I usually under-count this kind of puzzle, but I’ve looked at it 3 times and still get 12 (2 large, 4 medium, 6 small).
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I know that feeling well !!
Well done.
Relax.
I also found 12 but they are all the same shape. Maybe I am just being pedantic about the wording.
Agreed, the wording could have been more clear.
eg ‘How many triangles are there in total’
and definitely not 'How many different shapes/sizes of triangle are there ’
Fortunately, Steve provided answers to both those options.
I appreciate that the ‘6 small’ could be split into 4 + 2 depending on the ratio of the length and width of the enclosing rectangle.
I’m giving myself 5/10 for the question and Steve 10/10 for his answers (the sides of the rectangle look remarkably close to 2:1)
And another 10/10 to Ravvie for constructive criticism …
A friend set me this teaser a couple of years ago. I’ve never bee entirely convinced by his answer or his explanation. I keep wondering whether I should expand it a bit - see below.
Red or blue ? (original version)
I have three black, (opaque bags). Each bag contains two balls.
One bag contains two red balls. The second bag contains two blue balls. The third bag contains a red ball and a blue ball.
I select a bag at random and without looking inside, withdraw a ball. It is blue.
What is the probability that the other ball in that bag is also blue ?
I keep thinking I should expand it to read …
Red or blue ? (modified version)
I have three black, (opaque bags). Each bag contains two balls.
One bag contains two red balls. The second bag contains two blue balls. The third bag contains a red ball and a blue ball.
I select a bag at random and without looking inside, withdraw a ball.
If It is blue, what is the probability that the other ball in that bag is also blue ?
If it is red, what is the probability that the other ball in that bag is also red ?
Mrs R instinctively came up with a different answer from mine. Given that she teaches probability at degree level, this was a bit worrying.
After a bit of discussion and using two different methods, we agreed on an answer.
I’m saying 50%.
This teaser is the perfect opportunity to reappear here. I think I have the first answers by @Ravvie and Mrs. R. Now I have to figure out which one is the right one.
I think the answer to the first question, is there is a 2 in 3 probability that the other ball in the bag is also blue.
I can’t think of any reason why the second question doesn’t give the same probability for red.
The two initial answers in the R household were indeed 0.5 and 2/3.
I will wait for @Mulberry to give his deliberations first…
Does this teaser remind anybody of Monty Hall ?
I don’t think Monty Hall helps, but to my mind, there is frustrating similarity !!
PS. we had the Monty Hall teaser much earlier in this thread.
My logic is that the question of probability applies after a blue ball has been drawn, so there are only two options left for the other ball in the bag, blue or red, so a 50% chance.
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Hi Mike, that was also my first thought.
However, my logic is as follows:-
If the bags and balls are labelled as follows:
BagX contains R1, R2
BagY contains B1, B2
BagZ contains B3,R3
After the first draw, the ball is blue, so we can ignore BagX.
The first ball could be B1, B2 (from BagY) or B3 (from BagZ).
On drawing the second ball, it could be B2, B1 or R3
Therefore a 2 in 3 chance of drawing a blue on the second ball.
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Nicely explained Steve.
Mrs R’s initial thoughts were the same as @Mike_S.
@SteveD’s answer and explanation is spot on.
I suggested to Mrs R that she could consider using conditional probabilities. She quoted the relevant formula from memory and concluded that it leads to the answer of 2/3. I had long since forgotten the formula and I have again forgotten it since last night!
Don’s modified puzzle does lead to the same answer. It is worth observing that two thirds of the balls are paired with the same colour whilst only one third are paired with the opposite colour. Each ball has an equal chance of being picked so the chances of the second ball being a match is 2/3.
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Well done Steve,
That is the answer and your explanation is as good as any that I have seen.
And a lot better than most !
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The odd Ball.
I’ve got a set of old scales where you load one side, then load the other side to balance. This is called a weighing.
I’ve got 9 balls that look identical, but i’m told that one of the balls is actually heavier than the others, all of which are otherwise equal.
Using the old scales, can you describe how to identify the odd ball with the smallest number of weighings possible ?
The odd Ball - upping the anti !!
I’ve got 12 balls that look identical, but i’m told that one of the balls is either heavier or lighter than the others, all of which are otherwise equal.
Using the old scales, can you describe how to identify the odd ball and determine whether it is heavier or lighter using the smallest number of weighings possible ?
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I’ll hazard a guess without thinking too hard.
Start with four balls on each side of the scales. If it balances, then the ball not on the scale is the odd weight. Which is probably the answer, the minimum number of weightings to identify the odd ball is one. More weighings may be needed if the odd ball is not found at the first attempt.
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I think for 12 balls, the minimum number of weighings is three. 6 each to identify 6 balls, then 3 each to identify 3, then a third attempt which could identify the odd ball. But I’m having a whisky, so could be wrong.
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