Probability of a royal in texashold

[radek2166]

one in how many hands?

[elitegimp]

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one in how many hands?

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Assuming you have suited broadway cards? Assuming you have offsuit broadway cards? Assuming you have 2 random cards?

[radek2166]

I was playing b&m today chcik says 622k hands. i said no way.

Howmany hands best case secinerio?

[PygmyHero]

19,599 to 1 against as per Doyle Brunson's Super System, pg. 573.

Gotta be honest - I don't want to do the work for this problem given that it is SOOOO improbably that I can't believe knowing that statistic is useful. Its also on the late-ish side.

And besides, Doyle is NEVER wrong.

P.S. ALWAYS play T2...under any circumstances.

[PygmyHero]

Sorry, that statistic is only for FLOPPING the royal flush. I didn't look closely enough.

Well, that means your odds (if you stay through five cards) are BETTER than 1 in 19,599, so your friend is wrong.

I'll look a little more closely and post again if I find it. Otherwise I will (sigh!) work it out.

[PygmyHero]

Okay, I couldn't find your answer in SS, though I didn't look too closely. So here we go.

Best case scenario is that you hold two of the five cards you need - it doesn't matter which two you hold. For the sake of being concrete let's say you have AK suited in spades. So you need the queen, jack, and ten of spades, and the other two board cards can be any other cards in the deck.

The number of boards that will have all three of the cards you need are:
C(3,3) * C(47,2) = 3243

The number of possible boards is:
C(50,5) = 2,118,760

3243/2118760 is about 0.15% In odds this is 652.3:1 against.

Okay, this number seems WAY too low, though as I already posted, it can't be near the 600+K your friend said. I must be doing something wrong. I'm sure someone can correct me, but I just thought I'd get something out there for starters.

[PygmyHero]

Correction...sorry, it was late when I did this last night. I posted:

"The number of boards that will have all three of the cards you need are:
C(3,3) * C(47,2) = 3243

The number of possible boards is:
C(50,5) = 2,118,760

3243/2118760 is about 0.15% In odds this is 652.3:1 against."

The first equation is of course incorrect.

C(3,3) * C(47,2) = 1081

The C(50,5) is right, so the third step should be:

1081/2118760 = 0.051% This is 1959:1 against, which still seems a little high. Maybe I'm making another mistake.

[elitegimp]

[ QUOTE ]
Correction...sorry, it was late when I did this last night. I posted:

"The number of boards that will have all three of the cards you need are:
C(3,3) * C(47,2) = 3243

The number of possible boards is:
C(50,5) = 2,118,760

3243/2118760 is about 0.15% In odds this is 652.3:1 against."

The first equation is of course incorrect.

C(3,3) * C(47,2) = 1081

The C(50,5) is right, so the third step should be:

1081/2118760 = 0.051% This is 1959:1 against, which still seems a little high. Maybe I'm making another mistake.

[/ QUOTE ]


sorry, I meant to come back and post my numbers but yesterday was way too busy... anyway, here's what I get:

* If you have 2 Sooted Broadway cards, you can make a Royal if
(a) you hit the other 3 broadways of your suit, plus two random cards (there are 47*46/2 = 1081 different boards that look like that), or by having the board show a royal in one of the three other suits (3 different boards) => 1084 different boards will give you a royal. Further, there are C(50,5) = 2118760 unique boards given you hold two [specific] suited broadways, so you have a 1084/2118760 = 0.00051 chance (1/1954.58 or 1953.85 to 1) of hitting the royal.

* If you have two off-suit broadway cards, you can get a royal if you either hit a royal in either of your suits (need 4 other specific cards to show, plus one card can be anything) or if the board is a royal in one of the two suits you don't hold. So 2*1*46 =92 ways for you to play a card from your hand and have a royal, plus 2 ways for you to not play a card from your hand and have a royal => 94/2118760 = 0.0000437 chance (1/22540 or 22539 to 1).

* If you have only one broadway card in your hand, you can hit a royal in your suit 46 different ways (you need 4 specific cards, plus the 5th card can be anything left in the 46 card deck), or you need to hit one of the 3 royals that the board can show => 49/2118760 chance = 0.0000231 (1/43240 or 43239:1)

* If you have no broadway cards, then you're counting on the board to provide you a royal, and there are 4 ways for that to happen => 4/2118760 chance = 0.00000189 (1/529690 or 529689:1)

In general, if you have 2 random cards plus the board to make a royal with, you can treat it like a 7-card board. There are 4 sets of 5 cards to give you a royal, plus the remaining 2 cards can be anything => 4*47*46/2 = 4324 possible combinations and a total of C(52,7) = 133784560 ways to deal 7 cards => 4324/133784560 = 0.0000323 chance (1/30940 or 30939 to 1 against) that you hit a royal.

So to answer your question (I think) -- if you play 30940 hands, you expect to hit exactly one royal (but are guaranteed nothing, obviously). You also expect exactly one royal after being dealt suited broadway cards 1954.58 times, so hopefully the odds of being dealt suited broadway cards = 1954.58/30940 or 0.0632 (1/15.8ish).

I've got class, so I can't verify any of the claims I make in this post... hopefully there aren't any typos (all the methods are sound)

[elitegimp]

[ QUOTE ]

You also expect exactly one royal after being dealt suited broadway cards 1954.58 times, so hopefully the odds of being dealt suited broadway cards = 1954.58/30940 or 0.0632 (1/15.8ish).

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This may very well be the stupidest thing I've ever written on a public forum... please ignore this paragraph in my above post (it's too late to edit it out [img]/images/graemlins/blush.gif[/img] )

Pretty much, since you can get a royal without holding suited broadway cards, we should expect the number of hands before we get a royal [period] to be less than the number of hands before we get a royal with 2 suited broadway cards in our hand.

What I was actually trying to get at was that if you sum up the weighted average of the 4 values [i.e. P(2 suited broadways)*1/1954 + P(2 offsuit broadways)*1/22540 + ...] you should get the last value (P(royal if you have 2 random cards))

[radek2166]

So in 3rd grade english. That 1 royal in _______ hands?