/sci/ - Science & Math

>>12400298
>One is heads
Ok, let’s use your retarded version of logic.
you know the value of coin1: {Coin1-h, coin2-h.} or {Coin1-h, coin2-t}
Or you know the value of coin-2{coin1 head, coin2 h} or {coin1-t,coin2-h}
THIS IS THE SET of possibilities . It’s still 50/50.
You are trying to have it both ways where t,h is not the same as h,t but the coins are still indestent.
This is the wrong model you are using:
{Coin1-h, coin2-h.} {Coin1-h, coin2-t}{coin1-t, coin2-h}
^thats wrong wrong wrong. You don’t get it both ways, where h,t is not same as t,h but you still count them once