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## 3 Coins 10

Not sure what the answer is, ideas welcome!

You have 3 coins, one always comes up heads, the second always comes up tails, and the third is a fair coin. You select a coin at random. After selecting a coin, you flip it twice and get heads both times. What is the probability that your next flip is heads?

Follow up: Suppose you flipped twice and got the same outcome both times (either HH or TT), then what's the probability that your next two flips match the first two (e.g. if you saw HH, your next two are HH)

I'm honestly unsure...

surya

p(3h/2h) = p(3h,2h)/p(2h) p(3h,2h) = p(3h) = (1/3)1 + (1/3)(1/2)(1/2)(1/2) p(2h)= (1/3)1 + (1/3)(1/2)(1/2)

where the 1/3 is prob of each coin and 1/2 is prob of head and tail. answer is 9/10 according to me.

alvinfromva

This would be more helpful with an answer! spoilerI think it's .75.

Ravi

Jay Elliott

debashish ghatak

I agree with ravi , the answer is 9/10.

Tarun245

I think it is 25%

Gene

I think the answer is 75%. It doesn't matter what the odds were to get you to the final flip (this is not like the famous Monty Hall problem that requires you to make a choice at the beginning). We know the coin cannot be the tails coin. If it is the heads coin (odds of which being 50%), then of course the third flip will be heads. If it is the honest coin (also 50%), then 50% of the time it will come up heads, which means a composite probability of 25%. Thus, 3 out of 4 times it should be heads.

Asif Elahi

i could not understand

Guilherme

SPOILER DON'T READ THIS BEFORE TRYING TO SOLVE THE PUZZLE SPOILER ************** * * * * * * * * * *

First case. Since you are going to get heads twice it necessarily means you did not choose the Tails coin. This gives you two scenarios:

1 - 1/2 chance of choosing the fair coin times 1/2 chance of getting heads again. 2 - 1/2 chance of choosing the Heads coin times 1 chance of getting heads again,

(1/2 * 1/2) + (1/2 * 1) = 3/4 or 75%

For the second case. Three scenarios:

1 - 1/3 chance of choosing the Heads coin times 1 chance of getting Heads twice. 2 - 1/3 chance of choosing the Tails coin times 1 chance of getting Tails twice. 3 - 1/3 chance of choosing the fair coin, but since you've already got the same result twice, you only have the probability of getting the same outcome twice again which would be 1/2 * 1/2 = 1/4.

1/3 * (1 + 1 + 1/4) = 3/4 or 75%

Did this pretty quick, not sure if I'm right.

Ravi

The key is to infer the probability that each coin was selected based on the (heads,heads) outcome for the first two flips.

Label the three coins:

• C1 - TT
• C2 - HT
• C3 - HH

Data = HH. In general, as per Bayes' Rule, P(Ci | Data) is proportional to P(Data | Ci)*P(Ci), so:

• P(C1 | Data) proportional to 0*(1/3) -----> 0 (normalized)
• P(C2 | Data) proportional to (1/4)*(1/3) --> 1/5
• P(C3 | Data) proportional to (1/4)*(1/3) --> 4/5

Now, P(H | Data) = sum over Ci ( P(H | Ci) * P(Ci | Data ) = 0 * 0 + (1/2) * (1/5) + 1 * (4/5) = 9/10.

In summary, probability of heads on the third flip (given the first two flips being HH) is 9/10.

Jay Elliott

Nice! Solid explanation.

Bablu Dubey

can you explain P(C2) and P(C3) how it came 1/5 and 4/5 respectively...???

Kara Luke

That's Right man

BLRGH

I computed it by computing the probability that the coin was the fair coin given that HH appeared, getting the probability that it was the doublehead coin from that, and then computing the probability of a head in those disjoint cases.

Just used Bayes's theorem and some arithmetic. The probability that you have a fair coin given the HH appearance is only 1/5; hence 4/5 it's the unfair heads coin (it's never the tails one obviously). The first comes from Bayes's theorem giving that, prob(fair coin | HH) is equal to P(HH|fair) P(fair) / P(HH), which is (1/4)(1/3)/(1/3 1 + 1/3 1/4) when keeping in mind the background info. Then (1/5)(1/2) + (4/5)(1) = 9/10 = 0.90

Confirmed it with a simple python program. http://codepad.org/yl4Be298

city_slick

Good call applying Bayes' Theorem that way. My first thought was to use the HH information just to eliminate the all-tails coin. I think your logic is valid.

Sean Moran

If that's all true, which makes sense, so I'm not going to dispute it, then for the 2nd case it's the same for heads or tails as if you got 2 heads it's not the tails coin and if you got 2 tails it's not the heads coin so it's: Still 1/5 for the fair coin or 4/5 for the all-heads or all-tails coin then (1/5)(1/2)(1/2) + (4/5)(1)(1) = 1/20 + 4/5 = 1/20 + 16/20 = 17/20

Aakash

That sounds right to me.

Vishal barot

I think it is 0.25.

Suppose we get HH on first 2 flips then the nixt two flips can be HH,HT,TT,TH. therefore probability of HH =1/4 =0.25

Wolph42

its 25%.

Vishal barot explains it most intuitively and Guilherme most correctly:

The correct mathematical method is to calculate the probability (P) that its NOT heads (NH) on the third fli = Pnh. NH (=Tails) can only happen if the fair coin was used, which is 50% probability . (We've already deduced that i cannot be the Tails coin). The probability to flip a Tails on a fair coin is once more 50%. Hence Pnh = 1/2 * 1/2 = 1/4. Therefore the probability of throwing heads: Ph = 1 - Pnh = 1-1/4 = 3/4 or 75%.

Виктория Яцкова

75% 0.5 probability that the coin is only-heads coin, and in this case we have 1 probability that it would heads next 0.5 probability that the coin is fair and in this case we have 0.5 probability that it would heads next. By the rule of condition probability, we have as a result: 0.5 * 1 + 0.5 * 0.5 = 0.75

lee123

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Submitted by
city_slick
over 7 years ago
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Difficulty 3.0 ?

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