a hint please?
I’ll send a hint ..s
I really couldnt solve this one. Whats the solution?
Just divide the group of coins into two stacks. They will always have an equal number of heads because each coin will have one head and one tail.
Not quite: you don’t know which coins are heads and which are tails. You do know the total number of coins, call it N (if you want, make N = 2011, but N can be ANY number at least 99).
I am given 99 Heads up + Unknown Tails up. So will just flip given 1 Tail up coin to heads ( doesn’t matter how many) . it will be 100 now split in 50 each
anonymous: correct, well done! not posting your response as it’s the soln
glad you enjoyed //s
wow so cool
Fred: I don’t see how to do it this way, as now things are more indeterminate. Email me at [email protected] if you want to chat more about hints / soln.
Am I on the right track? You flip one coin which will give you either 98 or 100 heads. Then split the coins into 2 piles.
yes
Do you have to use all the coins?
sadly, you are blindfolded and can’t count like this
99 heads up, unknown heads down, and you are blindfolded
count the number of coins that are tails up. add 99, then divide that number evenly whether its 2 or 3, etc. then turn the appropriate number of heads coins to tails and move those to the tails pile. then flip half of the coins in the tails pile to heads. then flip half of the coins in the heads pile to tails thus giving you an equal amount of heads in each pile
is there 99 total coins or is there just 99 that are heads up?
Not quite — the problem is we don’t know which coins are heads and which are tails
When solving this question one must first understand that the number of tails is unknown i.e. it doesn’t matter. Therefore, if you are given an odd number of heads flip one of the 99 heads over and split 49 heads into two piles, arranging the tail coins as you wish. Voila!
Correct solution, explained nicely. Not posting as it’s the answer.
cute but there is a non-cute soln too
Put them on edge… 0 heads up in either group…
you start with one big pile with some number of tails and 99 heads and you must split into two piles such that….
are their 2 piles of coins , i.e., one for heads n one for tails(in the question)? or just one pile having 99 heads & unknown tails?
you’re on the right track but need a few more details
It’s not that hard, split the coins in 2 piles, because there is a head and a tail on both sides, then whatever one is with fewer coins add your “head” so then it is even
You may assume you have hands (actually, I think it’s safe to just assume one hand!)
you need hands to put them into piles…
hmmm tried but couldnt figure out! Please let me know the answer!
Glad you enjoyed it Joel; correct soln (not posting your note as it gives away the answer).
You want the same number of heads face up.
If we are talking about coins they each have one head so, just make two piles. Another way is make no piles no coins are in so there are the same ammount of heads.
They don’t have to be in two equal parts.
count the no of coins – assume that there are 100 coins. I know 99 of them are heads up, of these i have to turn 49 to tails.
99 cannot be divided equally in 2 parts
Did I mention you have no hands. Could’ve sworn I mentioned that…. Also have no eyes, no friends with eyes or hands, ….
You feel the ridges
Do not know which are heads.
i need one hint.
Do i know which 99 coins are heads or is there just a pile of coins where is 99 heads and unknown number of tails?
Sadly that’s cheating.
Feel the heads of the coin
Feel the tails of the coin
Later just sort them
Count the face ups by feeling the coins.
Count the face downs by feeling the coins.
Sort them.
This isn’t the correct answer. Try some special cases. Imagine you have say 6 coins and 2 are heads.
All the coins, no matter which way they face have boths heads and tails.
Thus if you want two piles with an equal number of heads, count all the coins and split the amount in two. And then you will have two piles with an equal amount of heads.
I’m not sure if I understood the question correctly, the answer seems too simple.
Not sure what this means.
You keep it up now, undrsteand? Really good to know.