How many times must you roll a die to get a six?

[@tonatiuhm.wiederhold1692]

The second solution was very cool. I'm not much into probability, but the idea of finding a subtree isomorphic to the total infinite tree seems to be present in many other areas of math! Nice video.

[@VisuallyExplained]

Glad you liked it!

[@mauisstepsis5524]

Same idea comes from calculating stopping time in DTMC.

[@luisguillermo6216]

This video is really cool, great job! For the follow up questions I got 1) 3 (estimating when total probability hits 0.5) 2) 42 (by recursion) 3) 3 (by recursion, analogous to a 3 sided dice)

[@VisuallyExplained]

1) is close, but not quite right. 2) is correct, nice! 3) is a very tricky one. When you condition by the event E = { all throws up to the first 6 are even}, it is not the same as throwing a 3-sided dice. One way to think about it is that a 6 is much more likely to occur than a 2 or a 4 when you condition by E.

[@raphaelpetrovich]

@@VisuallyExplained @Luis Guillermo ### 1) Median import random as rd import statistics as st lengths=[] for i in range (100000): a=0 p=[] while a != 6: a = rd.randint(1, 6) p.append(a) lengths.append(len(p)) st.median(lengths) ###Answer 4 ### 2) two consecutive 6s import random as rd import statistics as st lengths=[] for i in range (100000): a=0 p=[0,-1] while p[-1] != 6 or p[-2] != 6: a = rd.randint(1, 6) p.append(a) lengths.append(len(p)) st.mean(lengths)-2 ###Answer 42 ### 3) all throws up to the first 6 are even import random as rd import statistics as st lengths=[] for i in range (100000): a=0 p=[] while a != 6: a = rd.randint(1, 6) if a%2 == 0: p.append(a) else: break if len(p) != 0: lengths.append(len(p)) st.mean(lengths) ###Answer 1.5

[@peterhindes56]

I still got 36 with recursion and with the two rolls at once analogy :( How did you do it?

[@MathAndPhysics]

Nice video! I must admit I also used the brute force method using an infinite series, but I really enjoyed your more elegant solutions.

[@tejkiranv4056]

@VisuallyExplained Is the answer 2nd follow-up question (the median value) 2 throws? For example, take 100 throws, out of these, 16.6 throws would yield 6 in first throw. Around 41.6 throws would yield 6 in their second attempts. And since we want the 50th throw (or rather avg of 50th and 51st), it would be 2.

[@bharadwajeddanapudi8507]

Quick question: why is it for the 2nd solution that the formula has two E(T)s. Surely the 2nd E(T) would be one less than the initial E(T) because you have one less throw to deal with in comparison to the very beginning of the event.

[@cernejr]

Nice video. However the music is giving me a headache, I had to watch on mute with CC.

[@VisuallyExplained]

Sorry about that, I will keep your comment in mind for next videos.

[@darklordvadermort]

during first solution, why do you take the derivative of the geometric series formula or what's the justification for that? it needed to look more similar to what you had with the k out in front? is there a reason why taking the derivative preserves the identity of the formula?

[@VisuallyExplained]

Taking the derivative of a formula will of course change that formula, but if you have two formulas f(x) and g(x) that are equal to each other (i.e., f(x) = g(x)), and you take the derivatives of both, then you get two new formulas that are again equal to each other (i.e., f'(x) = g'(x)). This is exactly what's happening at 3:48. But maybe what you are asking is why do we even take derivatives in the first place. Then yes, you are right, taking the derivatives of both sides make the LHS more similar to the expression we want to reach ultimately. Hope that answers your question!

[@jakubjan44]

Simply beautiful

[@BrianAmedee]

3rd solution was my favourite. I froth the law of large numbers. Did you make this using manim?

[@VisuallyExplained]

The law of large numbers is truly magical! The video was made with manim for the most part. I have also used Blender for the 3d animation at the beginning.

[@weisanpang7173]

Is the answer to follow up question#1 = 4 ?

[@samirelzein1095]

Keep going...

[@eastcoasterner]

How many times do you throw the dice until the probability of not getting a 6 is 0? For example, if I throw the dice 10 times, there's still a pretty good chance of not getting a 6. But if I throw the dice 100 times, there's a very small chance of not getting a 6. If I throw the dice 1000 times, it's extremely unlikely that I won't get a 6 at least once. Also can this be used for other percentages? For example, if I have a 30% chance of getting a certain outcome, then after 3000 attempts, is it possible to only get that outcome 24% of the time?

[@aripocki]

The probability can never reach 0! The chances of not rolling a 6 would get very small as you roll more times, but there's always a possibility

[@peterhindes56]

@@aripocki but be careful not to use the gamblers fallacy, the probability is of rolling a 6 if you decide to roll n times, not the probability that the next roll will be 6

[@youtub3808]

Awesome work

[@VisuallyExplained]

Thank you! Cheers!

[@depressedguy9467]

How to find median of distribution

[@hamzaich7034]

maaan you are amazing

[@VisuallyExplained]

Thanks man! :)

[@zeroheisenburg3480]

For the 2nd solution, why aren't we multiplying 1/6 by 1 instead of zero? Isn't it supposed to be (1 throw) * (probability of getting 6)?

[@VisuallyExplained]

Good question. That's because we have already counted the first throw, it's the first term (" 1 + ...") in the right hand side). Alternatively, if you prefer, you can write E[T] = 1/6*(1+0) + 5/6 * (1+E[T])

[@zeroheisenburg3480]

@@VisuallyExplained Thanks for the quick reply. This form is easier to interpret for me. Since watching this video, I've been looking up why we can just stick (1 + E[T]) into the expectation equation. I know it make senses intuitively, but is it based on any formal properties? I've been googling terms like "Recursive expectation"/"recursive markov chain", this pattern of solving problems do appear, but they tend to not explain why it works, as it seems trivial. I do see some mention of "Law of total expectation" and "Law of iterated expectation", but I wasn't able to see any direct correlation to why we can just put (1+E[T]) there. Do you have any suggestions which topic should I search for to find out more? Thanks.

[@VisuallyExplained]

@@zeroheisenburg3480 Sure. Look up "Markov Chains" and "Hitting time". I think this will answer your questions.

[@theProf-xc5pe]

hmm close but no cigar

[@WellActualllyyy]

i don’t get where the 6 came from in the last solution

[@VisuallyExplained]

Start with the equation 100 = 1/6 * 100 * E[T], simplify both side by 100 to get 1 = 1/6 E[T], then multiply both side by 6 to get 6 = E[T].