This Is NOT 50/50

I can't sell the meme cards but I hope you enjoy Woven Math! Wrap your body in sweet, sweet knowledge. represent.com/store/vsauce2

It’s fair Michael: Or is it?

My opponent: chooses the Sequence RRR Me: Haha cards go BRR

I’m not trusting a game to be fair if the other person gets to go first.

Dream didn't cheat, he was just playing a non-transitive game

Vsauce 1 and 3: come back in 2019 Vsauce 2: Still been here this entire time

Every time I watch a vsauce vid: Me: No Kevin Yes! Me: Yes? Kevin: No!!

This game is like rock-paper-scissors, except you let your opponent reveal first...

I loved to say "breathtaking" at the same time that Vsauce did, I knew that he was going to say it. Perfect.

I‘ll try to introduce this as a new drinking game with my friends

I just watched a 12 minute video explaining if the second player gets to pick after the first in rock-paper-scissors, the game is unfair.

I appreciated the 30 seconds spent on explaining how rock-paper-scissors work. I needed an example of non-transitivity after having an example of non-transitivity.

Kevin: this is my dearest friend! Also Kevin: breaks his legs and then throws him off the table.

I mean real life is also like this game.

Plot twist Your friends says that he wants to go second or else he's not playing.

The simplest way to try to explain most of the scenarios is that you want your last 2 colors to be same as opponents first 2 colors. That way every time your opponent gets their 2nd color correct, there is about 50% chance that it finishes your sequence.

For those wondering a more simplified logical explanation as to why this certain strategy works for this game. This strategy pretty much maximizes P2 to be able to stay one step ahead P1 when P1 is doing good. When P1 is doing good, P2 is likely doing better. (1). Why is the last two sequence of P2's = the first two of P1's? Suppose P1 chose the sequence BRB. Now, consider when the most recent two cards drawn out the deck are BR, that is, P1 now has a 50% chance of winning the game depending on whether the next card drawn is B or not. P1 has seemed to have gotten through the more difficult part of the game of hitting the first two cards (25%), and now just needs that final measly 50%. Good for P1, right? Well, maybe. Given that the last two sequence of P2's choice were the same as P1's first two, this would imply that P2 would have either picked R[BR] or B[BR]. Meaning that depending on what was pulled prior to that BR (the one that P1 is currently so happy about), there is a 50:50 chance that P2 has already won. So any time that P1 gets to the point where they have a 50:50 chance of winning (25% probability), P2 would've already won 50% of the time. (2). Why is P2's first sequence = the opposite of P1's second? This is essentially to counter what was said above in (1). To basically guarantee to never be behind P1 whenever P2 is starting to heat up (gets their first card selected). To expand a bit more, anytime P2 gets a hit on their first card, IF P1 hadn't won already, P1 would either be on their zeroth or first card. Because P2's first sequence is the opposite of P1's second, it'll never be the case that P1 is one draw away from winning (50%) while P2 is two draws away (25%). This part is a bit trickier to explain, I think applying this logic to (1) will help. So there you have it. Tl;dr, Whenever P2 loses, they were just simply unlucky. When P1 wins, they were simply lucky.

When I see Vsauce and Phil Swift, I click.

Kevin: new math merch is avaiable Me: how much for the meme cards ?

Once you brought up the idea of (non)transitive games I was hoping you would use Rock Paper Scissors as an example. Did not disappoint! This was the best way I found to explain this idea to students when I tutored