You can make this game foolproof by starting with 1, 3, 5, 7 or something else with "nim sum" of zero and let the opponent go first. You can also fool them into thinking the game is fair by letting them choose whether you win or lose when you take the last coin.
Dean Dowling That's the great part - you can try for the next pattern. I've been trying this for a couple of times now - it's pretty easy to hit the next "pole position", since the hope is always that they either don't know the pole positions, or aren't savants. Oooh I'm sooo gonna try this on my office buddies soon.
this game should not be too hard, at best time consuming, to figure out. Going backwards and assuming a capable opponent, the process should be clear. (let x be anything, X be anything >1) you win one turn ago: opponent took last coin (1 0 0) you removed all other coins from the last pile | you removed another pile and the last pile had 1 coin (X 0 0) | (1 x 0) however one of (X X 0), (X 1 0), and (X 0 0) can lead to (X 0 0) you opponent did not remove an entire pile to get to (X 0 0) as that is suicide, and if the other pile is 1 he could have just won. He also could not have taken some coins and left you with more than 1 because that would be handing you victory. So (X 0 0) is not possible. so it was (x 1 0) before. another turn ago: (X 1 0), (x X 0), (x 1 x) could have led to (x 1 0). you did not force a (X 1 0) because your opponent can win there. in (x X 0) if x!=X your opponent could have won by taking the difference from the larger pile, so x=X. (X=X 0) is possible. in (x 1 x) if x=x, your opponent could have won by taking the extra 1 then keeping the piles equal from there unless x=x=1 so only (X!=X 1) or (1 1 1) you much have forced a (X=X 0), a (X!=X 1) or a (1 1 1) a (1 1 1) only comes from a (X 1 1) a (X=X 0) can come from a (X!=X 0) or a (X=X 1) a (X!=X 1)can come from a (X=X 1) or a (X!=X 0) so you could have gotten a (X 1 1), a (X!=X 0), or a (X=X 1) another turn ago: you opponent never had the chance to force you into a (1 1 1) so your (X 1 1) did not come from another (X 1 1). It must come from a (X X 1). Your opponent can win with a (X=X 1), so this much be a (X!=X 1) you can force that (X!=X 1) with either (X=X 1). so if you won against a perfect opponent, you must have gotten either (X=X 1) or (X!=X 0) from here on you actually kind of have to discuss each individually (that or im missing something obvious), and this comment is long enough as is...
Actually you don't even need to memorize anything. There's a solution that works with all starting configurations, and to use it, you just need to know how to count in binary. www.archimedes-lab.org/How_to_Solve/Win_at_Nim.html
Grulps Can you please explain - I've been losing my mind since last night with this. When seeing the video, it seems so easy and obvious, but when I play with my sis, I ALWAYS keep losing. Start: 3 5 7 Me: 2 5 7 Sis: 0 5 7 What now? Whatever I play, I end up losing. HALP!
Lakshminarayanan Krishnan Next steps would be something like this: You: 0 5 5 Sis: 0 3 5 You: 0 3 3 Sis: 0 1 3 You: 0 1 0 Or you can solve it with binary numbers: 3, 5 and 7 are 11, 101 and 111 in binary. Put them in a grid like this: 111 101 011 Now your goal is to make sure that each column has an even number of ones. In this case the 3rd column has 3 ones so you have to start by taking away one of them. For example, 2 5 7 in binary is: 111 101 010 This strategy works until there's only one stack larger than 1 in the game. After that you win by leaving an odd number of stacks of 1 to your opponent.
Hey Grulps - thanks for that explanation. Actually, when Brian says "even number in both piles", I thought he meant it to be multiples of 2. Then, in further gaming with my sis, I now realize it meant same number of coins in each pile... Also, I was playing the game wrongly, thinking I could pull out only 1, 2 or 3 coins at any time from any of the piles.. Now it's a little bit clearer... And your binary method makes it easier to map the moves... Gonna try that one also.. Thanks so much
@@lnkrishnan that's the way I interpreted "two even piles" I thought you had to get 2-2 or 4-4. but if he means two equal piles, things are a bit easier now. the grouping method he teaches with the matchstick version works for this version too. it's easier
The first time i tried this with my dad I lost horribly when he went first. Not sure if I screwed up but I could never make one of the checkpoints. Game breakdown below Setup: 3, 5, 7 Dad: 3, 3, 7 Me: 3, 3, 6 Dad: 3, 3, 3 Me 3, 3, 2 Dad: 1, 3, 2 *Insert loss here* He didnt hit the 1,2,3 until the end but I dont know what i should have done on my first or second turn to pull it in my favor. And no he did not know the game. Any Help?
Keegan T couldn't you have gone for 2 pairs in your first move? 3, 3 would have given you the win. (forgive me... I'm half-remembering this from almost a decade ago)
Scam School Now that you mention it yes 3, 3 would be a win. I was under the impression that "two even piles" meant two even numbers in two piles. Such as 2, 2 or 2, 4, etc. Does it mean even as in matching?
