- Jun 24, 2006
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One of the coins is a trick coin that has both sides heads. You pick a random coin from the jar and flip it 4 times, and each time it comes up heads. What are the chances that the next flip will come up heads?
Originally posted by: KoolAidKid
Using Bayes theorem, the probability that you are holding the trick coin given 4 consecutive heads flips is about 0.457. Therefore, the probability of a heads outcome on the next flip is 0.457(1) + (1 - 0.457)*0.5 = 0.7285.
Originally posted by: chuckywang
I'm gonna guess 51/70.
EDIT for rationale:
Write down all possibilities of the first four flips:
The 19 regular coins will have 32 possibilities (2 with first four being heads).
For the sake of consistency, if the flip the 1 trick coin 32 times, you are obviously going to get all heads 32 times.
So therefore, in total you have 32 (from the one trick coin) + 2*19 (from the 19 regular coins) where the first four flips are heads. Of those 70, 32 + 19 = 51 of them have the 5th flip being heads as well.
Hence 51/70.
Originally posted by: TuxDave
Originally posted by: KoolAidKid
Using Bayes theorem, the probability that you are holding the trick coin given 4 consecutive heads flips is about 0.457. Therefore, the probability of a heads outcome on the next flip is 0.457(1) + (1 - 0.457)*0.5 = 0.7285.
This
Originally posted by: rocadelpunk
Originally posted by: TuxDave
Originally posted by: KoolAidKid
Using Bayes theorem, the probability that you are holding the trick coin given 4 consecutive heads flips is about 0.457. Therefore, the probability of a heads outcome on the next flip is 0.457(1) + (1 - 0.457)*0.5 = 0.7285.
This
unless i'm reading the question wrong - i'm pretty sure it states you select one coin and then you flip that coin 4 times. Not you select a new coin each time - if that were the case wouldn't you have to know with without replacement? Although with bayes' theorem could you just massage your condition or something? I dunno - been forever since I had stats : (
plz correct if wrong
Originally posted by: rocadelpunk
Originally posted by: TuxDave
Originally posted by: KoolAidKid
Using Bayes theorem, the probability that you are holding the trick coin given 4 consecutive heads flips is about 0.457. Therefore, the probability of a heads outcome on the next flip is 0.457(1) + (1 - 0.457)*0.5 = 0.7285.
This
unless i'm reading the question wrong - i'm pretty sure it states you select one coin and then you flip that coin 4 times. Not you select a new coin each time - if that were the case wouldn't you have to know with without replacement? Although with bayes' theorem could you just massage your condition or something? I dunno - been forever since I had stats : (
plz correct if wrong
Originally posted by: Praxis1452
I'm gonna guess 21/40. The previous four flips do not affect the probability that the next flip will be heads or tails. Heads and tails are determined by the coin, and if we are unsure of the specific coin, then we can view the jar as a whole. Well that's my best guess.
Originally posted by: rocadelpunk
can you break it up as P(A or B) where A is the probability of picking a normal coin (19/20) * first flip (1/2) * Second flip (1/2) * third flip (1/2) *fourth flip (1/2)*fifth flip(1/2) and B is you pick the two headed coin (1/20) * first flip (1)*second flip(1)*third flip(1) * fourth flip (1)*fifth flip(1)
and then since events are mutally exclusive P(A and B) = 0 since you can't pick a normal coin and a two headed coin
P(A or B) = P(a)+P(b) so about 7.97%
keep your fazers on stun
You mean you don't know?Originally posted by: KingGheedora
Ugh. I'm seeing multiple answers. Which of these is correct?