masterj
masterj
Joined: Dec 19, 2018
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January 11th, 2019 at 11:59:28 PM permalink
Can anyone tell me if this assumption is correct? I am aware that you will never know which players will reach these numbers at the beginning. But if all of them play until reaching 37 different numbers, then the biggest part of them should be close to the average numbers.
7craps
7craps
Joined: Jan 23, 2010
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January 16th, 2019 at 10:30:33 PM permalink
Quote: masterj

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?

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%
more precisely
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.
winsome johnny (not Win some johnny)
EvenBob
EvenBob
Joined: Jul 18, 2010
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January 16th, 2019 at 11:28:04 PM permalink
Quote: Keyser

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
that, though.
"It's not enough to succeed, your friends must fail." Gore Vidal
masterj
masterj
Joined: Dec 19, 2018
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February 21st, 2019 at 8:46:34 AM permalink
Hello 7 Craps,

sorry for the long wait.
In your calculation, the last open number should show up on average after 133 spins?
So after getting 36 numbers in 118 spins (on average) it takes only on average 15 spins to get the last open number? I don't understand this.


If this is true, then we should wait until 36 numbers show up, bet 15x the open number. If we hit, we win, if we don't hit, we stop after 15 bets and observe a different table and start again.
If 133 is the most likely occurence, then we should hit every other time in the long run while betting 15x. The average win ist higher then the loss of 15 bets. So we are profitable.

But I can not believe that this is correct.
OnceDear
Administrator
OnceDear
Joined: Jun 1, 2014
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February 21st, 2019 at 10:35:23 AM permalink
Quote: masterj

Hello 7 Craps,
it takes only on average 15 spins to get the last open number? I don't understand this.

Don't try to: It's wrong.
Quote:

If this is true, then we should wait until 36 numbers show up, bet 15x the open number.

It's not true, so don't
Quote:

The average win ist higher then the loss of 15 bets. So we are profitable.

But I can not believe that this is correct.

Nope. The maths is squiffy. It's not correct.
If you are enjoying the game, you're already winning.
michael99000
michael99000
Joined: Jul 10, 2010
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February 21st, 2019 at 8:15:16 PM permalink
Quote: EvenBob

Maybe it will. I've seen the ball fly
out of the wheel many times and
end up on the floor. No bet for
that, though.




You should be able to bet on that happening

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