bonanova

Betty bought a box when she was 20. Betty puts $250 in the box on every birthday. Each year, on her own birthday, Betty's evil sister Veronica steals $50 from the box. Betty died when she was 60. When the box was opened it was found to contain only $500. Who is to blame for this shortfall, Betty or Veronica?

The notation convention 0.999 ... denotes an endless string of 9's. You can't write the decimal representation of 1/3, but you can write 0.333 ..., Thus, given the meaning of that notation, one can meaningfully write 1/3 = 0.333 ... Similarly one can meaningfully write 1 = 0.999 ...

Here's one

I think you're correct in your last statement. You can, as far as I know. guarantee 50% but not more.

The classic solution I saw a while back is clever, but not as neat as Rainman's:

Agree! Another Aha puzzle. But still ... if I had to pick the most satisfying approach, I would define (zero-width) line segments that run midway from the edges of the 1-inch line, and then sum their lengths. I think this was the approach taken in previous posts. Or, take the average of the left and right edges of the 1-in-thick line.

The puzzle states: "There is a spy who informs the PO of the strategy. So in placing the hats, the PO does his best to foil it." I interpreted this to mean the PO placed the hats himself, as diabolically as he could. I follow you. And the OP, again, to be fair to you, does not preclude your comment. The difference I drew was between knowing the strategy in general terms and being privy to the details of its implementation. And that is a finer distinction than the OP makes. BTW, This may distinguish between Phil's and Rainman's (post 11) approaches. I don't think your PO tactic can defeat Rainman's solution. Or can it?

Was this result calculated? The first two cards didn't seem to have that much effect in simulations. The code calculates the exact probability for the average payoff with the best strategy (unless I messed up.) Both simulation and code must take into account the cards leaving the deck. Also, must discard invalid (in terms of strategy) variations. E.g., a draw like (13, 13, 8). The criteria for the staying card on the third turn can be established analytically (see the spoiler.) Very nice Prime, as usual. And no, I don't think anything is messed up - nothing I see suggests that 10.2070 is incorrect. It would be interesting to see your code's four payoff values for 10 9 7 and 10 9 8 when the first two cards are above and below 14. The code would have to be modified from "search" mode to where 10 9 are given, and results sorted into two categories with respect to 14. I think I see how to calculate it directly, at least the EV for 10 9. The EV contribution for the third draw, in the two cases is a little complicated, and I didn't finish it last night.

To make it solvable, change the random scale into an alternating scale.

I did, I think, last night... check your mail.

Good observation TC. But in fairness to Phil, the PO, knowing that half will make blue=1, etc., may not know which half that is. The PO, I think, would only choose the distribution of red/blue, not the individuals that receive a certain color. I'm still thinking through Phil's solution before I declare the puzzle solved. In addition to the solutions posted (Rainman's #11 is very concise, and I'm wondering whether it's equivalent to Phil's?) there is a another clever one, not yet posted.

Was this result calculated? The first two cards didn't seem to have that much effect in simulations.

I'd like to see the spreadsheet. Thx.

First, the good news: Nobody gets killed. 100 prisoners are out on parole, and the those who guess wrong will simply have their parole revoked. No blood. What a relief. The children will sleep well tonight. But a lot of the usual stuff in the red hat / blue hat genre still applies: The Parole Officer (PO) will fit each parolee with a red hat or a blue hat. Each parolee can see all the other hat colors, but not his own. No communication is permitted. Just looking at hat colors. Seriously, no shouting, pausing, singing, pointing, facial expressions. If anything like that happens, then the PO will get PO'd and shoot everyone. On a common signal, each parolee will simultaneously guess the color of her own hat. Parolees initially meet, to formulate a strategy that dictates what color to guess, based on what she sees. The object is to guarantee the most successful guessers, regardless of the hat distribution. And here's a wrinkle. There is a spy who informs the PO of the strategy. So in placing the hats, the PO does his best to foil it.* How many parolees can be saved? * For example, if the strategy is to "guess the majority color" the PO will place 50 red and 50 blue hats, and no parolee will be saved.

It works in 2D. The smallest square just can't be on an edge.

Can you ask about the other people?

Nice.