A probability theory problem - page 6

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Prival:
komposter:

By the way, a 100% probability is achieved on a company of only 28 people.

not 32 ?

With 28 it is already a little more than 100%.

 
I guess I don't really understand the condition, the same birthday, i.e. born on the same day. Let's say the 1st. There is always a variant of 31 people with birthdays from the 1st to the 31st, but 32 people exclude this combination. I find it hard to imagine a probability of more than 1.
 

Of course probability > 1 does not exist in nature, it is figuratively expressed when 1 is reached by a margin.

In order to get a probability of at least 1, you need to have at least 365 different pairs.

It takes 28 people to make these 365 pairs: 28 people make 378 pairs.

Recall combinatorics - the number of combinations of 28 by 2.

 
I understand combinatorics, I know the formula by heart. I used to be into sports bingo :-). It's just if the variant is 28 people, no combination will fall under the condition of being born on the same day, if their birthday is from the 1st to the 28th.
 

Don't go all Taleb on me, Taleb has studied well, ouch! He's got it right.


First, consider the probability that none of the 23 DRs will coincide with any other. Put the first one in one of the 365 cells, and then try to put the second one. What is the probability that his DR is different from the first one? 364/365. OK, put the third one in. The probability that his DR is different from the other two is 363/365. And so on, the last one will have 343/365. As a result, we get the probability that all 23 will have different DRs:


p = 364*363*...*343 / 365^22.


This stuff can be calculated with higher mathematics, or you can just prolagarithm it and calculate it in XL in a minute:


ln(p) = ln(364)+ln(363)+...+ln(343) - 22 * ln(365)


The result is -0.70785. Potentiate it and you get 0.492703. So, the probability that at least two people will match is equal to 1 - p = 0.507297.


P.S. Well, probability 1 and higher :) is reached only at 367 people.

 
Mathemat:

Don't get tough on Taleb here, Taleb studied well, oy malaica! He's got it right.

That's what I've been waiting for. Thank you, Alexei.


ps: Privat, DR refers to a specific day of the year, not the number of the month, i.e. 1 in 365.

 
SergNF писал (а)

(The combination of "8 identical candles in a row on EUR/USD and simultaneously on GBP/USD." True, not identical, as the number will be almost zero.

A maximum of 10 times (i.e. 0.08%) we encountered 8 bar combinations "on EUR/USD and simultaneously on GBP/USD". Moreover, it was

EURUSD=01001001

GBPUSD=01001001

Almost according to the request ("8 identical candlesticks in a row on EUR/USD and simultaneously on GBP/USD."), but only 10 times a year. i.e. it's not a question of any system/repeatability etc.

Actually, why did I start this answer - I personally - a trader - am not interested in "at the same time ".

What an amazing result! Thank you for your work.

I, too, am not interested in "at the same time". There must not be a system/repeatability. And although I theoretically assumed such a result, practical confirmation is always beneficial.

Although I do not really understand combinations and what does not equal mean? (....Pruth not equal, as the number will be practically zero.)

Did I understand correctly that 8 consecutive identical candlesticks on both pairs did not occur at all on the studied interval (even on M30)?

 
Mathemat:

Don't go all Taleb on me, Taleb has studied well, ouch! He got it right.

Indeed. We're the underachievers :(

Thank you, Alexey!

 

Here, I came across it, liked it:

Imagine that you are a participant in a game in which you are in front of three doors. The host, who is known to be honest, has placed a car behind one of the doors and a goat behind the other two doors. You have no information about what is behind which door. The presenter tells you: "First you must choose one of the doors. Then I will open one of the remaining doors, behind which there is a goat. Then I will suggest that you change your initial choice and choose the remaining closed door instead of the one you chose first. You may follow my advice and choose another door, or confirm your original choice. I will then open the door you chose and you will win whatever is behind that door."

You choose door number 3. The presenter opens door number 1 and reveals that there is a goat behind it. The presenter then invites you to choose door number 2. Will your chances of winning the car increase if you follow his advice?

You can easily find the answer on the internet, but don't post it here - let the pundits think first ;)

 
komposter:

Here, I stumbled across it, liked it:

Imagine you are a participant in a game in which you are in front of three doors. The host, who is known to be honest, has placed a car behind one of the doors and a goat behind the other two doors. You have no information about what is behind which door. The presenter tells you: "First you must choose one of the doors. After that I will open one of the remaining doors, behind which there is a goat. Then I will suggest that you change your initial choice and choose the remaining closed door instead of the one you chose first. You can follow my advice and choose another door, or confirm your original choice. I will then open the door you chose and you will win whatever is behind that door."

You choose door number 3. The presenter opens door number 1 and reveals that there is a goat behind it. The presenter then invites you to choose door number 2. Will your chances of winning the car increase if you follow his advice?

Of course you should choose door number 2. The odds double or triple. I don't remember the exact calculations....))))))))))))))

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