Coin toss - part 1

Probability Level 3

I have two coins, one of them is fair and has a 50% probability of being either heads or tails, the other is biased having a 90% probability of being heads and a 10% probability of being tails.

You take one of the coins at random and toss it twice. If in the first toss you get tails, the probability of getting tails in the second toss increases, decreases, or stays the same?

Part 2 here

Increases Decreases Stays the same

João Areias by João Areias , 24, Brazil

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1 solution

João Areias
Jan 28, 2021

The coin can be the biased coin or the fair coin and each of them will give you heads or tails with their respective probability, so we can write the probability of a coin toss coming out as tails by writing:

P ( T ) = P ( C = F ) P ( T ∣ C = F ) + P ( C = B ) P ( T ∣ C = B ) P(T) = P(C=F)P(T|C=F) + P(C=B)P(T|C=B)

Without the information of which coin you have, there is a 50% chance that you have any of those coins, so this comes down to:

P ( T ) = 0.5 ∗ 0.5 + 0.5 ∗ 0.1 = 0.3 P(T) = 0.5*0.5 + 0.5*0.1 = 0.3

If we know that the first toss came out tails, we can use the Bayesian rule for updating the probabilities, so:

P ( C = F ∣ T ) = P ( T ∣ C = F ) ∗ P ( C = F ) P ( T ) = P ( T ∣ C = F ) ∗ P ( C = F ) P ( C = F ) P ( T ∣ C = F ) + P ( C = B ) P ( T ∣ C = B ) = 0.5 ∗ 0.5 0.3 = 0.83 P(C=F|T) = \frac{P(T|C=F)*P(C=F)}{P(T)} = \frac{P(T|C=F)*P(C=F)}{P(C=F)P(T|C=F) + P(C=B)P(T|C=B)} = \frac{0.5*0.5}{0.3} = 0.83

P ( C = B ∣ T ) = P ( T ∣ C = B ) ∗ P ( C = B ) P ( T ) = P ( T ∣ C = B ) ∗ P ( C = B ) P ( C = F ) P ( T ∣ C = F ) + P ( C = B ) P ( T ∣ C = B ) = 0.5 ∗ 0.1 0.3 = 0.17 P(C=B|T) = \frac{P(T|C=B)*P(C=B)}{P(T)} = \frac{P(T|C=B)*P(C=B)}{P(C=F)P(T|C=F) + P(C=B)P(T|C=B)} = \frac{0.5*0.1}{0.3} = 0.17

This means that we are more confident that we have the fair coin instead of the biased coin, so now, the probability that the second toss will be tails is given by:

P ( T ) = 0.83 ∗ 0.5 + 0.17 ∗ 0.1 = 0.43 P(T) = 0.83*0.5 + 0.17*0.1 = 0.43

2 Helpful 3 Interesting 3 Brilliant 0 Confused

More simply, P ( T ) = 1 2 ( 1 2 + 1 10 ) = 3 10 P ( T T ) = 1 2 ( 1 4 + 1 100 ) = 13 100 \begin{aligned} P(T) &= \; \tfrac12\big(\tfrac12 + \tfrac{1}{10}\big) \; = \; \tfrac{3}{10} \\ P(TT) & = \; \tfrac12\big(\tfrac14 + \tfrac{1}{100}\big) \; = \; \tfrac{13}{100} \end{aligned} so that P ( T T ∣ T ) = P ( T T ) P ( T ) = 13 30 > P ( T ) P(TT|T) = \frac{P(TT)}{P(T)} = \frac{13}{30} > P(T)

Mark Hennings - 4 months, 2 weeks ago

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