A Fistful of Brain Teasers
Posted by: Don Atkinson on 13 November 2017
A Fistful of Brain Teasers
For those who are either non-British, or under the age of 65………. The UK used to have a brilliant system of currency referred to as “Pounds, Shillings and Pence”. Simplified to £ ״ s ״ d. No! Don’t ask me why the “Pence” symbol is a “d”, just learn it and remember it !
A £ comprised 20 Shillings and a Shilling comprised 12 Pence. Thus a £ comprised 240 Pence. I reckon that both Microsoft and Apple would have difficulty with these numbers in their spreadsheets, more so if we included Guineas, Crowns, Half-Crowns and Florins. However, I digress..............
The purpose of the explanation is to assist with the first two or three teasers that follow. So just to ensure a reasonable comprehension has been grasped…. ….. if each of three children has £3 − 7s − 9d, then collectively they have £10 − 3s − 3d Got the idea ? Good ! Just try 5 children, two each with £4 − 15s − 8d and three each with £3 − 3s − 4d. How much do they have between them ? (this isn’t the first brain teaser, just the basic introduction with some “homework”, the Teasers follow)
Ok, we can all look up the "Birthday" answer on the internet !
I have just looked at four sites.
Three of them give 23 as the answer and show their logic and calculation.
One of them gives 24, and whilst the logic is nicely (and correctly) explained there is no calculation.
I have based my logic and calculation on the basis of how many people in a group are likely to NOT have the same birthday. The probability drops to < 0.5 at a count of 23. This corresponds to a probability of > 0.5 that two people in that group WILL share a birthday.
My probability with 23 people is 0.500477 and I did a check with a second (different route) calculation and with 23 people got 0.507297. Both these figures appeared in the sites I visited on the internet.
The third internet site gave a probability of 0.509 with 23 people.
I'm reasonably confident that 23 is the correct answer............
...........but if someone has the calculation to show otherwise, I'm all eyes and ears !
Fair enough!
23 and a goat . (Goats have a different run of life of years compared to human years ).
hungryhalibut posted:Fair enough!
You were close !
TK421 posted:23
Spot on !
David Hendon posted:I think it’s 23...
you think right !
TOBYJUG posted:23 and a goat . (Goats have a different run of life of years compared to human years ).
You overlooked the Gold/Silver coins and the black/white cards.......
Don Atkinson posted:Open the Box !
I have three unmarked, identical boxes containing coins.
One box contains two gold coins, the second contains two silver coins and the third, one of each.
Each box is divided and has two lids such that one coin can be seen at a time.
I select one box at random.
I open one lid of said box and reveal a gold coin.
What is the probability that this box is the one with the gold and silver coins
Returning to the problem involving the three boxes containing 2xGold coins; 2xSilver coins and 1xGold/1xSilver coin.
I have noticed on the internet that this problem manifests itself in two, slightly different, forms.
First, as the form I posted, in which the question is “what is the probability that this BOX is the one with the gold and silver coins ?”.
Secondly in which the question is “what is the probability that the other COIN in this box is Silver”
Some of the internet solutions appear to drift from one form to the other without making it clear what is going on. I have therefore searched our loft and found the book from which I recalled this Brain Teaser. Fortunately my post was an accurate rendition of the book, which was written by a well respected professor of mathematics (who also got the Monty Hall problem correct – so I trust him !)
Just to be confident that my question is the same as the professor’s question, here are his words to the relevant part ..........……
”Pick a box at random. What if I open one of its lids and find a gold coin inside ? What are the chances of this box being the G/S one ?”
A summary of his explanation follows in my next post :
The professor's explanation is as follows.............
"What is the probability of picking (at random) the box with the gold and silver coins in it ? The answer is simple of course: one in three. But that is NOT the puzzle.
If you choose a box at random and, before opening one of the lids, you know there is a one in three chance that it is the G/S box, then how, by seeing one of the coins inside, and gaining no information at all from this, since you know you are certain to find either a gold coin or a silver coin anyway, does the probability switch from one in three to one in two (or any other probability) ?
