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Mathematical Puzzles
A jar contains 3 coins. 2 are heads and tails.

 
Thu Jan 17, 2013 11:30 pm  by tartle

A jar contains 3 coins. 2 are heads and tails. 1 is heads and heads.

You pick out at coin and toss it 3 times. You get three heads.

Question ; What is the probability of getting a head on the fourth toss ?
 
   
Sun Sep 08, 2013 9:18 pm  by L

3/8 using a probability tree.
Could not be bothered to try the binomial theorem as I have no tables.
 
   
Tue Dec 24, 2013 6:49 pm  by Twerp

Me thinks it be 2/3.
 
   
Thu Jul 14, 2016 6:35 am  by Dzallen

Probability of getting 4 heads from the start:
P(A)=2*(1/3)*(1/16)+1/3=9/24

Probability of getting 3 heads from the start:
P(B)=2*(1/3)*(1/8)+1/3=10/24

Probability of getting 4 heads given that 3 heads flipped already:
P(A given B)=P(A)/P(B)=9/10

The probability of another head is 9/10


Last edited by Dzallen on Mon Oct 31, 2016 9:08 am; edited 1 time in total
 
   
Thu Jul 14, 2016 2:30 pm  by bathfilms

No need for probability trees, or even looking at the history of the first 3 throws!

The question only asks about the fourth toss.

There is
1/3 probability of coin A (with p(heads) = 1/2) being used
OR
1/3 probability of coin B (with p(heads) = 1/2) being used
OR
1/3 probability of coin C (with p(heads) = 1) being used

P(A OR B OR C) = P(A) + P(B) + P(C)
= (1/3 * 1/2) + (1/3 * 1/2) + (1/3 * 1)
= 2/3
 
   
Thu Jul 14, 2016 2:30 pm  by bathfilms

No need for probability trees, or even looking at the history of the first 3 throws!

The question only asks about the fourth toss.

There is
1/3 probability of coin A (with p(heads) = 1/2) being used
OR
1/3 probability of coin B (with p(heads) = 1/2) being used
OR
1/3 probability of coin C (with p(heads) = 1) being used

P(A OR B OR C) = P(A) + P(B) + P(C)
= (1/3 * 1/2) + (1/3 * 1/2) + (1/3 * 1)
= 2/3
 
   
Fri Jul 15, 2016 1:29 am  by Dzallen

bathfilms, yes you do need to look at the first three throws, because the first 3 heads tell you that it is more likely you are holding a head and head coin rather then a head and tail coin. In other words, the probabilities of having the three coins are not 1/3 each.

For example, if you started with 1 head and head coin, but 2 tail and tail coins, and you throw three heads, the probability of a 4th head is clearly 1, but by your calculation it would be 1/3.

This is a question in conditional probability, the solution is as I posted above.
 
   
Fri Jul 15, 2016 8:25 am  by bathfilms

Dear Dzallen,

Yes, thank you, I see your point of view, which is of course correct.
So, if the 1st three tosses were heads, but you had your eyes closed while so-doing, that would change the probability. It's a bit like Monty Hall's dilemma.
 
   
Mon Oct 31, 2016 2:37 am  by Sagitaur

Surely the answer is 2/3. Label the coins A and B (each has a head & tail) and C for the two headed coin.
The probability of selecting coin A is 1/3. The probability of throwing a head regardless of previous throws is 1/2. The probabilit of a Head if Coin A is 1/3 multiplied by 1/2 = 1/6. The same applies for Coin B.

For coin C the probability is 1/3 to select it and then 1 to throw a head = 1/3. So the probability of a head is 1/6 or 1/6 or 1/3. For 'or' results you add the probabilities ie 1/6+1/6+1/3 =2/3
 
   
Thu Apr 26, 2018 8:48 pm  by bedead

2/3
 
   
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