17 Jun 03
Originally posted by royalchickenI'm not great at statistics but:
129 coins lie on a table. Of these coins, 128 are fair and the other has heads on both sides. After picking a coin at random, ignorant of its nature, you flip it eight times, turning up eight consecutive heads. What is the probability that the ninth is a head?
You have a 128/129 chance of getting a fair coin.
A fair coin would give you 1/2 of getting a head.
It doesn't matter that you got 8 heads already it's still 1/2.
So, 128/129 x 1/2 = 128/258 chance of it being a head?
But you have 1/129 of getting the bent coin.
This would give you a probability of 1 of a head.
1/129 x 1 = 1/129.
So the answer is 128/258 because this is more likely than picking the bad coin?
I told you I was no good at probability 😳
17 Jun 03
Originally posted by royalchickenhmm, I forgot all these formula's.
129 coins lie on a table. Of these coins, 128 are fair and the other has heads on both sides. After picking a coin at random, ignorant of its nature, you flip it eight times, turning up eight consecutive heads. What is the probability that the ninth is a head?
My guess: 5/6 (83.33333...😵?
Gil.
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17 Jun 03
129 coins lie on a table. Of these coins, 128 are fair and the other has heads on both sides. After picking a coin at random, ignorant of its nature, you flip it eight times, turning up eight consecutive heads. What is the probability that the ninth is a head?Hmm, is this right?:
There are 2^8 = 256 ways of tossing 8 heads with the unfair coin as each of the heads could be from either of two sides.
There is only one way of tossing 8 heads with a single fair coin. However there are 128 fair coins to toss them with - therefore there are 128 ways of throwing heads with the fair coins.
Therefore the probability that the actual coin chosen is unfair is 256/(256+128) = 2/3, and so the probability that it is a fair coin is 1/3
The probability of the next toss being heads is the probability of EITHER the actual coin being fair (1/3) AND tossing a head with it (1/2) OR the actual coin being unfair (2/3) AND tossing a head with it (1).
So the total probability of throwing a head next is 1/3*1/2 + 2/3 = 5/6
(I think, if you have F fair coins, U unfair coins and have already thrown H heads then the probability of throwing another head is:
(2^H*2U+F)/(2^H*2U+2F)