Gizmodo Monday Puzzle: How to Solve a Diabolical Hat Trick

It has been a axenic delight to melt your brains each week, but today’s solution volition beryllium the past installment of the Gizmodo Monday Puzzle. Thank you to everyone who commented, emailed, oregon puzzled on successful silence. Since I can’t permission you hanging with thing to solve, cheque retired immoderate puzzles I made precocious for the Morning Brew newsletter:

• An unconventional mini crossword 
• A full-sized crossword with a tricky theme 
• A caller codification breaking puzzle called Decipher

Thanks for the fun,
Jack

Solution to Puzzle #48: Hat Trick

Did you past last week’s dystopian nightmares? Shout-out to bbe for nailing the archetypal puzzle and to Gary Abramson for providing an impressively concise solution to the 2nd puzzle.

1. In the archetypal puzzle, the radical tin warrant that each but 1 idiosyncratic survives. The idiosyncratic successful the backmost has nary accusation astir their chapeau color. So instead, they volition usage their lone conjecture to pass capable accusation truthful that the remaining 9 radical volition beryllium capable to deduce their ain chapeau colour for certain.

The idiosyncratic successful the backmost volition number up the fig of reddish hats they see. If it’s an unusual number, they’ll outcry “red,” and if it’s an adjacent number, they’ll outcry “blue.” Now, however tin the adjacent idiosyncratic successful enactment deduce their ain chapeau color? They spot 8 hats. Suppose they number an unusual fig of reds successful beforehand of them; they cognize that the idiosyncratic down them saw an adjacent fig of reds (because that idiosyncratic shouted “blue”). That’s capable accusation to deduce that their chapeau indispensable beryllium reddish to marque the full fig of reds even. The adjacent idiosyncratic besides knows whether the idiosyncratic down them saw an adjacent oregon unusual fig of reddish hats and tin marque the aforesaid deductions for themselves.

2. For the 2nd puzzle, we’ll contiguous a strategy that guarantees the full radical survives unless each 10 hats hap to beryllium red. The radical lone needs 1 idiosyncratic to conjecture correctly, and 1 incorrect conjecture automatically kills them all, truthful erstwhile 1 idiosyncratic guesses a colour (declines to pass), past each consequent idiosyncratic volition pass. The extremity is for the bluish chapeau closest to the beforehand of the enactment to conjecture “blue” and for everybody other to pass. To execute this, everybody volition walk unless they lone spot reddish hats successful beforehand of them (or if idiosyncratic down them already guessed).

To spot wherefore this works, announcement the idiosyncratic successful the backmost of the enactment volition walk unless they spot 9 reddish hats, successful which lawsuit they’ll conjecture blue. If they accidental blue, past everybody other passes and the radical wins unless each 10 hats are red. If the idiosyncratic successful backmost passes, past that means they saw immoderate bluish chapeau up of them. If the second-to-last idiosyncratic sees 8 reds successful beforehand of them, they cognize they indispensable beryllium the bluish chapeau and truthful conjecture blue. Otherwise, they pass. Everybody volition walk until immoderate idiosyncratic towards the beforehand of the enactment lone sees reddish hats successful beforehand of them (or nary hats successful the lawsuit of the beforehand of the line). The archetypal idiosyncratic successful this concern guesses blue.

The probability that each 10 hats are reddish is 1/1,024, truthful the radical wins with probability 1,023/1,024.