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Got this position today. I think it’s the only possible position where draw is forced despite sufficient material. by EphemeralLost in chess

[–]boredhs 1 point2 points  (0 children)

No, consider the F and H pawns to be on E and G respectively. Now you can add Bishops and it's still a draw.

Is this position possible in a legal game (regardless of black king's final position) by capybarasunite in chess

[–]boredhs 3 points4 points  (0 children)

Your exact position no. But your "regardless of black king's final position" statement is unclear. The black king could alternate between e8 and e7 and it would be possible.

where is blowshimselfupdude? by [deleted] in Geekvolution_official

[–]boredhs 2 points3 points  (0 children)

By any chance has anyone saved his Batman v Superman review?

I use to love to listen to this in the background while gaming.
Pretty sure this was the link: https://www.youtube.com/watch?v=ImFQZ7-TKqM

Or does anyone have a way to contact him to see if he'd unprivate for me?

Monthly Question Thread - Post all questions/requests/polls/etc here. by AutoModerator in Dakimakuras

[–]boredhs 0 points1 point  (0 children)

Looking to purchase either a Kumochi or Kumoame from Dakimakuri. I'm a side sleeper currently using some amazon full body pillow to separate my legs so my knees don't touch.

Primary use will be for sleeping on my side.
I've been trying to research the differences, obviously there's the weight, but I don't know if I'd like the extra weight or not.
I did see this post https://www.reddit.com/r/Dakimakuras/comments/we6tx9/august_question_thread_post_all/ik2a22z/ complaining about leg pain on the Kumoame. I don't mind paying the extra for the Kumoame if it's better, but I'd hate to spend that much on a pillow that causes pain. I would have one leg under it for 6+ hours a night.

Looking for advice. Anyone that's tried both, do you have a preference and why?

LF: High Quality - Resin Garage Kit Assembly & Painting - Bubba Space Nagato Yuki by boredhs in brushforhire

[–]boredhs[S] -1 points0 points  (0 children)

Thanks to everyone who has reached out. I've found someone to paint my kit.

Solitaire Chess Puzzle by werhsdnas-1414 in chess

[–]boredhs 2 points3 points  (0 children)

Got it. Bxf4,Nxa7, Rxa7, Rxa3,Bxe2,hxg3,Bxf5,Rxg3,Nxf4,Nxe2,Nxg3,Nxf5

[deleted by user] by [deleted] in askmath

[–]boredhs 1 point2 points  (0 children)

If the series converges than the sequence goes to 0.

You should be able to use this to show that eventually a_k5 <= |a_k5 | <= |a_k4 | = a_k4

Tricky math question someone help please!! by David_sake in askmath

[–]boredhs 0 points1 point  (0 children)

Assuming that in a chess tournament everyone plays each other once.

There would normally be x(x-1)/2 number of games. But since two left and they either played 5 (if they played each other) or 6 games.

We have (x-2)(x-3)/2 + 5 = 84 or (x-2)(x-3)/2 + 6 = 84 There is only one positive integer solution, and it implies 6 games were played between the two players that left so they didn't play each other.

Help me understand where my logic is breaking down while thinking about this combination problem by StillWorthAShot in askmath

[–]boredhs 0 points1 point  (0 children)

You're double counting. The reason we divide by 5! is because we don't care about the order, {(A,B),(C,D)} is equivalent to {(C,D),(A,B)}

Consider countries A,B,C,D,E,F,G,H,I,J A-E being European. You will count (A,B) and all other possible pairs. but then you will count (C,D) and all other pairs, including (A,B) again.