There appear to be multiple valid solutions to the Picross, but all take you to the same url

How do you know you got 1-4? You don't know until you get all 6

Nah but I *know* I've got them. Pretty stuck on 5.

Got all 6, onto the 3rd part now. Seems a lot harder...

Nice one! Good luck

Pretty sure I'm stumped now

Director GCHQ's Christmas card puzzle - how did you do?

600,000 people successfully attempted the first stage. Seems no one solved the whole thing, but 3 people got close.

There is a road that is one mile long. A car is at the beginning of the road and it will travel to the end of the one mile road. At x=0, the car’s initial speed is 30mph. It will continue at 30mph until the road’s halfway point. At the halfway point (1/2 mile mark), the car can instantly change to any other speed it chooses for the second half of the road. The car is not accelerating at any time during its travel, there is only a discontinuity in its speed at the halfway point. What speed must it select at the halfway point to have an average speed over the full one mile road of 60mph?

This can't be done. 60mph is the same as 1 mile per minute. If the car is going 30mph it will take the full minute to get to the halfway point. I guess if you want to argue about relativity, then the car could go the speed of light for the 2nd half of the trip and this would effectively accomplish an average speed of 60mph for the full mile.

I suppose the purely mathematical answer is INF. Oh how I miss Algebra 2.

I think the poster's above have misread the question slightly. There's a next-level trick to this one. Correct answer here:

Buckle this!

bump

Not really a brain teaser per se, as I don't know the answer, but I'm playing a game and I can't quite figure out the math needed to determine if my outcome is 'good' or not.
So in short:

I'm playing a game, and I gain tokens (for the sake of the conversation). I use those tokens to buy bonuses at random.
Some bonuses are good, some suck. I get them at fairly predictable percentages.

I can calculate the chances of any particular configuration of bonuses given repeats and the fact that order doesn't matter.
But I don't really know how to combine that with the fact that some bonuses are good and some aren't to determine if I should 'accept' the bonus, or 'try again'.

Real world example:
There are 7 bonuses.

Bonus	% Chance to Get	Benefit
A	20%	300
B	20%	80
C	20%	400
D	20%	40
E	6%	20
F	6%	800
G	6%	300
H	2%	0

So I know that for any combination of A though H, it has an XXXX chance of happening. And I can then multiply it by the 'benefit' of that configuration to get a sense of the weighted value of that configuration. But I don't know how to compare that against the likely outcomes I should be able to expect given the conditions.
Any help? / Anything I can clarify that didn't make sense?

I'm not sure what you are looking for, the only thing I can add to the above table is that the EV of this game = 60 + 16 + 80 + 8 + 1.2 + 48 + 18 + 0 = 231.2

Based on the listed probabilities and benefits you want to pay less than 213.2 to play and come ahead in the long term.

Not sure I fully understand what you are asking

You know to use weighted average to calculate your ev so you can just compare that to the ev if you don’t spend the token on buying bonuses and determine which decision is higher ev

Another way to look at it is you have a 52% chance to get a 300+ bonus


If there are a set number of rounds and bonuses work after purchase, then obv don’t buy a bonus at the very end, but how far before the end to stop buying depends on the ev and how many tokens it take to buy a bonus (I’m assuming you want to optimize final token count)


Keep it simple:
Use ev of bonus buy based on number of rounds left to play and compare that to the tokens it costs to buy the bonus; whichever is higher do that

Of course, if bonuses stack on each other, then the ev of buying a bonus now given you expect to buy another bonus later is higher than initial calculation bc this bonus will make the next bonus even higher ev

I missed the “accept or try again” part the first time

Same logic tho…compare ev of trying again to ev of the bonus you drew the first time…If higher, try again

If a cost to try again, then be sure to discount the ev of trying again to account for that added cost

How do you win? Sometimes the goal shouldn’t be to maximize EV but rather to maximize the probability that you get at least X points or to maximize the probability of winning in the next Y turns or something like that. Does the strategy change if you are ahead or behind an opponent (are there opponents in the game?). So I think knowing what your goal is with these bonuses is pretty important.

If it is more complicated than just a straight EV calc then a simulation can sometimes be one of the simplest ways to understand something more complicated.

Yeah you should be able to code up a quick simulator and run it a few million times to see the best options

What is the next number in the below sequence

1
11
21
1211
111221
??

You knew it or you figured it out?

I have seen it before, but even then I figured it out.

This is a good puzzle. Not sure if seen before but I know the answer (a 6 digit prime).

I think numberphile has done this problem before (and showed how it was a closed sequence)