Extremely Lucky Lucky (Blackjack Side Bet), what are the odds?
March 7th, 2013 at 3:25:17 PM
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I was staying at the M Resort in Henderson over the weekend and had an extremely fortunate run of cards in Blackjack in which I was playing the Lucky Lucky side bet on all my hands. In addition to a bunch of the smaller payouts (which we won't worry about), I hit one 777 and two suited 678s in the course of about 2-3 hours (I don't know the exact time).
Assuming an average amount of hands dealt per hour (it was about 4 people at the table if that matters), what are the odds of actually getting that Lucky Lucky?
Assuming an average amount of hands dealt per hour (it was about 4 people at the table if that matters), what are the odds of actually getting that Lucky Lucky?
March 8th, 2013 at 8:05:43 AM
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Assume average 40 hands dealt per hour
3 hours of play = 120 hands total = N
Assume only first 3 listed events are what you are interested in since the chance of getting at least one of all of the listed Lucky Lucky events in 120 hands is nearly 100%:
p = 0.000016+0.000172+0.000388 = 0.000576
P(LL=0) = (1 - p)^N
= (1 - 0.000576)^120
= 93.3% chance of not getting a Lucky Lucky in 120 hands
P(LL>0) = 1 - P(LL=0)
= 6.7% chance of getting at least one Lucky Lucky in 120 hands
3 hours of play = 120 hands total = N
Assume only first 3 listed events are what you are interested in since the chance of getting at least one of all of the listed Lucky Lucky events in 120 hands is nearly 100%:
p = 0.000016+0.000172+0.000388 = 0.000576
P(LL=0) = (1 - p)^N
= (1 - 0.000576)^120
= 93.3% chance of not getting a Lucky Lucky in 120 hands
P(LL>0) = 1 - P(LL=0)
= 6.7% chance of getting at least one Lucky Lucky in 120 hands
March 8th, 2013 at 8:14:01 AM
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Maybe I'm not understanding what you posted or perhaps you misunderstood the question. If I played 120 hands total let's say, I'm trying to figure out what the odds are of getting 2 suited 678s (0.000172) and 1 unsuited 777 (0.000388) in the course of those 120 hands. I know for sure that it is much, much worse than 6.7% just by having played the game before and knowing the payouts.
March 8th, 2013 at 8:21:37 AM
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The chances of getting those 3 events to occur exactly (no more no less) in 120 hands is p = (0.000172)^2 (0.000388) (0.999424)^117 COMBIN(120,3)
= 0.00030%
= 0.00030%
March 8th, 2013 at 8:37:01 AM
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Quote: paisielloThe chances of getting those 3 events to occur exactly (no more no less) in 120 hands is p = (0.000172)^2 (0.000388) (0.999424)^117 COMBIN(120,3)
= 0.00030%
I think you are missing a *3 in there. combin(120,3) ways for the 3 events, but they can be in 3 different orders.
“Man Babes” #AxelFabulous
March 8th, 2013 at 8:38:50 AM
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I'm going to expand on what paisiello has already done:
This probability is the chance of getting any of the following: suited 777, unsuited 777, suited 678
p(0 wins) = (1-0.000576)^120 = 0.933196104
p(1 win) = (1-0.000576)^119 * 0.000576^1 * combin(120,1) = 0.06453969
p(2 wins) = (1-0.000576)^118 * 0.000576^2 * combin(120,2) = 0.002213179
p(3 or more wins) = 1 - p(0 wins) - p(1 win) - p(2 wins) = 5.10279E-05 = 0.0051%
Indeed, very lucky; nearly a 1 in 20,000 chance of getting 3 or more of the wins listed above.
Quote: paisielloAssume average 40 hands dealt per hour
3 hours of play = 120 hands total = N
Assume only first 3 listed events are what you are interested in since the chance of getting at least one of all of the listed Lucky Lucky events in 120 hands is nearly 100%:
p = 0.000016+0.000172+0.000388 = 0.000576
This probability is the chance of getting any of the following: suited 777, unsuited 777, suited 678
p(0 wins) = (1-0.000576)^120 = 0.933196104
p(1 win) = (1-0.000576)^119 * 0.000576^1 * combin(120,1) = 0.06453969
p(2 wins) = (1-0.000576)^118 * 0.000576^2 * combin(120,2) = 0.002213179
p(3 or more wins) = 1 - p(0 wins) - p(1 win) - p(2 wins) = 5.10279E-05 = 0.0051%
Indeed, very lucky; nearly a 1 in 20,000 chance of getting 3 or more of the wins listed above.
I heart Crystal Math.
March 8th, 2013 at 8:41:08 AM
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In my response, most of the time, this will be 3 suited 678 pays, so you were even luckier than this.
I heart Crystal Math.
March 8th, 2013 at 8:52:24 AM
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Quote: mipletI think you are missing a *3 in there. combin(120,3) ways for the 3 events, but they can be in 3 different orders.
COMBIN(120,3) = 120! / 3!(120-3)! = 280,840 possible combinations should also take into account 3 different orders.
March 8th, 2013 at 9:24:55 AM
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Before I take my shoes off and dig into the math that goes beyond ten, I just want to comment on your post's non-math aspects.
Its good that you were alert to how many people were at the table since you are playing at under or over the assumed rate of play depending upon number of players and number of them who were taking advantage of the side bets available to them.
If you are playing slower than the Rate of Comps that is marginally good for you.
Its good that you were alert to how many people were at the table since you are playing at under or over the assumed rate of play depending upon number of players and number of them who were taking advantage of the side bets available to them.
If you are playing slower than the Rate of Comps that is marginally good for you.
March 8th, 2013 at 9:34:34 AM
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Quote: paisielloCOMBIN(120,3) = 120! / 3!(120-3)! = 280,840 possible combinations should also take into account 3 different orders.
Actually, miplet is right.
You have one set of 2 matching items and 1 unique item. To use the combin function, it should be combin(120,2)*combin(118,1), which is identical to combin(120,3)*3.
I heart Crystal Math.
March 8th, 2013 at 12:27:32 PM
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Makes sense. I didn't understand miplet's orignal comment.
March 10th, 2013 at 12:52:05 PM
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Thanks guys for the math work and responses. The odds weren't as astronomical as I thought, this was more probable then ending with a royal flush in video poker (1 in 40,000), but still quite a hit!
