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The big GambleJoe Advent Calendar 2023 starts today!
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The big GambleJoe Advent Calendar 2023 starts today!
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i only calculated something for muschmusch, I didn't complain in any way, sorry if it came across that way! I was already among the winners of this calendar....
i'm mathematically a little zero but I'll give you a 10 if you manage to calculate it
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The big GambleJoe Advent Calendar 2023 starts today!
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Was once so free...😉
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Can you actually find out how often you have been drawn in the calendar in recent years or is that too much effort? I'd be really interested to know how many times I've been drawn as I'm pretty sure there's never been a day here that I wasn't in the Pot
I'd be really interested in my odds if I'm honest haha
The monthly raffle is really nice and clear
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The big GambleJoe Advent Calendar 2023 starts today!
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gagapapamama
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Ditto unfortunately
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not bad Hanshanshans !!! did you google how to calculate it or did you know?
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Got a little help from chatgpt 😉
I already figured that the non-winner rate was low, but 4% is of course a great figure. Why can't it be so "difficult" not to win everywhere?
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Oh dear... no offense, but ChatGTP is not recommended for something like this.
You can also look for yourself. The result says 0.0004 and at the end it says it's 4%. But 4% is 0.04.
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In my opinion, this is correct. To calculate the percentage value here, the 0.0004 (decimal value) is multiplied by 100. This gives you 0.04, which is also 4 percent.
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Since it always depends on how many people qualify, we have to take an average value. It always deviates, that's clear, but there's no other way to calculate it.
Let's assume that an average of 670 people qualify and 40 win. To simplify the example, let's leave out the 2 days where "special qualifications" apply and the normal conditions also apply there. The result won't change much, but I just don't feel like doing an exact calculation. I hate the subject too much for that, sorry
How to calculate something like this is usually explained at school using a dice. What is the probability of rolling a 6 2 times in a row?
On the first roll, it's clear: a die has 6 sides, one side is shown, so 1:6 or 1/6 as a fraction. In the second roll, this probability also applies because the dice is the same, but since you have to fulfill a condition (roll a 6 twice in a row), you cannot add the two probabilities together. You have to multiply them together. 1/6 * 1/6 equals 1/36.
This is roughly how you can calculate the probability here, as a condition must also be fulfilled here (you don't have to be drawn 24 times). So with 40/ 670 you have a 6% chance of being drawn and 94% of not being drawn.
You just have to multiply this 94% btw 0.94 by 24 times. I did it and came up with something like 22% (maybe I calculated too much or too little).
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As I said, I hated statistics, but to the best of my knowledge (which may of course be wrong) the probability is a little higher. Maybe I have a logic error, idk and tbh idc xD
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Also did the math with "0.94 24 times with yourself". Also comes to approx. 22%. Approach and calculation makes sense.
However, my or Chatgpt's 4% would be closer to the "final result" from last year...
https://www.gamblejoe.com/forum/regeln-und-hinweise/news-und-hinweise/der-groszlige-gamblejoe-adventskalender-2022-startet-heute-314682/38/#p319389
Not proof, of course. I'm not a mathematician myself. I do deal with math for my job, but this is more about calculating angles, etc.
Maybe we'll pass the question on to Counter.
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