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What is the probability of being dealt 3 cards to a SeqRF and then hitting it?
I was dealt As x x Js Ts and caught the Ks Qs in the correct order for the one and only recognized sequential A->T. T->A was not considered seq. This was on an ill-paying deuces wild game where 5k=15, 4k=4, fh=3, fl=2, str=2. The seqRF=60,000 with 5 coins. And yes, I did have max coins in.
Sequential Royal Flush
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Eduardo
- Video Poker Master
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- Joined: Thu Aug 31, 2006 7:19 pm
Some here would say it's pretty much impossible and suggest that something was fishy about your machine, given the odds.
What a great hit! I look forward to a serious answer.
What a great hit! I look forward to a serious answer.
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Mutton412
- Forum Rookie
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I meant "..one of 4 recognized sequentials..". It was not spade-specific.
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shadowman
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There are, I believe, 10 ways to be dealt a sequential RF (AKQ, AKJ, AKT, AQJ, AQT,AJT, KQJ, KQT, KJT, QJT). Now for some off-the-cuff estimates ...
With 4 suits that gives you 40 total. Now, the other 2 cards can be anything but the missing royal cards. That means 47 cards arranged two at a time. Hence, there are 21620 possible arrangement of cards with exactly 3 to a sequential RF which means you are dealt it about once every 14,425 hands.
As for the draw, you have 1081 possible draws and needed combo is just one of those. Of course, half the time you'll have the wrong order. So, 14,425*1081*2 = 31.2 million.
Seems a little high, I'll have to recheck this later when I have more time.
With 4 suits that gives you 40 total. Now, the other 2 cards can be anything but the missing royal cards. That means 47 cards arranged two at a time. Hence, there are 21620 possible arrangement of cards with exactly 3 to a sequential RF which means you are dealt it about once every 14,425 hands.
As for the draw, you have 1081 possible draws and needed combo is just one of those. Of course, half the time you'll have the wrong order. So, 14,425*1081*2 = 31.2 million.
Seems a little high, I'll have to recheck this later when I have more time.
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Lucky Larry
- Video Poker Master
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Congratulations Mutton
Welcome to the forum.
Welcome to the forum.
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New2vp
- Video Poker Master
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- Joined: Mon Sep 11, 2006 4:02 am
I think shadow man has the draw numbers correct. I also agree with his logic for the dealt combinations, but they multiply as 40 * 1081 * 2 = 86480. With that modification the odds drop to 7,796,879 to 1. These numbers are off somewhat because the extra 2 cards on the deal cannot be 2 deuces to form a wild royal flush or a combination that makes a pat straight flush. We would throw 1 deuce away with a 3 card sequential royal, but not 2 deuces. This would make the odds 7,855,743 to 1. If this were a non wild game the odds would be different and possibly depend on the specific pay schedule.
