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# Probability without equally likely events

Up until now, we've looked at probabilities surrounding only equally likely events. What about probabilities when we don't have equally likely events? Say, we have unfair coins? Created by Sal Khan.

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• How would you calculate, given 13 flips of a fair coin, the probability of getting a palindromic flip?

A palindrome reads the same both ways, forwards and backwards. Examples of 13-flip palindromes are:

TTTTTTTTTTTTT
TTTHHHTHHHTTT
HHHHHHHHHHHHH
HHHTTTHTTTHHH

...

Is there an easier way to do this than calculating the probabilities of all 128 possibilities? I know you can write out and calculate for smaller numbers of flips, but what about much larger ones?

• if you took the possibilities for each spot it would be
2*2*2*2*2*2*2*1*1*1*1*1*1
because the first 6 can be either heads or tails and the back six must copy the first six to stay a palindrome. you then know the middle one can be either. so it would be
2*2*2*2*2*2*2*1*1*1*1*1*1/2*2*2*2*2*2*2*2*2*2*2*2*2 = 2^7/2^13 = 1/64
• Assuming every time you get a job interview, your chance of getting an offer is 25%, how many interviews must you get before you can be certain of getting one job offer?
• You can never be certain. Since we can never achieve total certainty, we instead arbitrarily assign a 'degree of confidence' where we say the chance that we're wrong is small enough that we don't care about it.

An example for a commonly chosen degree of confidence (5%):

Change of an offer = 1/4
Chance of no offer = 3/4
Chance of no offer after n interviews = (3/4) to the nth power.
As n -> infinity, the chance of no offer tends toward 0, but never reaches it, so we can see that we never reach total certainty of having had an offer, because we can't go to infinity interviews.

If we'd been to 11 interviews, we'd have a (3/4)^11 = 0.04.22 = 4.22% chance of not having an offer. We've met our chosen standard, so we would say that we were certain of getting at least job offer after 11 interviews with a 5% degree of confidence.

Hope this helps.
• At , he says that the probability of getting heads is 60%. How does he get the 60%? Or is it just an example?
• It's just an example of an unfair coin. He could have easily made it 51% or 70% or 12% - anything except 50%, which is a fair coin.
• The Gambler's fallacy states that in a large number of throws, the probability for each event remains the same, but a gambler will tend to believe that after let's say 3 heads, the probability of getting a tail increases (which is wrong). On the other hand, if we calculate the probability to get the same thing many times in a row, this probability 1/(2^n). Therefore the probability for the other event to happen increases. An argument would be this happens because the Gambler's fallacy assumes the events are correlated when they are not. But calculating the probability for the tenth throw in a row assumes the same thing, correlated events. How does one get past this paradox?
• That's a very interesting question. I'll try explaining as best as I could. Let's assume that we've tossed a fair coin five times already and that all were heads. Now, the paradox for the sixth try is between
a) that we tend to believe that it is more likely to get a tails now as we've got a series of heads already and because the probability of getting an all heads in a row is always less
and
b) the probability for each event remains the same (Gambler's fallacy)

Now, to get past this paradox, instead of thinking that the probability of getting all heads 6 times is very less(1/64 to be exact) and it's likely to be a tail the sixth time, we must think that we've already gotten 5 heads in a row, the probability of which is 1/32 and irrespective of the sixth try being a tails or a heads, the probability will be 1/64. (ie), the most unlikely event component of the very small probability has already occured.

"But calculating the probability for the tenth throw in a row assumes the same thing, correlated events"

This assumption is wrong. The probability for the tenth throw in a row does NOT assume correlated events. If you see the sample space, the probability of any combination of tails and heads for ten times is 1/512. (ie) P(HHHHHHHHHH) as well as P(HHHHHHHHHT) or even P(HTHTHHTTHT) is the same 1/512. Hope this answers your question.
• I have been looking for a good mathematical explanation as to why these probabilities are multiplied. The two answers are always; "If you add, the answer eventually is greater than 1", or the other answer is, "because that's how you do it". Now, there are all sorts of formulas and functions in probability class, and some are very complex and well explained here, but the simple multiplication of probabilities is opaque. Why? It must be harder than it looks.
• What would the probability of a coin landing on it's side be? Let's say the coin is a quarter, with a thickness of 1.75mm and a diameter of 24.3mm. Would you calculate surface area or would this involve some physics? The stickiness of the flat surface it lands on would also affect it landing on it's edge. Thanks in advance!
• Mathematically speaking, a coin is nothing but a cylinder. The surface area of a cylinder - excluding the top and bottom - is the circumference of the top circle (or bottom, they're equal of course) times the height of the cylinder.
So, for your coin, the area of the side of the coin is pi*24.3 mm * 1.75 mm = 133.6 mm^2. The area of its top and bottom is pi*R^2, with R = radius = half the diameter: pi*12.15^2 = 463.8 mm^2.
To calculate the actual probability of the coin landing on this side would take some fairly complicated physics though. A naive approximation would be this:
The coin has a top and bottom, each of 463.8 mm^2, and a side area of 133.6 mm^2. The chance of landing on the side area is 133.6 / (2*463.8+133.6) = 0.1259, or 12.59%. Of course the real probability is much less, since this completely disregards things like equilibrium, kinetic energy, and all that fun stuff.
• At around , I got a little bit confused. All probabilities should add up to 1, right? Is the reason that one specific example (such as the one at ) doesn't add up to 1 (or equal one on its own) because there is more than one combination in the sample space?
• Here 3 coins are being flipped. Since they are unfair the calculation is slightly complicated.
P(H) = 60% = 0.6 and P(T) = 40% = 0.4
Possibilities are
HHH P(HHH) = 0.6 x 0.6 x 0.6 = 0.216
TTT P(TTT) = (0.4)^3 = 0.064
HHT, HTH, THH P(2 heads and a tail) = 3 x (0.6)^2 x (0.4) = 0.432
TTH, THT, HTT P(2 tails and a head) = 3 x (0.4)^2 x (0.6) = 0.288
Add all the probabilities = 0.216 + 0.064 + 0.432 + 0.288 = 1
We have to know which probabilities when added = 1
Here we are flipping 3 coins or the same coin 3 times so the events and the sample space is different.
• between -19, sal talked about flipping the coin many times as a trial. but, even if he did so and got more heads in the trial,what is the guarantee he would get more heads in the real event. i mean, a coin doesn;t have any memory of coming heads many times in a trial
• Supernova: There is a statistical concept known as "The Law of Large Numbers". In layman's terms, essentially that in this case if you were to flip this coin 1,000,000 times and it came up heads 60% of the time, you could be VERY confident that this coin was biased towards heads and that the probability of flipping a heads is 60%.

Think of a baseball player at the beginning of a season. Let's say he gets 10 hits in his first 20 at-bats. Would you say he is a .500 hitter? No way -- the sample size is way too small. When the number of at-bats starts getting large, you can then make the type of determination of what type of average this hitter truly is.

It's the same with the coin. The coin may be biased where it will fall on heads 60% of the time. But if you flip it 10 times, it could reasonably fall on tails 7 times. 10 times is not enough. How much is enough? Well, the more the better, but you can usually say that 1000 or more is enough to give you a true picture of probability.