mtobeiyf
Joined: Jan 27, 2013
• Posts: 1
January 27th, 2013 at 9:48:19 AM permalink
My local casino runs a special prize for making a Royal Flush (RF) hand in Hold Em poker. The hand does not have to go to showdown, but both your hole cards must play (i.e. they must be 2 of the 5 RF cards). I was discussing the odds of making such a hand with other players and I got a lot of different feedback, none of which I felt was correct. Here is the scenario:

You have 2 of the 5 RF cards in the hole. Doesn't matter if they are As-Ks or Js-Ts, etc. What are the odds you will get a Royal Flush by street:

a) make RF by the Flop
b) make RF by the Turn
c) make RF by the River

My calculations were as follows:

a1) 19,599 to 1 on the flop
b1) about 5,000 to 1 on the turn
c1) about 2,000 to 1 on the river

The general consensus was the true value by the river was either 60,000 to 1 or 30,000 to 1 by the river. This seems totally nonsensical to me but I was in the distinct minority (i.e. it was only me!). Can someone with more probability know-how step up and provide a definite answer to this question?

Thank You.
rdw4potus
Joined: Mar 11, 2010
• Posts: 7237
January 27th, 2013 at 9:57:17 AM permalink
I think I agree with your flop math: (3/50)*(2/49)*(1/48)=1 in 19600.
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
miplet
Joined: Dec 1, 2009
• Posts: 2092
January 28th, 2013 at 9:39:52 AM permalink
Quote: mtobeiyf

My local casino runs a special prize for making a Royal Flush (RF) hand in Hold Em poker. The hand does not have to go to showdown, but both your hole cards must play (i.e. they must be 2 of the 5 RF cards). I was discussing the odds of making such a hand with other players and I got a lot of different feedback, none of which I felt was correct. Here is the scenario:

You have 2 of the 5 RF cards in the hole. Doesn't matter if they are As-Ks or Js-Ts, etc. What are the odds you will get a Royal Flush by street:

a) make RF by the Flop
b) make RF by the Turn
c) make RF by the River

My calculations were as follows:

a1) 19,599 to 1 on the flop
b1) about 5,000 to 1 on the turn
c1) about 2,000 to 1 on the river

The general consensus was the true value by the river was either 60,000 to 1 or 30,000 to 1 by the river. This seems totally nonsensical to me but I was in the distinct minority (i.e. it was only me!). Can someone with more probability know-how step up and provide a definite answer to this question?

Thank You.

If you have 2 royal flush cards as hole cards the odds that you will make a royal flush by the river using those hole cards are combin(47,2)/combin(50,2) or 1081/2118760 or 1 in 1960. 1/10 the time you flop it. 3/10 it will be made on the turn and 6/10 on the river.
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tringlomane
Joined: Aug 25, 2012
• Posts: 6278
January 28th, 2013 at 11:03:10 AM permalink
Quote: mtobeiyf

The general consensus was the true value by the river was either 60,000 to 1 or 30,000 to 1 by the river. This seems totally nonsensical to me but I was in the distinct minority (i.e. it was only me!). Can someone with more probability know-how step up and provide a definite answer to this question?

Thank You.

Because in terms of it just generally happening they were correct. It's only about one in 2000 to happen by the river AFTER you get Royal holecards dealt to you, unfortunately 97% of starting hands aren't two Royal cards.

The probability of getting 2 Royal cards to start: 4*C(5,2)/C(52,2) = 40/1326 = 0.030166

The probability of the board containing the other 3 Royal cards: C(47,2)/C(50,5) = 1081/2,118,760 = 0.000510204

The probability of both events happening for you to win the high hand jackpot: 0.030166*0.000510204 = 0.00001539 = 1 in 64,974.

The one in 30,000 number tossed around is the chances of getting any Royal Flush with zero, one, or two hole cards:

4*C(47,2)/C(52,7) = 1 in 30,940.