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#12
04-08-2005, 10:00 AM
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Re: Probability of a royal in texashold
If you hold two suited broadway cards (this is the best case scenario), then you should hit a royal flush 1 in 1959 hands (on average), assuming you stay for all 5 board cards. Please be aware that this is simply an average. You might hit the next time you have suited broadway cards. You could easily go several thousand hands and not hit.
elitegimp seemed to agree with my answer and also gave estimates for situations where you only hold one card, etc. Of course, those scenarios aren't as favorable. But if you played every hand you had at least one broadway card, then the number of hands you'd have to wait through would be less (be warned that if you're trying to hit a 'royal' bonus on a poker sites most sites only allow it if you use BOTH of your hole cards). When the odds are listed as x:1 against (as I did in my previous posts), this just means that you will (on average) hit the desired hand 1 time in x tries. The reason odds are listed in such a way is that it puts the information is a useaful format. We can compare the odds against making the hand to the pot odds we're being offered. Hope that clears it up. 
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#15
04-08-2005, 11:39 AM
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Re: Probability of a royal in texashold
To clarify, the number I have been giving is the average number of hands you should need before hitting a royal flush ASSUMING that you already hold TWO suited broadway cards. This number was around 1950 - I can't remember it exactly right now.
elitegimp's answer is the number of hands you need to be dealt before hitting the royal flush. So it includes all the times you will be dealt 72o and other such junk and obviously not hit the royal no matter what you do. I have not verified this figure, but it seems reasonable. This answer is probably closer to what you wanted from your question in the OP, but it is not completely clear to me.
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#16
04-08-2005, 11:52 AM
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Re: Probability of a royal in texashold
[ QUOTE ]
[ QUOTE ] So in 3rd grade english. That 1 royal in _______ hands? [/ QUOTE ] 30940 [/ QUOTE ] Correct! That's any suit royal, given 7 cards out of 52 in no particular order. If you need a specific suit (for a jackpot or something), it would be: 123,670 Specific odds for one suit/jackpot royal: 1. If you hold no hole cards for the royal: 2,118,670 (5/5 cards out of 50 cards left) 2. If you have 1 hole card: 46,060 (4/5 cards out of 50 left) 3. 2 hole cards: 1,960 (3/5 cards out of 50 left) So if the spade royal jackpot is up to $50,000 should you see the flop with any 1 spade 10 or higher?
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#17
04-08-2005, 05:15 PM
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Re: Probability of a royal in texashold
How it came up is. I was playing I floped the nut straight. With a gutshot to the royal. I missed the royal. It pays 1k to hit it. I made the comment that I have hit 2 royals in 200k hands online. a big debate followed afterwards as to how offten a royal is hit in hold'em. Lady told me 1 in `650k. I said it has to be more often that that. If i remeber correctly thats how often it is hit in draw poker. it happens more often in holdem due to the 7 cards.
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#19
04-12-2005, 09:30 PM
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Re: Probability of a royal in texashold
[ QUOTE ]
C(50,5) = 2,118,760 [/ QUOTE ] Could someone inform me if the quoted equation or whatever you call it means fifty, pick two, and how one would solve such an equation?
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#20
04-12-2005, 10:36 PM
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Re: Probability of a royal in texashold
[ QUOTE ]
[ QUOTE ] C(50,5) = 2,118,760 [/ QUOTE ] Could someone inform me if the quoted equation or whatever you call it means fifty, pick two, and how one would solve such an equation? [/ QUOTE ] "50 choose 5" or "combinations of 50 things taken 5 at a time". It's the number of 5 card hands you can make out of 50 cards. Compute it as 50*49*48*47*46/(5*4*3*2*1). The numerator has 5 terms since we want 5 cards. There are 50 ways to choose the 1st card, 49 ways to choose the second card, etc. Then we divide by 5! = 5*4*3*2*1 because we don't care about order. For example, we consider A [img]/images/graemlins/spade.gif[/img]2 [img]/images/graemlins/spade.gif[/img]3 [img]/images/graemlins/spade.gif[/img]4 [img]/images/graemlins/spade.gif[/img]5 [img]/images/graemlins/spade.gif[/img] to be the same hand as 5 [img]/images/graemlins/spade.gif[/img]3 [img]/images/graemlins/spade.gif[/img]A [img]/images/graemlins/spade.gif[/img]2 [img]/images/graemlins/spade.gif[/img]4 [img]/images/graemlins/spade.gif[/img]. The number of ways to arrange each combination of 5 cards is 5! = 5*4*3*2*1, so we divide by this number so that each combination is counted only once. You can use the COMBIN function in Excel to compute this, or use google and type in, for example, "50 choose 5" (without the quotes).
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