Consider the case where you find a gold coin inside the box. There are three gold coins in total – let us call them G1, G2 and G3 and let us say that the GG box contains G1 and G2 while G3 is the gold coin in the G/S box. If you open one of the boxes and find a gold coin inside, there is a 2 in 3 chance of your having picked the GG box since the coin you are looking at could be either G1 or G2. There is only a one in three chance that it is the G3 coin and therefore that the box you have chosen is the G/S one."
So the correct answer to the problem that I set is 1/3.........unless you know a better Professor than I do
PS, that second paragraph of the professor's is one long, difficult-to-read sentence ! The bit from ".....then how, by seeing.......to .......(or any other probability) ?" is the important bit.
Congratulations to Dozey, HH and Mulberry !
Commiserations, but great thanks, to the others who took part ! (unless your professor is better than mine )
What is the next number in the series 3, 4, 8, 17, 33........?
58
I concur, 58. The next would be 94
58 here too.
Yes ! 58 it is.
A bit more straight forward than Monty Hall and the Gold/Silver coins ?
The difference is that it’s much easier to see the reasoning and it leaves no room for disagreement.
best
David
Now, this one appeals to me both as a frequent passenger flyer these days and also because of my profession...
100 passengers board an airplane with exactly 100 passenger seats. Everyone has a ticket with an assigned seat number.
However, the first passenger has lost his/her ticket and takes a random seat. Every subsequent passenger attempts to choose their own seat, but takes a random seat if their assigned seat is taken.
Suppose you are the very last passenger to board the plane. What is the probability that you will get your assigned seat?
David Hendon posted:The difference is that it’s much easier to see the reasoning and it leaves no room for disagreement.
best
David
Looks like I missed quite a bit the last few days. Would you mind to giving the solution for the 3, 4, 8… teaser? It’s probably harder for me than coins, goats and cards .
4 is 1² more than 3.
8 is 2² more than 4
17 is 3²...
Thanks, I can see it now. The probabilities are really easier to get for my mind. Clearly more the analytical type, not a creative one.
Mulberry posted:Thanks, I can see it now. The probabilities are really easier to get for my mind. Clearly more the analytical type, not a creative one.
I’m the reverse - but I thought that wss me having an analytical not creative mind!
Mulberry posted:Thanks, I can see it now. The probabilities are really easier to get for my mind. Clearly more the analytical type, not a creative one.
Hi Mulberry, we all have a mixture of strengths and weaknesses. Perhaps in return you could help IB with the gold/silver coin teaser
Also see the comment made by David Hendon above.
It takes all sorts to make the world go around !
Don Atkinson posted:Now, this one appeals to me both as a frequent passenger flyer these days and also because of my profession...
100 passengers board an airplane with exactly 100 passenger seats. Everyone has a ticket with an assigned seat number.
However, the first passenger has lost his/her ticket and takes a random seat. Every subsequent passenger attempts to choose their own seat, but takes a random seat if their assigned seat is taken.
Suppose you are the very last passenger to board the plane. What is the probability that you will get your assigned seat?
Hmm. 50%. As it’s the same for two seats or three seats, I’m assuming it’s the same for 4 seats or more, including 100 seats.
Don Atkinson posted:Mulberry posted:Thanks, I can see it now. The probabilities are really easier to get for my mind. Clearly more the analytical type, not a creative one.
Hi Mulberry, we all have a mixture of strengths and weaknesses. Perhaps in return you could help IB with the gold/silver coin teaser
Also see the comment made by David Hendon above.
It takes all sorts to make the world go around !
The coin one was straightforward for me. I think I get it with the card trick. But I have no idea where to start with the other probability ones...
Innocent Bystander posted:Don Atkinson posted:Mulberry posted:Thanks, I can see it now. The probabilities are really easier to get for my mind. Clearly more the analytical type, not a creative one.
Hi Mulberry, we all have a mixture of strengths and weaknesses. Perhaps in return you could help IB with the gold/silver coin teaser
Also see the comment made by David Hendon above.
It takes all sorts to make the world go around !
The coin one was straightforward for me. I think I get it with the card trick. But I have no idea where to start with the other probability ones...
Mulberry got the coin one right.
Hopefully my explanation of the G/S coin one has helped and also my explanation of the Monty Hall one