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Wizard
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Wizard
Joined: Oct 14, 2009
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Thanks for this post from:
camapl
August 24th, 2022 at 9:06:54 AM permalink
Quote: camapl


Divide the 25 racehorses into 5 groups. Each group in turn will comprise 5 heats. The winner from each heat will comprise the 6th race. The 7th, and last, race will include 2 horses from race 6 (the second and third fastest), 2 horses from the heat of the fastest in race 6 (the second and third fastest from that heat), and 1 horse from the heat of the second fastest in race 6 (the second fastest from that heat). The fastest 3 horses, in order are (1) the winner of race 6, (2) the winner of race 7, (3) the runner up of race 7.

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Correct!
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
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Wizard
Joined: Oct 14, 2009
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camapl
August 24th, 2022 at 9:08:15 AM permalink
Quote: camapl

Who is she, and where shall I inquire as to hire her to carry my bags?
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She is a Victoria's Secret model.
It's not whether you win or lose; it's whether or not you had a good bet.
charliepatrick
charliepatrick
Joined: Jun 17, 2011
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September 7th, 2022 at 2:03:18 AM permalink
Rolling dice and either keeping odd or even.

Sorry I don't know the answer to these puzzles, the basis of it came in a dream last night and I thought it an interesting puzzle.

Dream Version
The initial version had a pack of 100 cards numbered 1 to 100. The aim was to collect cards with the largest total but they either had to be all odd or all even. You drew four random cards and could elect to keep any or none. If you kept any odd cards, all cards kept had to be odd, you could not keep any even ones. Similarly for even cards. Also once any cards were selected you had to keep them, unlike Yangzte.

The game consisted of two draws. Obviously in the second draw, if you had kept some from the first draw, you would pick either all the odd or even cards as appropriate. If not you would keep the highest total. You add up all the cards you managed to collect.

Dice Version
I think it makes the maths easier, since each roll can be independent from the others and there's no question about replacement for the second round, if you assume you roll 100-sided dice (or 6-sided dice) i.e. numbered 1 to N. (Sadly this makes the maths slightly difficult as the average of the even numbers is greater than the odd ones, so if it's easier assume an infinite dice where the average is N/2 for both!)

So the question is what is your best strategy/approach and the expected outcome?
charliepatrick
charliepatrick
Joined: Jun 17, 2011
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September 12th, 2022 at 1:57:01 PM permalink
Quote: charliepatrick

...Rolling dice and either keeping odd or even...

Sorry for such a hard problem, please ignore it as I've given in, especially if the discarded cards stay out (as it's rather like trying to work out all the combos for draw poker). I have worked out the average before the draw and that if they're all the same type you keep any 74 or greater. Also if they all add up to more than 148 and are all larger than 25, I think you keep them all (it happens to beat the average from redrawing the lot).

Here's an easier one from today's paper.
If you reverse the two digits of Adam's age, add one and divide by two, you'll end up with Adam's age. How old is Adam?
ksdjdj
ksdjdj
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September 12th, 2022 at 4:35:52 PM permalink
37
DogHand
DogHand
Joined: Sep 24, 2011
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Thanks for this post from:
camapl
September 12th, 2022 at 5:19:08 PM permalink
Quote: Wizard

Quote: camapl

Who is she, and where shall I inquire as to hire her to carry my bags?
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She is a Victoria's Secret model.
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Well, considering what Victoria's Secret clothing looks like, couldn't she get by with just a large business envelope? ;-)

Dog Hand
charliepatrick
charliepatrick
Joined: Jun 17, 2011
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Thanks for this post from:
ksdjdj
September 13th, 2022 at 12:27:46 AM permalink
Quote: ksdjdj

37

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Correct - well done!

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