A scam from the archives
I posted this scam on my previous blog over a year ago. While I had a lot of guesses, no one found the answer. Let’s see if the new year has some new answers…
The con man and his victim each put five coins of varying amounts into a pile. The victim then lines the ten coins up in a row, in any order he likes. The two players each take turns taking a coin from either end of the row until they have five coins each. Whoever gets the most money, wins the kitty.
The con man picks first and ends up winning.
The con artist, keen to make more money, asks the victim if he’d like to play again. This time, to sweeten the deal, the con artist puts in six coins to the victim’s five. This time, the con artist lines up the coins and let’s the victim go first. Since the victim is going first he gets more choices and will always end up with one more coin than the con man.
The con man still wins.
Why does the con man always win?
argh my brain.
Something about how with 11 coins the conman is still in the same place when the vic takes the first coin?
-Kate
Okay, I’ve been thinking about this, so I’ll bite.
First, I thought “The scammer only puts in five coins of minimum denomination (i.e. in Australia, 5c pieces), so whatever coins he/she takes is going to add up to at least the amount they put in.” Then I re-read it and saw the winner takes all, not just the coins they pick up. Also the mark is willing to play a second time. So that’s not it.
To avoid this problem of unfair contributions making the game unbalanced, I assumed both players put in the same coins - $2, $1, 50c, 20, 10c (assuming Australian currency). I note that the maximum value coin is worth more than all the others put together, so realistically, the strategy should be to get both of the $2 coins - the rest is irrelevant.
In game 1, as the mark, I thought the best strategy would be to pair up the coins, and put them down in a in a palindrome: e.g. 2-1-50-20-10-10-20-50-1-2.
I would always draw from the opposite side of the scammer, meaning I always got the same value. The result would always be exactly a draw.
Then I re-considered it, and put the coins in pairs: 2-2-1-1-50-50-20-20-10-10. I would always draw from the front. In this case, the worst case for me is still a draw, but if the scammer makes a mistake (by drawing from the back) I could do better.
So, I reckon, in the first game, I can at least neutralise the scammer. Because I am dealing with a known scammer, though, I wonder where I have missed the trick!
(Does the scammer makes sure that there aren’t an equal number of each denomination when they contribute their coins?)
In the second game, if I was the scammer, I would again use a palindrome approach, but I would make sure the highest valued coins were in the even positions - e.g. 10-2-20-1-50-2-50-1-20-2-10 (the scammer has added an extra $2 coin - they could make it a $100 note for added interest, because they always get it back!) This time, as scammer, I would always take a coin from the same side as the mark. I’d get all the even positioned coins, and the mark would get all the odd positioned coins.
So, I can see the second game is rigged against the mark, but the first game seems defensible.
I have given this more thought. Same conclusion as last time. Please put me out of my misery…
I thought about this and I don’t think this is possible for the conman to win no matter what, which is how I understood the description. If the mark understands the game, and can alighn the coins anyway he wants to start with. This is the most logical string of events as I see it using American coins. I will use the value of the coins in cents to show their order.
Before I start let me say that if the coins are arranged symetrically, then there will be a draw. This is a third possibility (along with the mark winning), the fact that this possibility was not mentioned leads me to believe there may be missing information.
Also, lets assume that the con and the mark use the exact same coins. If the con used five coins of the lowest value (in American coins that would be five pennies or five cents), the mark would simply say the game isn’t fair because the con had less to lose and stands to win more. Think about it, would you play a game with $185 (that’s a five, a ten, a 20, a 50, and $100 bill) if your opponent only played for $5? No. You have too much at stake.
The reverse is also true, if someone wanted to play with $185 and all you had to put in was $5, you would know it was a scam, because no one would risk almost 200 to win five unless there was some kind of trickery. For those reasons lets assume they both play with equal coinage.
The mark would logically arrange the coins in the following way: 1,25,1,5,10,10,5,1,25,1. Why? I’ll show you.
If the con MUST go first, and can only pull a coin from the end of the line, then the con MUST then pull a 1, because there is a 1 on either side.
The mark then takes a 25, because it is the largest possible coin.
The con then takes another 1, because that’s his only option. He would likely take a 1 from the right side because if he took the 1 from the left he would allow me to take the second 25, and there’s then no hope of him winning.
The mark naturally takes the 5, because it’s higher than a 1 and if he took the 1 he’d allow the con to take the 25 on his next turn.
The con takes a 10, and the mark takes the second 10. Both wanting to avoid the 1 at the far left.
The con takes the 5.
The mark a 1
The con a 25,
the mark a 1.
Anyone would take this series of turns, which obviously ends in a draw. Only the mark intentionally losing would allow the con to win. Try it with Austrailian coins, arrange them in the pattern shown with coins of the lowest value on the outer most sides, and coins of the highest value immediately inside of those coins. You’ll see that the most likely possibility is a draw, without the mark intentionally losing, or arranging the coins without thinking it through.
Maybe this is a trick with only works with Austrailian coins. Or there is missing information.
In the first game, look at every second coin. Then look at every first coin. See which adds up to more.
20-20-10-20-10-20-10-20-10-10
In this case, the EVEN coins (from left to right) add up to more.
On your first go - take an even coin. (in this case, the 10th coin)
Allow the mark to take his second go. Then take your second coin from same side as him. This will ensure you always end up with the even coins.
In the second game, there are 11 coins.
Julian’s suggestion for the second game is spot on. You place winning coins in the even positions and force them to always take odd.
Ah, so Ian and I have taken the same approach here, for the first game.
By placing the coins as a palindrome, we have made it so that the odd coins and the even coins add up to exactly the same amount. There’s no preference between odd and even coins. The mark won’t *win* in that case, but will break even.
They could break even on the first game but not the second (unless you wanted it to)
OK I think i have it (over a year late).
The conman must pick 3 alternating coins in succession of the highest value and must make sure he picks those coins. In my example those coins will be the 2nd,4th and 6th coins. Once those coins are chosen the conman can’t take the coins touching them (1st, 3rd, 5th, 7th) so will begin with the 10th coin. This will cause the mark to have to take either the 1st or 7th coin eventually. Meaning that the 2nd, 4th and 6th coins are the conman’s for the taking.
eg
1-2-50-1-2-1-50-50-10-2
In the second game the conman sets it up so that there are 3 coins in alternate succession of the highest value and one of those coins is not one of the coins on the end. And yes, you can work it out from here again.
How did I go?