I'm getting a bit dumb on the probabilities. - page 9

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Mathemat:

2 Dersu: But what is the overall balance, I don't understand shit. What do you mean by that?

Sorry, I meant: 1/6 the probability of a six in one shot.

And funnily enough, 0.16666666 multiply by 6 and you get the total balance, i.e. one.

But how do you get one from 0.517747 ?

 

Why would you want to get a unit out of it? It's not a problem here. This is not accounting, where you have to reconcile credit and debit.

Read our conversation with tara, all the logic is there.

 

I'm a careful person, so I'm asking.

Here's the thing (I don't know if you'll understand): I am neither a mathematician nor a programmer.

I'm a "maverick" and an accountant. Here and there a little bit, here a little bit.

Surprised, interested, memorized. I went on. Logical flowcharts.

And so the time goes by, I endure. Solution is getting saturated, but time will tell if it will help.

But that's all first lyric.

Regarding probabilities: Surprised, interested, but no block yet.

The probability of the event is 50 to 50. Even an encounter with a dinosaur in the street.

Even a nine hundred and ninety-ninth flip of a coin, if the previous ones were the same.

That's where it gets me. I don't get it at all. Maybe I'm just dumb.

Elliot's got a chance of turning a three into a five.

And no sevens.

The dinosaurs are extinct.

But the next throw is 50-50.

 
Mathemat:

This is your problem. As you can see, it didn't have what you just wrote, but was more like a "it will rain only on one day out of three" condition.

Now to the point: you did your calculations correctly in the first post.

If directly, the reasoning is as follows: count separately the probability of the events "rain on one day only", "rain on exactly two days", "rain three days out of three" and sum up.

C(3,1)*p^1*(1-p)^2 + C(3,2)*p^2*(1-p)^1 + C(3,3)*p^3*(1-p)^0 =

3*0.1*0.9^2 + 3*0.1^2*0.9^1 + 1*0.1^3*0.9^0 =

0.243 + 0.027 + 0.001 = 0.271.

But it is easier to do it the first way, because the sum of all probabilities is 1.


much easier:

if it rains on the first day, all is ok)) exit

else if it rains on the second day also ok ext

else if it rains on the third day also ok exit

else not ok

0.1 + 0.9*0.1 + 0.9*0.9*0.1=0.271

 

Dersu: Я такссать "бродяга" и бухгалтер. Там чуть, здеся чуть.

That's how I knew you were an accountant :)

You've been in this thread. At least someone there is trying to explain something on their fingers.

Of course, there's also a "balance" in tervers: the sum of the probabilities of all possible outcomes is always 1.

In this case, 1 - (5/6)^4 = 0.517747 is the probability of hitting at least one six when 4 dice are rolled simultaneously. To balance, it is necessary to calculate the probabilities of all other outcomes (here - "no sixes") and add them to this one. Then the total would also be 1.

The probability of the event "zero sixes" is exactly (5/6)^4, so the balance is trivial here.

 

Okay, you got it. Thank you.

I need to calculate the probabilities of all the other outcomes (here - "no sixes") and add them to this one.

Somehow the series reminds me of Renko. Everyone wants to know the height of the brick, but no one knows.

 
Avals:

is much simpler:

[...]

else not ok

0.1 + 0.9*0.1 + 0.9*0.9*0.1=0.271

And all this equals 1 - 0.9*0.9*0.9. Well, yes, right even in the general case, for any number of days, if you replace 0.1 with p.

So where to strain the brain more: with five arithmetic operations for you - or with three for me?

 

Cool topic: almost 27 hours of non-stop discussion was enough :)

 

2 Mathemat: wonderful proof right in the terminal unbeliever - bravo!

There's an interesting question about probabilities, I've been wondering how to substantiate it for a long time - can you help?

The bottom line - many poker novices, playing cards among themselves with a real pack of cards, get into an online room where up to 20 million people play simultaneously and begin to wonder why combinations fall out so often at the table, which in real life are very rare ... For example - in real life I fell flush straight once in 5 years of playing, and online 5 times in 2 years ... So my question is - may this increased probability be explained by the fact that online CRT deals hundreds of deals per second? Or I play at the table I need to count only the distribution of my table?

S.U. 1. 2 years online I played twice as many games than for 5 years, approximately ... 2. Let's assume that the CRT is perfect...

 
moby_dick: So the question is - can this increased probability be justified by the fact that the CRC deals hundreds of hands per second online? Or should I play at the table and only count my table's hands?
Nah. It's more likely because the speed of online play is pretty decent. And there's not a lot of players with that kind of experience who only play one table online. At least 2 at a time. Some even up to 8 or more.
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