Looking for the name of a trick/game

[TomMarv]

hi - looking for a bit of help!

So I've been in Amsterdam this weekend (feel like I'm still recovering!), and met a Canadian while playing pool in a bar. We were playing some bar games with others in the place, and also showed eachother some of the tricks we knew.

He showed us a trick/scam type game where he set a pyramid of 10 cards all face down. It was up to the other person if they wanted to go first or second, and on each turn each player could remove up to three cards from any single row (so they had to be all from the same row). The person that took the last card left lost, and he won every single time.

So I kinda figured out how he was doing it, but now I'm at home I can't find it on the net anywhere - does any have any idea what this game is called??

Thanks!​

 

[Maaz Hasan]

@TomMarv I think this is just a variation of the game Nim. You can look it up on YouTube - Scam School and Numberphile have great videos on it, and I believe the Scam School Remix uploaded this Monday (yesterday for me) also features a similar game.
I'm assuming the pyramid was like this?
---*---
--*-*--
-*-*-*-
*-*-*-*

The object of the game is to not get stuck with the last card, right? And since you have the option of taking away just one card, you can't leave them with two on the last turn. There fore, you must force them to choose the last card by not making it possible for you to not select it. Basically, you have to take the 9th card. Their are a few of ways to do this, but only one way will ensure you win (the others will rely on if the other guy catches on or not).

You're allowed to take 3 cards right? and 3 is a multiple of Nine. Therefore, every time the 3rd card is taken the game basically "resets", since the 9th card is just one of the 3rd cards secretly. So your job, is to complete each set of 3, ensuring you take the 9th card, and forcing them to take the 10th. If they take 1 on the first turn, you take 2. If they take 2, you take 1.

If they took 3, one set of 3 is complete, so you would assume you should complete the next set by taking away 3. However, this would turn the tables. Instead, you should take one card. This is because the amount of sets of 3s is not even, meaning if they take the last set in their 2nd turn, you lose. But, if they take the 1st set, they are set to also take the 3rd one. However, if you take two on your turn here, they must chose to take one, two, or three cards. If they take one, it completes the set, you take 3, and they lose. If they take, it completes the set and adds one to the next one, so you complete the last set with a 2, forcing them to lose. If they took 3, you take the 9th card, and they lose.

Basically, you need to force yourself to take the 9th card, and you do this in sets of 3 by simply completing each set they start, or forcing them to allow you to chose the 9th if they take the first set.

Hope that wasn't too confusing.

 

[TomMarv]

Thanks so much! You're right and it does seem to be a variation on this game... I've found the videos on Youtube which I'll be watching once I'm back from work.

In the moment after a few beers and other Amsterdam delights, I was getting so frustrated as I couldn't work out how to beat him! Will be taking this one to the next party!

Thanks again