Puzzle/logic question

[KingGheedora]

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?

 

[KLin]

42

 

[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

 

[Epic Fail]

21/40

 

[Matthiasa]

Random guess places it at around 52.5% but I'm not really sure... I hate stats...

 

[chuckywang]

I'm gonna guess 51/70.

EDIT for rationale:

Write down all possibilities of the first five flips:
The 19 regular coins will have 32 possibilities (2 with first four being heads).
For the sake of consistency, 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.

 

[rocadelpunk]

for people that are just 'guessing' you should at least give an explanation of how you arrived at that answer.

and where's dr. pizza...only 2:30am on eastcoast

 

[gdextreme]

Shouldnt the nmber of coins in the jar be given?

Oh sorry I didn't read the title description completely

 

[Matthiasa]

Fine... So theres 19 coins at 50% chance of it happening and 1 coin with 100% so guess places it at [.5*19+1*1]/20.

That is not taking into account however that the probability of your randomly selected coin having been the trick coin increases the longer the streak goes on due to diminishing odds of that combination.
At some point it would be a safe assumption that you had picked the trick coin in which case you would know the odds off another head coming up approaches 100%
/worded badly

 

[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.

 

[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

 

[rocadelpunk]

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.

just to point out...
ignoring a trick coin - say there are 20 regular coins

so it doesn't really matter which coin you pick.

Chances of getting heads on first flip (1/2)
chances of getting heads on first flip and second flip (1/2)*(1/2)
chances of getting heads on first flip and second flip and third (1/2)*(1/2)*(1/2)

so chances of getting five heads in a row is roughly 3%

I'm pretty sure adding just one trick coin won't raise it odds of you getting five heads in a row after selecting a coin out of the 20 will get you anywhere near 70+%

 

[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

 

[gdextreme]

Lets calculate the probability of 4 consecutive heads

(1/20)(1) + (19/20)(1/{(2)^4}) = 7/64

Now we calculte probabilty of a trick coin (event A) given that 4 heads in a row (event B) has occured

We need P(A/B) = P(B/A) {P(A)}/P(B) = {(1) (1/20)} / (7/64) = 16/35


Now to calculate probabilty of head in 5th try

(16/35) (1) + (1-(16/35)) 1/2 = 51/70

The answer is 51/70.

Whew I hope there are no calculation mistakes!

 

[gdextreme]

Oh no! already solved by somebody .

 

[gdextreme]

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

Yeah that has been assumed to be true in his calculation as well as mine. Checkout my somewhat detailed calculations.

 

[Xylitol]

draw an extensive tree diagram of every possible situation to get the 100%, easy to understand, painful way

 

[GodlessAstronomer]

The first four throws are a red herring. They don't factor into the odds of the final throw. The fact that you got 4 heads in a row doesn't give any extra information, unless I've got this all screwed up in my head.

 

[TuxDave]

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

You need to use conditional probability. What's the probability of having a fair coin given that he flipped 4 heads in a row. It actually matters.

 

[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.

 

[gdextreme]

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.

In fact they do - thats why this is a puzzle! If there was no trick coin then it would not have not made a difference. With all regular coins the answer would be 1/2.

 

[CatchPhrase]

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

Got it^

 

[her209]

Probability of picking a normal coin * probability of flipping heads:
19/20 * 1/2

Probability of picking the abnormal coin * probability of flipping heads:
1/20 * 1

19/40 + 2/40 = 21/40 = 52.5%

 

[KingGheedora]

Ugh. I'm seeing multiple answers. Which of these is correct?

 

[her209]

Originally posted by: KingGheedora
Ugh. I'm seeing multiple answers. Which of these is correct?

You mean you don't know?