+royklein12 That's the part where the one who goes first has the upper hand. After that you could try and go for the other positions. Take all the 7 and then if he takes from the 2 pile take everything but one and if he takes from the 7 pile just take the rest of the 7 pile to balance it out or if you are left with 2-1 you just remove the 2 and you done won.
+royklein12 I think remove le pile with 2 coins is better... ADV 2.5.7 ME 0.5.7 Then the other has 12 possible configurations : 0.7 I win 1.7 I win 2.7 I win via a 2.2 3.7 I win via a 3.3 4.7 I win via a 4.4 0.5 I win 1.5 I win 2.5 I win via a 2.2 3.5 I win via a 3.3 4.5 I win via a 4.4 5.5 I can Lose : then search for 4.4 or 3.3 or 2.2 to win 6.5 I win via a 5.5 In general with two piles I think you have to search for : 6.6 / 5.5 / 4.4 / 3.3 / 2.2 What do you think ?
@@Xandawesome I think like I was, a lot of people are confused about the "two even piles". it sounds like you want two piles of 2-2 or 4-4. but what he meant was two equal piles. 2-2, 3-3,4-4,5-5
Ok Scam School - a few hours in, my head hurts. Maybe I'm overthinking this, but what do I do in this scenario playing against my sister: Start: 3 5 7 Me: 2 5 7 Sis: 0 5 7 What now? Whatever I play, I end up losing. HALP!
Hey - thanks. Yeah - I finally figured out what Brian meant when he said "two even piles"... Now I can almost play this game and always win, except in the one time when the other guy starts, and pulls out 1 coin from the middle pile - then, you have to play a little carefully, and hope they don't reach one of the pole positions unwittingly.
You can make this game foolproof by starting with 1, 3, 5, 7 or something else with "nim sum" of zero and let the opponent go first. You can also fool them into thinking the game is fair by letting them choose whether you win or lose when you take the last coin.
2 5 7 is also a pole position, so if two experts went against each other the 1st one would win.
Ditto for 3 4 7 and 3 5 6. Again, P1 would always win provided she selected only 1 coin from the initial 3 5 7.
I instead use the thing about binary number pairs, because this means that I can adapt to other combinations.
Wow, I just stumbled across this video today, and holy shit his hair is... something.
Zhu Li, Do The Thing.
Brian what do you do if they make one of the patterns
Dean Dowling That's the great part - you can try for the next pattern. I've been trying this for a couple of times now - it's pretty easy to hit the next "pole position", since the hope is always that they either don't know the pole positions, or aren't savants. Oooh I'm sooo gonna try this on my office buddies soon.
Thanks for the info
if there is the 4 and 5 pile left i think you take one from the 5 pile to make it even with the other pile.
This one was extra spiky XD
nice~
this game should not be too hard, at best time consuming, to figure out. Going backwards and assuming a capable opponent, the process should be clear. (let x be anything, X be anything >1)
you win
one turn ago:
opponent took last coin
(1 0 0)
you removed all other coins from the last pile | you removed another pile and the last pile had 1 coin
(X 0 0) | (1 x 0)
however
one of (X X 0), (X 1 0), and (X 0 0) can lead to (X 0 0) you opponent did not remove an entire pile to get to (X 0 0) as that is suicide, and if the other pile is 1 he could have just won. He also could not have taken some coins and left you with more than 1 because that would be handing you victory. So (X 0 0) is not possible. so it was (x 1 0) before.
another turn ago:
(X 1 0), (x X 0), (x 1 x) could have led to (x 1 0).
you did not force a (X 1 0) because your opponent can win there.
in (x X 0) if x!=X your opponent could have won by taking the difference from the larger pile, so x=X. (X=X 0) is possible.
in (x 1 x) if x=x, your opponent could have won by taking the extra 1 then keeping the piles equal from there unless x=x=1 so only (X!=X 1) or (1 1 1)
you much have forced a (X=X 0), a (X!=X 1) or a (1 1 1)
a (1 1 1) only comes from a (X 1 1)
a (X=X 0) can come from a (X!=X 0) or a (X=X 1)
a (X!=X 1)can come from a (X=X 1) or a (X!=X 0)
so you could have gotten a (X 1 1), a (X!=X 0), or a (X=X 1)
another turn ago:
you opponent never had the chance to force you into a (1 1 1) so your (X 1 1) did not come from another (X 1 1). It must come from a (X X 1). Your opponent can win with a (X=X 1), so this much be a (X!=X 1)
you can force that (X!=X 1) with either (X=X 1).
so if you won against a perfect opponent, you must have gotten either (X=X 1) or (X!=X 0)
from here on you actually kind of have to discuss each individually (that or im missing something obvious), and this comment is long enough as is...
yes u can......your move would be to take 3 coins from the 7 pile leaving you with the 1,4,5 target
MATHeMAGIC
If I got 1,4,5 then they pick 1, what do I do?
Take 1 from the 5 making the piles equal
Actually you don't even need to memorize anything. There's a solution that works with all starting configurations, and to use it, you just need to know how to count in binary. www.archimedes-lab.org/How_to_Solve/Win_at_Nim.html
Grulps Can you please explain - I've been losing my mind since last night with this. When seeing the video, it seems so easy and obvious, but when I play with my sis, I ALWAYS keep losing.
