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This technique is called maximum likelihood estimator and is used to estimate parameters of a given probability distribution!

That's an awesome take on Bernoulli's trials! Never thought of it as a function of p, always as a function of k instead. Awesome video as usual Sir! I love how simplistic the result is in the end, no nCr's etc. and just overall awesome.

This problem seems a truism. In order to get heads 7/10 times, we would better take a coin that has a probability of 7/10 of getting heads. Unbelievable. But the calculations needed to confirm this obvious conclusion are indeed interesting. ;-)

I didn't understand it but I feel there is something furiously interesting about it. Thanks.

When I see a simple (intuitive) final result like this, I wonder if there isn’t a (provable) more general way of looking at the problem. If there is such a way, it could probably be used to solve much more complicated versions of the problem. [Maybe looking at the continuum version, with an arbitrary probability distribution would help.]

This would make a great calc 1 problem

but what are crooked coins? which denominations? and i would imagine there are variances of crookedness. how would you measure that to get to this point? is there a table of this data?

A simpler approach would be the using odds, which doesn't require calculus at all. You want p/(1-p) = k/(N-k)

3:47 Chen Lu! I've missed her so....

An interesting question is the following: someone gives you a coin and says: "This is a false coin, having a certain probability p for throwing heads on each throw. You can do 10 throws and then give me your best estimate (by which I mean the expectation value) of the p of the coin. The coin was taken randomly from my very large collection of coins, which is uniformly distributed, so a priori every p between 0 and 1 is equally likely." You then toss it 10 times, and get 7 heads and 3 tails. Now, what is the expected value of p (denoted )? Hint: it is not exactly 7/10. (Is it possible to throw 0 heads with a coin with p>0? Further hints below.) Let = ʃ p• P(7 H, 3 T | p) dp / ʃ P(7 H, 3 T | p) dp where integration bounds are from 0 to 1 and P(n-k H, k T) = (n choose k) pᵏ(1-p)ⁿ⁻ᵏ. Show by repeated integration by parts that ₁ ʃ xᵏ(1-x)ⁿ⁻ᵏ dx = 1/ (n choose k) ⁰ Use this formula in both numerator and denominator of the expression for to get = 1/(n+1 choose k+1) /(1/(n choose k)) = (k+1)/(n+1) So the answer is 8/11.

This appears to me as a way of deducting the widely known theoretical probability formula p=k/n. So, more like a math lesson than a probability lesson, right?

Great problem and such a simple solution.

very interesting indeed!!!

Here’s another fairly easy one I came up with that ChatGPT o1-preview can’t solve (correctly))) There are N identical black balls in a bag. I randomly take one ball out of the bag. If it is a black ball, I throw it away and put a white one back into the bag instead. If it is a white ball, I simply throw it away and do not put anything back into the bag. The probability of getting any ball is the same. Questions: 1. 1. How many times will I need to reach into the bag in order to empty it? 2. What is the expected maximum number of white balls in the bag in the limit as N goes to infinity ?

really cool!

Nice little video, you should show the graph you mentioned at the end.

How do you do clean the board by finger snapping? I've practised a lot, but to no avail.

You're great

:o

the probability of getting head is 1/2

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