The paradox of changing odds
 Bill Stokes
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The paradox of changing odds
Three sets of two playing cards  you have:
[A][A]  [A][K]  [K][K]
All 3 sets are placed face down randomly. Your job is to select (guess!) the pair [A][K].
Now, at this stage, the chances of getting the correct pair is 3 to 1 ~ that is not the question.
You select a pair.
I turn over one card of that pair, and it is a [K].
Now what are the chances you have the [A][K] pair? And why, nothing has changed?
Nick
[A][A]  [A][K]  [K][K]
All 3 sets are placed face down randomly. Your job is to select (guess!) the pair [A][K].
Now, at this stage, the chances of getting the correct pair is 3 to 1 ~ that is not the question.
You select a pair.
I turn over one card of that pair, and it is a [K].
Now what are the chances you have the [A][K] pair? And why, nothing has changed?
Nick
Re: The paradox of changing odds
some beans
Re: The paradox of changing odds
something has changed....you can now disregard the [A][A], so the odds are 2  1
 Bill Stokes
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Re: The paradox of changing odds
Nothing has changed  why do the odds change just by showing you a card you already have? If I didn't show you one of the cards, are the odds still 3 to 1?Parts wrote:something has changed....you can now disregard the [A][A], so the odds are 2  1
Nick
Re: The paradox of changing odds
but you did show me...so the odds go down

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Re: The paradox of changing odds
As you don't have knowledge of where the A A combo is that doesn't no good for you in the odds, so I think it is still 31.
Re: The paradox of changing odds
That's not the point....the AA set is disregarded....you either have AK or KK, so there is a 21 chance that you have the AK
 Bill Stokes
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Re: The paradox of changing odds
OK, think about this. You choose your two cards. Now then, without looking, the card on the left is either [K] or [A]. That means you either have ( [K][K] or [K][A] ) ~ OR ~ ( [A][K] or [A][A] ).
There, I didn't show you any cards, yet the odds seem to change from [3 to 1] to [2 to 1]. Impossible!
Nick
P.S. I do know the solution and can show the proof.
There, I didn't show you any cards, yet the odds seem to change from [3 to 1] to [2 to 1]. Impossible!
Nick
P.S. I do know the solution and can show the proof.
 Bill Stokes
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Re: The paradox of changing odds
Well, seeing as everybody has given up, here is the proof.
The problem is you don't know what [K] (or maybe an [A]) you have, as there are three each.
The possible combinations of the sets of two cards are:
[A1][A2]
[A2][A1]
[K1]K2]
[K2][K1]
[A3][K3]
[K3][A3]
Now, at first sight, and correct, the odds are 3 to 1. Pick one set, and it's a [K] ~ but which one? That means you either have:
[K1][K2]
[K2][K1]
[K3][A3]
...still 3 to 1.
Nick
The problem is you don't know what [K] (or maybe an [A]) you have, as there are three each.
The possible combinations of the sets of two cards are:
[A1][A2]
[A2][A1]
[K1]K2]
[K2][K1]
[A3][K3]
[K3][A3]
Now, at first sight, and correct, the odds are 3 to 1. Pick one set, and it's a [K] ~ but which one? That means you either have:
[K1][K2]
[K2][K1]
[K3][A3]
...still 3 to 1.
Nick
Re: The paradox of changing odds
That's cheating...you added a shit load more cards!
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Re: The paradox of changing odds
That makes sense now. I had to reread the original topic the key being "3 sets" of cards.
 Bill Stokes
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Re: The paradox of changing odds
No  there are still only 3 sets of two cards [K1][K2] ; [A1][A2] ; [K3][A3] ; but do you see [K1] or [K2]?Parts wrote:That's cheating...you added a shit load more cards!
Nick
Re: The paradox of changing odds
yes .... before the cards were [A][A], [K][K] etc, now they all have numbers. but if you have any [K] you can still safely say you don't have the [A][A] set.
I rest my case m'lud
I rest my case m'lud
 Bill Stokes
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Re: The paradox of changing odds
Then you are going down for a very long timeParts wrote:
I rest my case m'lud
You could have either of the [K]'s  you don't know which one. So you still have a one in three chance that you got the set [K][A].
Nick (P.S., this isn't my paradox  it's well known in the maths circles).
Re: The paradox of changing odds
Let's look at your 1st post....
Three sets of two playing cards  you have:
[A][A]  [A][K]  [K][K] no numbers here...just 3 K's and 3 A's
All 3 sets are placed face down randomly. Your job is to select (guess!) the pair [A][K] ok, I have to find the only set out of the 3 that has 1 of each
Now, at this stage, the chances of getting the correct pair is 3 to 1 ~ that is not the question.
You select a pair.
I turn over one card of that pair, and it is a [K]. I now know that I don't have the only pair of A's and can have either the pair of K's or the set with 1 of each  the set I was asked to guess
Now what are the chances you have the [A][K] pair? 21And why, nothing has changed? yes it has, you turned over 1 of my pair and it was a K
Don't trust these mathy people, they want to trick you all the time with their sneaky numbers things.
Three sets of two playing cards  you have:
[A][A]  [A][K]  [K][K] no numbers here...just 3 K's and 3 A's
All 3 sets are placed face down randomly. Your job is to select (guess!) the pair [A][K] ok, I have to find the only set out of the 3 that has 1 of each
Now, at this stage, the chances of getting the correct pair is 3 to 1 ~ that is not the question.
You select a pair.
I turn over one card of that pair, and it is a [K]. I now know that I don't have the only pair of A's and can have either the pair of K's or the set with 1 of each  the set I was asked to guess
Now what are the chances you have the [A][K] pair? 21And why, nothing has changed? yes it has, you turned over 1 of my pair and it was a K
Don't trust these mathy people, they want to trick you all the time with their sneaky numbers things.
 Bill Stokes
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 Posts: 748
 Joined: Sun Aug 17, 2003 9:36 am
 Location: Pompey, England.
Re: The paradox of changing odds
From the original question, the odds (probabilty) is 3 to 1. But the chance is 2 to 1 against.
That means you still have a 2 to 1 against chance by being shown one card of the pile you picked.
I know John, it is doing my head in, but the maths is correct.
Links to read:
https://en.wikipedia.org/wiki/Monty_Hall_problem
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
Nick
That means you still have a 2 to 1 against chance by being shown one card of the pile you picked.
I know John, it is doing my head in, but the maths is correct.
Links to read:
https://en.wikipedia.org/wiki/Monty_Hall_problem
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
Nick