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Random maths


davenportram

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You have 3 cards (let's say 2 diamonds - red and 1 club - black)

The cards are face down, from the same pack.. You want to choose the black club..

You choose a card, one of the 3.. I then reveal another card of the 3 cards which is a red diamond.. Leaving 2 cards, one red diamond and one black club..

I give you the option to change cards from your original pick.. Do you change? And why?

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You have 3 cards (let's say 2 diamonds - red and 1 club - black)

The cards are face down, from the same pack.. You want to choose the black club..

You choose a card, one of the 3.. I then reveal another card which is a red diamond.. Leaving 2 cards, one red diamond and one black club..

I give you the option to change cards from your original pick.. Do you change? And why?

Do you know where the black card is?

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No.

Then I wouldn't switch because the chance if my card being the black one is still 1 in 3

And the chance if the other face down card is also 1 in 3( the fact that you don't now where the black card is means it has no effect on the probability of the other cards being the black one)

If you knew where the card was I would switch because the probability of the card you didn't turn over being the black one is two in 3 where as the probability that my card is black is still 1 in 3.

(See the three doors problem with a quizmaster)

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Then I wouldn't switch because the chance if my card being the black one is still 1 in 3

And the chance if the other face down card is also 1 in 3( the fact that you don't now where the black card is means it has no effect on the probability of the other cards being the black one)

If you knew where the card was I would switch because the probability of the card you didn't turn over being the black one is two in 3 where as the probability that my card is black is still 1 in 3.

(See the three doors problem with a quizmaster)

You quoted before I edited.. Well done sir you are correct.. It's amazing how many people wouldn't change still beliveing there is a 50/50 chance when it comes to the final 2..

My girlfriend whose in her 3rd year studying economics at the university has advanced maths as a part of her studies and she still doesn't believe me when I tell her she's wrong and it's not a 50/50 percentage

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You quoted before I edited.. Well done sir you are correct.. It's amazing how many people wouldn't change still beliveing there is a 50/50 chance when it comes to the final 2..

My girlfriend whose in her 3rd year studying economics at the university has advanced maths as a part of her studies and she still doesn't believe me when I tell her she's wrong and it's not a 50/50 percentage

There's a good explanation on Wikki using tree diagrams and conditional probability there's also an online simulation that shows the experimental probability too.

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I tried explaining it using larger numbers.. I told my girlfriend to write down a number 1-10 and I'll do the same.. (she writes 5 I write 7 for example).. I then told her I'll remove all the other numbers and leave 2, (5 and 7)..

I then ask her if she'd like to change number.. She refuses, still beliveing that with the final 2 numbers it's a straight 50/50 chance..

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I have 9 tins that appear identical. One is ever so slightly heavier than the rest.

I have a set of balance scales to use to find the tin that is heavier.

What is the smallest number of times I need to use the scales to find it?

Oohh... Maths questions! So I'm guessing the question is the minimum number of times you have to use the scales to be guaranteed to find the heavier one. Hmmm...

- Divide the tins into 3 groups of 3 denoted A, B and C

1. Weight A against B

- If A and B are of equal weight then the heavy tin is in C

- If A is heavier than B it is in A

- If B is heavier than A it is in B

- Now we only use the group that has the heavier tin in it

- Now we have 3 tins, lets call them Tin 1, Tin 2 and Tin 3

2. Weight Tin 1 against Tin 2

- If Tin 1 and Tin 2 are of equal weight, then the heavier tin is Tin 3

- If Tin 1 is heavier than Tin 2, then the heavier tin is Tin 1

- If Tin 2 is heavier than Tin 1, then the heavier tin is Tin 2

So 2 uses of the scales can find the heavier one every time. Someone else has probably already gotten that, but hey.

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