# More Puzzles

 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 9/10Last 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 All times are GMT Page 1 of 1