February 1st, 2014 at 5:35:15 PM
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If I press the 4,
What is the probability/chance that I would be able to press the 4
5 times
4 times
3 times
before a 7 hits?
The streak calculator (http://www.pulcinientertainment.com/info/Streak-Calculator-enter.html) doesn't help me.
The binomial formula doesn't help does it? (http://stattrek.com/online-calculator/binomial.aspx)
What is the probability/chance that I would be able to press the 4
5 times
4 times
3 times
before a 7 hits?
The streak calculator (http://www.pulcinientertainment.com/info/Streak-Calculator-enter.html) doesn't help me.
The binomial formula doesn't help does it? (http://stattrek.com/online-calculator/binomial.aspx)
February 2nd, 2014 at 12:33:02 AM
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I have seen someone power press the 4 from $5 up to $1000 (table max) and have it hit. Funny thing is that it was a relatively short roll. The guy made one pass but threw more 4's than anything else. Entire roll lasted maybe 15 minutes. I think (if I calculated correctly) that he threw seven 4's. I was on the 6 and 8 and the point and made a little money, nothing compared to the guy that went $5 across and power pressed!
I think you just do 2*x power to figure out the odds. So 5 times would be 2 to the 5th which is 32 to 1 against.
I think you just do 2*x power to figure out the odds. So 5 times would be 2 to the 5th which is 32 to 1 against.
February 2nd, 2014 at 9:03:46 AM
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I read an older post by the wizard that answered a similar question. It would seem the answer is this:
P(A being thrown before B x number of times) = (P(A)/P(A+B))^x
X being the number or series in question - so in my case, 3,4, and 5.
So I get
5 times = 0.004115 or 0.4%
4 times = 0.012356 or 1.2 %
3 times = 0.037 or 3.7%
I hope that is correct.
So in about 8 hrs of play, you could expect a run of 5 4s four times.
Well I answered my own question. But i need help with this one.
During that single run of hitting a 4 3 times before a 7, how likely is it that I hit my other place bets (5,6,8,9,10) at least 8 times?
What about on a run if hitting a 4 4 times before a 7, hitting those place bets at least 13 times?
P(A being thrown before B x number of times) = (P(A)/P(A+B))^x
X being the number or series in question - so in my case, 3,4, and 5.
So I get
5 times = 0.004115 or 0.4%
4 times = 0.012356 or 1.2 %
3 times = 0.037 or 3.7%
I hope that is correct.
So in about 8 hrs of play, you could expect a run of 5 4s four times.
Well I answered my own question. But i need help with this one.
During that single run of hitting a 4 3 times before a 7, how likely is it that I hit my other place bets (5,6,8,9,10) at least 8 times?
What about on a run if hitting a 4 4 times before a 7, hitting those place bets at least 13 times?
February 2nd, 2014 at 6:55:54 PM
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Of course you don't need the 4 to come 5 times before a 7 you just need it to come 5 times before the shooter throws a 7 out. A 7 on a come out roll wouldn't hurt you at all.
February 2nd, 2014 at 8:03:50 PM
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Quote: slackyhackyI read an older post by the wizard that answered a similar question. It would seem the answer is this:
P(A being thrown before B x number of times) = (P(A)/P(A+B))^x
X being the number or series in question - so in my case, 3,4, and 5.
So I get
5 times = 0.004115 or 0.4%
4 times = 0.012356 or 1.2 %
3 times = 0.037 or 3.7%
I hope that is correct.
So in about 8 hrs of play, you could expect a run of 5 4s four times.
Well I answered my own question. But i need help with this one.
During that single run of hitting a 4 3 times before a 7, how likely is it that I hit my other place bets (5,6,8,9,10) at least 8 times?
What about on a run if hitting a 4 4 times before a 7, hitting those place bets at least 13 times?
Parley $25/Buy on the 4 - 4 times and you are at max bet of 1k and also will have 900 in your rack. :)
8 more years till retirement.