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July 28th, 2015 at 11:23:21 AM
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In video poker we hear of long streaks without hitting a royal flush. This got me wondering. How many hands must be played before achieving a 95 percent chance of hitting a royal? How about for a 98 percent chance? A 99 percent chance?
July 28th, 2015 at 11:48:45 AM
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here is what i get for 9/6 JoB (perfect play required)
120,999.12: 95.0%
130,012.00: 96.0%
141,631.63: 97.0%
158,008.57: 98.0%
186,005.13: 99.0%
279,007.69: 99.9%
99% = 1 in 100 on average will exceed that value
so 100 million vp players
expect 1 million to have a drought at least that long
1 in 1000 on average exceed 279,000 hands (big time loser i would imagine, imo of course)
how i do this for 95.0% example
LOG(1-0.95) / LOG(40390/40391)
http://www.wolframalpha.com/input/?i=LOG%281-0.95%29+%2F+LOG%2840390%2F40391%29
i think i did this right
120,999.12: 95.0%
130,012.00: 96.0%
141,631.63: 97.0%
158,008.57: 98.0%
186,005.13: 99.0%
279,007.69: 99.9%
99% = 1 in 100 on average will exceed that value
so 100 million vp players
expect 1 million to have a drought at least that long
1 in 1000 on average exceed 279,000 hands (big time loser i would imagine, imo of course)
how i do this for 95.0% example
LOG(1-0.95) / LOG(40390/40391)
http://www.wolframalpha.com/input/?i=LOG%281-0.95%29+%2F+LOG%2840390%2F40391%29
i think i did this right
I Heart Vi Hart
July 28th, 2015 at 4:35:50 PM
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It's just a shame that I don't have the time to set down at one machine and play it out multiple times for some real-world data.
"Those who have no idea what they are doing, genuinely have no idea that they don't know what they are doing." - John Cleese