A thread started in another section here, earlier today, about “mathematical skills”, has just reminded me that I meant to post this little puzzle a while ago, to see whether anyone was interested, and promptly forgot …
A coin (“coin A”) is weighted so that it comes up heads 25% of the times it’s tossed, and another (“coin B”) is weighted so that it comes up heads 75% of the times it’s tossed. I have both coins in my pocket, and take one out and toss it (i.e. the same one) twice, without telling you which one it is. It comes up heads both times. What’s the probability that it’s “coin B”?
It’s [B]not[/B] a trick question of any kind, or a semantic question, or anything like that: just an honest little puzzle with (perhaps) slightly more to it than you might expect, at first glance.
As the saying goes, “give reasons for your answer”!
OK, so i’m going against the grain here. Regardless of them being weighted or not, you are choosing this ‘coin’ at random from your pocket, which is a 50/50 chance of either coin A or Coin B. So the probability that it happens to be coin B is 50%.
If you’re not taking it at random then let me know!
You will have less than 12% chance by the second flip of the coin that it is coin B as stated by the definition of what B will do:
You have a 50-50 shot right out of the pocket that it is B,
Then only 25% or 1 in 4 that it is still B when it comes up heads,
And even less when it comes up heads twice (More likely it was A)-
Otherwise known as Hindsight is 100%, paraphrased. : )
I’m going to say that it’s 50/50 as to which coin it is.
Like our trades, we have a random distribution of winners & losers, the same goes for whether a coin lands on heads or tails. A coin could be weighted to have a 1% chance of landing on heads but it’s not to say that it wouldn’t land on heads twice in a row if it was tossed twice.
This is a video of Derren Brown flipping a normal 10p coin & returning a heads result 10 times in a row.
If you watched the full programme (it’s on YouTube: Derren Brown The System), he explains that he flipped a coin for 9hrs in order to generate those 10 heads in a row; which links back to our random distribution of winners & losers.
Long story short: there’s a 50% chance of it being either coin.
I may have ‘liked’ someone’s post here by mistake .The answer’s not 50% and definitely not 100%…I can pretend to be smart and provide the solution but I’ve .spotted a near identical question on Quora so that would be cheating… Nowunless you have at the very least a basic understanding of probability theory (Bayes theorem to be precise) it’s highly unlikely that’s you’ll get the right answer…that said, you can however still get the right answer without using BT… But lexy this is babypips for goodness sake - have mercy! *Originally posted in wrong thread\
[B]Here’s my first answer, which must be wrong:[/B]
[B]15/16 = 0.9375[/B]
It was easier for me to frame the question as “what’s the probability that it is [B]NOT[/B] coin A?”. If the coin pulled were indeed coin A, there would be only a [B]0.25 ^ 2 = 0.625[/B] chance of flipping heads twice with it. We subtract that probably from 1 and get our answer.
[I]I realized, though, that the probability of flipping coin B for two heads in a row is [B]0.75 ^ 2 = 0.5625[/B] [/I]
It’s been a while since I’ve done any serious discrete math, but my naive guess is to take the mean of the two values (although my gut tells me it’s more complex than this).
I really do not know, nice puzzle though Lexys. I wonder simply what is the probability of flipping coin A and having it land on heads twice in a row? If I read google right it is simply the original chances 1/4 (as it is weighted as so) squared in this case because it happened twice. so 6.25%.
I don’t believe the chance of 50/50 of picking the coin has any relevance on the actual question. Therefore, if it lands on heads twice, it has a 93.75% chance of being coin B. (like the guy before me said.)
The answer is 90%. It’s one of those “once you see it, it’s totally logical” things.
The chances of coin A coming up heads twice are 1/4 x 1/4 which is 1/16, and the chances of coin B coming up heads twice are 3/4 x 3/4 which is 9/16.
9/16 is 9 times as much as 1/16, so the chances of it being coin B are 9 times as high, hence 90% for B and 10% for A.
No “advanced probabilistic concepts”, no trickery, so not that complicated, really. On the other hand, only one person got it right, so not that easy, either.
Duly impressed…but you make no mention of the probability of randomly selecting either coin from your pocket - an important detail. So you lose one mark) And Tommor didn’t show his workings and for this reason he cannot claim the full prize which means he wins only 1 lump of coal instead of two.
