0
$\begingroup$

So this is an analogy for a real world example but for simplicity. So if I were to flip a normal coin ten times I would expect heads 50% of the time or 5 head results. I could then compare this to the actual results after flipping.

Now imagine I have 10 different coins, the coins have varying odds of getting a head, it might only be 40% it might be higher at 60%. I would know the odds of each coin before any flips.

How would I calculate the equivalent 50% of heads expected this time when the odds are different each flip. The coin flips do not affect each others.

For arguments sake lets have the odds as 0.5, 0.2, 0.25, 0.3, 0.4, 0.35, 0.6, 0.7, 0.7, 0.8

$\endgroup$
  • $\begingroup$ I think this may help you: en.wikipedia.org/wiki/Binomial_distribution $\endgroup$ – Brian Cheung Aug 28 '15 at 14:31
  • $\begingroup$ What would you expect if one coin had heads on both sides, and one had tails on both sides ? $\endgroup$ – true blue anil Aug 28 '15 at 14:31
  • $\begingroup$ My initial thoughts on this were to take the mean of all the probabilites (0.48) in this case. This was a guess on my part though. I've tried to do some worked examples which seem to reinforce this but that might be because I have the logic backwards from the start. @ true blue anil, if the coin had both heads the odds would be 1.0. That could be a valid unfair coin. I will check the wiki article user3313320 Thank you. $\endgroup$ – Jon Smith Aug 28 '15 at 14:38
0
$\begingroup$

Hint: Have you learned about expected value in a probability or statistics course?

Let's make $H$ the random variable that takes on the value of the number of heads in 10 flips. Then $E(H)$ is the expected number of heads in 10 flips. You calculate this using the formula $$E(H) = \sum\limits_{i=0}^{10}P(H = i)\times i.$$

Verify that if you are flipping a fair coin, this formula gives you $E(H) = 5$, which is 50% of 10. In this case, $P(H = i)$ can be calculated using the binomial distribution formula: $E(H) = p \times N$, where $p = 0.5$, the probability of a success, and $N = 10$, the number of attempts.

Now use the same formula when the probabilities are different. Here, $P(h) = i$ is a bit uglier to calculate.

$\endgroup$
  • $\begingroup$ Wow, that looks great. I'm self taught in all things everything so will do my best to get my head around this. It seems like this is exactly what I need though. $\endgroup$ – Jon Smith Aug 28 '15 at 14:45
  • $\begingroup$ @JonSmith If you google "expected value" there will be tons of results. I'm sure you can find an introduction on the topic that will be much more in-depth than the short explanation I gave here. $\endgroup$ – camel_case Aug 28 '15 at 14:53
  • $\begingroup$ Awesome. I think I have got it, if I were to feed in the values I suggested the equation given would E(H)=0.48 which is 0.48% of 10. This is the same as the Mean. Am I correct? I'm assuming the Mean is always the same in my case as I am always checking for the same result. $\endgroup$ – Jon Smith Aug 28 '15 at 15:32
  • $\begingroup$ Whoops I made a mistake; let me fix my answer $\endgroup$ – camel_case Aug 28 '15 at 16:26
  • $\begingroup$ @JonSmith How did you calculate P(H = i)? For example, to get P(H = 3), you have to look at all possible sets of 3 coins, calculate the probability those are all heads, and the probability all the rest are tails. Kind of ugly, but definitely doable $\endgroup$ – camel_case Aug 28 '15 at 16:30

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.