Board Game Arena

Meeplelowda wrote: ↑05 May 2025, 20:35

Mathew5000 wrote: ↑05 May 2025, 19:50 If we're on the page, it's the same as the probability of getting exactly 7 of the same face when you throw 10 standard dice, which is:
(6*(10 C 7)*5^3)/6^10 = 15000/6^9 = 1/671.85

I'm getting something pretty different using the binomial probability formula: probability of exactly k successes is P(X=k) = (n C k)*(p^k)*(q^(n-k)), where n=10, k=7, p=1/6, and q=5/6

Works out to about 1/4032

Mathew5000 wrote: ↑05 May 2025, 19:50

euklid314 wrote: ↑05 May 2025, 16:19 Looking at the turn based games of OP, he is playing different games (Splendor, Through the Ages, Great Western Trail, Azul, Mischwald,....) with his 3 friends.

Thanks; that's how I originally understood what OP said. I guess Mikkashar interpreted it the other way.

euklid314 wrote: ↑04 May 2025, 22:51 Still possible that it is random and pure coincidence (as it should be to my knowledge), since there are only 6 turn orders.
But it is approx. 1:45.000 to get same turn order in 7 consecutive 4-player games, thus quite unlikely to happen. If your 7 games were not started consecutively (as I said, you have more than 7 games with these players) then it immediately gets much more likely. 7 same turn orders out of 10 games is perhaps a 1:1000 event (I am too lazy to calculate).

If we're on the page, it's the same as the probability of getting exactly 7 of the same face when you throw 10 standard dice, which is:
(6*(10 C 7)*5^3)/6^10 = 15000/6^9 = 1/671.85

For the probability of getting at least 7 of the same face when you throw 10 dice:
((10 C 7) * 5^3 + (10 C 8) * 5^2 + (10 C 9) * 5 + (10 C 10)) / 6^9
= 337/209952 ≅ 1/623

I also asked ChatGPT, but it went wrong somewhere and told me the answer is 3.85%: https://chatgpt.com/share/68190835-ab68 ... 86f0c3da1f

Edited to add: I eventually got ChatGPT to agree that my answer is correct: https://chatgpt.com/share/68190835-ab68 ... 86f0c3da1f

Mathew5000 wrote: ↑05 May 2025, 20:37

Meeplelowda wrote: ↑05 May 2025, 20:35

Mathew5000 wrote: ↑05 May 2025, 19:50 If we're on the page, it's the same as the probability of getting exactly 7 of the same face when you throw 10 standard dice, which is:
(6*(10 C 7)*5^3)/6^10 = 15000/6^9 = 1/671.85

I'm getting something pretty different using the binomial probability formula: probability of exactly k successes is P(X=k) = (n C k)*(p^k)*(q^(n-k)), where n=10, k=7, p=1/6, and q=5/6

Works out to about 1/4032

That's the probability of getting exactly 7 threes when rolling 10 dice. You have to mulitply by 6 to get the probability of getting exactly 7 of the same face.