7
$\begingroup$

We have 3 coins, of which 2 are fair and the 3rd is biased (gives heads with probability 4/7). Two friends throw, in turn, all coins together, one time each, and the winner is the one who gets more heads. What is the probability of a draw?

My first attempt: In order to have a draw, both flips must give the same output: Either

TTT or HTT or THT or TTH or THH or HTH or HHT or HHH, where the first two letters correspond to the fair coins and the 3rd letter to the biased.

So we have 8 possible outputs for EACH draw and we calculate the individual probability for each of them: 1st: 1/2*1/2*3/7=3/28 2nd, 3rd, 7th: same All the rest: 4/28 = 1/7. All of them must add to 28/28, which indeed is OK.

But then, in order to have a draw in both flips, we must have:

TTT with TTT (zero heads) or

HTT or THT or TTH with either of HTT or THT or TTH

or THH or HTH or HHT with either of THH or HTH or HHT

or HHH with HHH.

Probability for first is 3/28 * 3/28

Same also for the last one, 4/28 * 4/28

I am a bit confused as regards the other two cases and also how to calculate the total probability for a draw.

(Also, I am not sure for what I have done so far!)

I am not at all familiar with probabilities and combinatorics :(

$\endgroup$
  • $\begingroup$ You did very well! Take the case: "HTT or THT or TTH with either of HTT or THT or TTH". What is the probability that you throw "HTT or THT or TTH"? Since the outcomes are non-overlapping, you can just add up the probabilities: $\frac3{28} + \frac3{28} + \frac4{28} = \frac{10}{28} = \frac5{14}$. What is the probability that your friend throws "HTT or THT or TTH"? It's the same, obviously. Now what's the probability that you both throw "HTT or THT or TTH"? Since these are independent events, it is the product: $\frac{25}{196}$. Can you continue with the other open case? $\endgroup$ – Ingix Jun 18 '18 at 7:51
2
$\begingroup$

For an individual person the probabilities are:

  • zero heads $\frac3{28}$
  • one head $\frac3{28}+\frac3{28}+\frac4{28}=\frac{10}{28}$
  • two heads $\frac4{28}+\frac4{28}+\frac3{28}=\frac{11}{28}$
  • three heads $\frac4{28}$

making the probability of a draw

  • $\left(\frac{3}{28}\right)^2+\left(\frac{10}{28}\right)^2+\left(\frac{11}{28}\right)^2+\left(\frac{4}{28}\right)^2 = \frac{246}{784} \approx 0.3138$

If all three coins had been fair, the answer would have been $0.3125$ so not much different

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.