Start: 3 5 7
Me: 2 5 7
Sis: 0 5 7
What now? Whatever I play, I end up losing. HALP!
*
Lakshminarayanan Krishnan Next steps would be something like this:
You: 0 5 5
Sis: 0 3 5
You: 0 3 3
Sis: 0 1 3
You: 0 1 0
Or you can solve it with binary numbers:
3, 5 and 7 are 11, 101 and 111 in binary.
Put them in a grid like this:
111
101
011
Now your goal is to make sure that each column has an even number of ones. In this case the 3rd column has 3 ones so you have to start by taking away one of them.
For example, 2 5 7 in binary is:
111
101
010
This strategy works until there's only one stack larger than 1 in the game. After that you win by leaving an odd number of stacks of 1 to your opponent.
Hey Grulps - thanks for that explanation. Actually, when Brian says "even number in both piles", I thought he meant it to be multiples of 2. Then, in further gaming with my sis, I now realize it meant same number of coins in each pile... Also, I was playing the game wrongly, thinking I could pull out only 1, 2 or 3 coins at any time from any of the piles.. Now it's a little bit clearer... And your binary method makes it easier to map the moves... Gonna try that one also.. Thanks so much
@@lnkrishnan that's the way I interpreted "two even piles" I thought you had to get 2-2 or 4-4. but if he means two equal piles, things are a bit easier now. the grouping method he teaches with the matchstick version works for this version too. it's easier
The first time i tried this with my dad I lost horribly when he went first. Not sure if I screwed up but I could never make one of the checkpoints. Game breakdown below
Setup: 3, 5, 7
Dad: 3, 3, 7
Me: 3, 3, 6
Dad: 3, 3, 3
Me 3, 3, 2
Dad: 1, 3, 2
*Insert loss here*
He didnt hit the 1,2,3 until the end but I dont know what i should have done on my first or second turn to pull it in my favor. And no he did not know the game. Any Help?
Keegan T couldn't you have gone for 2 pairs in your first move? 3, 3 would have given you the win. (forgive me... I'm half-remembering this from almost a decade ago)
Scam School Now that you mention it yes 3, 3 would be a win. I was under the impression that "two even piles" meant two even numbers in two piles. Such as 2, 2 or 2, 4, etc. Does it mean even as in matching?
+Keegan T
What Brian meant to say and write in the film was "Two EQUAL Piles" not "Two EVEN Piles".
Thanks. Hahaha finally😁😁😁
is this brian's channel ?
what if the other guy goes first and take one of the 3 pile, so i have 2,5,7 ?
+royklein12 That's the part where the one who goes first has the upper hand. After that you could try and go for the other positions. Take all the 7 and then if he takes from the 2 pile take everything but one and if he takes from the 7 pile just take the rest of the 7 pile to balance it out or if you are left with 2-1 you just remove the 2 and you done won.
+KerrankiSuomee > "Take all the 7"
i'M nOt SurE : Set : 3.5.7 - Adv : 2.5.7 - Me : 2.5.0 - Adv : 2.2.0 - Me : i'm lost
+royklein12
I think remove le pile with 2 coins is better...
ADV 2.5.7
ME 0.5.7
Then the other has 12 possible configurations :
0.7 I win
1.7 I win
2.7 I win via a 2.2
3.7 I win via a 3.3
4.7 I win via a 4.4
0.5 I win
1.5 I win
2.5 I win via a 2.2
3.5 I win via a 3.3
4.5 I win via a 4.4
5.5 I can Lose : then search for 4.4 or 3.3 or 2.2 to win
6.5 I win via a 5.5
In general with two piles I think you have to search for :
6.6 / 5.5 / 4.4 / 3.3 / 2.2
What do you think ?
yeah that seems to be the best option, unless they're on to the 5.5. nice job man!
then you just have to take coins to make the other two piles equal
the most annoying part is that they immediately grab the whole seventh pile
Then just take 2 from the 5 pile making it 3 and 3
@@Xandawesome I think like I was, a lot of people are confused about the "two even piles". it sounds like you want two piles of 2-2 or 4-4. but what he meant was two equal piles. 2-2, 3-3,4-4,5-5
go down to 4,4
@kroasana no
Ok Scam School - a few hours in, my head hurts. Maybe I'm overthinking this, but what do I do in this scenario playing against my sister:
Start: 3 5 7
Me: 2 5 7
Sis: 0 5 7
What now? Whatever I play, I end up losing. HALP!
Lakshminarayanan Krishnan you make 2 even piles, you make it 0 5 5.
Hey - thanks. Yeah - I finally figured out what Brian meant when he said "two even piles"... Now I can almost play this game and always win, except in the one time when the other guy starts, and pulls out 1 coin from the middle pile - then, you have to play a little carefully, and hope they don't reach one of the pole positions unwittingly.
Don't forget homophobic.
any configs on 1 3 5 7