Let's say 1000 Players start immediately, then the biggest part of these group will get 36 different numbers around 118 spins and 37 different numbers around 155 spins. Right or wrong?
because the actual distribution of how many spins it takes is NOT a normal distribution
133 is the most likely spin (the mode)to collect the last of all 37 numbers
this would be correct but not answer your question
the biggest part of these group will get 36 different numbers BY 118 spins (118 or less)
and 37 different numbers BY 155 spins(155 spins or less)
the biggest part
is only over 50%
0.576996723 is the probability for one player to get ALL by and including the 155th spin
(that is just over 50% and nowhere near even 99%)
on average, we would expect only 577 (out of 1,000) to collect all the numbers before the 156th spin
not drawn Probability cumulative
u=32 5 5.792644277504847e-05 6.034500425211893e-05
u=33 4 0.0009855780328734401 0.001045923037125559
u=34 3 0.01121913540428637 0.01226505844141193
u=35 2 0.08080438752582932 0.09306944596724125
u=36 1 0.3299338309788963 0.4230032769461376
u=37 0 0.5769967230538622 1
in summary, they won't be close to the 'average number' (except by luck)
but add up all the players spins and divide by 1000 and there will be close to the average number.
seems you got the average to be the most common probability and for this type of distribution it is not
I feel you are trying to do something with this information?
thank you for the share.
In other words, simply count the number of pockets on the wheel and realize that the ball can land in any one of them.
Maybe it will. I've seen the ball fly
out of the wheel many times and
end up on the floor. No bet for