# Probabilities of VP

#### retlaw

##### Meister Member
Most of the time I play Single hand VP on MG casino's with a 1.25 euro bet.

During the last 4 months I got approximately 200 times "4 to a royal flush"
without getting the royal flush.
I find this is not normal ,although I know that statistically it is possible, because each time again I have a probability of 1/47.

However I got 2 royals from other combinations in this period of time.
I think all together I will be not so far away from the 1/40000 probability.

I think that there is "some system" which prevents me from hitting the royal in the case of "4 to a royal flush".

I know all RNG believers will not agree with my theory, nevertheless this is my opinion.

Most of the time I play Single hand VP on MG casino's with a 1.25 euro bet.

During the last 4 months I got approximately 200 times "4 to a royal flush"
without getting the royal flush.
I find this is not normal ,although I know that statistically it is possible, because each time again I have a probability of 1/47.

However I got 2 royals from other combinations in this period of time.
I think all together I will be not so far away from the 1/40000 probability.

Probablity for missing a 1/47 chance 200 times in a row is 1.4%.

Unlucky, but nothing remarkable. Most VP players uses #hands to see if they have been unlucky with the RF. For JoB VP you should hit a RF about
every 40000 hand is you play optimal strategy. So if you played less than 80K hands, you have actually been lucky.

I once ran a simulation (single hand also) to test where the RF's came from (#cards in starting hand) when playing JoB because I seemed also to hit most RF when keeping 3 cards, and I wanted to test of this was normal. It was.

This is the result:

Iterations:173000000, running for 57957.4 seconds

#Cards held:#RF
0 held:14 (0.35%)
1 held:153 (3.78%)
2 held:826 (20.4%)
3 held:1670 (41.3%)
4 held:1125 (27.8%)
5 held:257 (6.35%)

As you can see the most common source of RF's actually comes from situations where you kept 3 cards.

Edit: I remember having a quite related experience. I had 3 dueces on 100-play Dueces Wild (MG). Looking forward to unspeakable richies I pressed the deal button with excitement. I hit 4 dueces ZERO times. Amoung the 100 cards I was dealth in replacements there was not a single duece. I got a bit pissed and forgot to take
screenshot for the 'screenshit that sucks' thread. Later I also calculate the probability to be around 1.4% like yours, if I remember the probability correct.

Zoozie

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Amoung the 100 cards I was dealth in replacements there was not as single duece.
Zoozie

It was actually 200 cards since it was 100-line. I just calculated the probability again, it is 1/6592. It was damn freaky unlucky. Wish I had taken
a screenshot. (It was just at 5\$/spin though)

Zoozie

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did you try covering your eyes and screaming "C'MON YOU MOFO!" at the top of your lungs? no. i didn't think so. if you had you would have hit that card at least 5 of those times.

It was actually 200 cards since it was 100-line. I just calculated the probability again, it is 1/6592. It was damn freaky unlucky. Wish I had taken
a screenshot. (It was just at 5\$/spin though)

Zoozie
Perhaps I am missing something. I'd expect the probablity to be:

Per hand odds of not getting the final duece on first card: 48/49
Per hand odds of not getting the final duece on second card: 47/48
Probablity of not getting deuce in any of the 100 hands {(48/49)*(47/48)}^100 ~= 1 in 65

Perhaps I am missing something. I'd expect the probablity to be:

Per hand odds of not getting the final duece on first card: 48/49
Per hand odds of not getting the final duece on second card: 47/48
Probablity of not getting deuce in any of the 100 hands {(48/49)*(47/48)}^100 ~= 1 in 65

You are right, this is the way I tried to calculate it, but dont know what went wrong. But remember 3 cards are discarded.

Per hand odds of not getting the final duece on first card: 46/47
Per hand odds of not getting the final duece on first card: 45/46
Probablity of not getting deuce in any of the 100 hands {(46/47)*(45/46)}^100 = 1 to 77 or 1.3%.

So I remembered it almost correct (1.3%) in the first post actually, but somehow miscalulated when I tried again. But I am more pleased now, the other result was a little to unlucky.

Zoozie

Royal from 3

I am sure this reflects the far greater likelihood of starting with 3 to a royal rather than 4 to a royal.
When dealt 4 to a Royal in the 50 and 100 hand versions I have got far more Royals than when starting with 3 to a royal. Having 50 or 100 attempts all at once makes it clearer.
Certainly, it makes 3 to a royal a good hand to hold, and I once obtained a RF by keeping 3 to a royal over a pair Jacks or Better!
I have had very few RF's recently, however, earlier I seemed to be getting them all the time (I am getting the freaky slot hits now!).

Certainly, it makes 3 to a royal a good hand to hold, and I once obtained a RF by keeping 3 to a royal over a pair Jacks or Better!

I also keep "3 to a royal " instead of "4 to a flush" .I never got the royal but lost many flushes

I read that according to the Wizard, it is sometimes better to hold the "4 to a flush", but I would feel very bad if the 5th card which is drawn,forms part of the royal.

Re

I have hit about mabey 12 royals total. I find that I have hit when I held 2 or three in the flush. When its 4 I never get it. I dont think even once.
Or it just gets dealt to me.

There are more times than I can count though, where I lost royals where there is only one jack lets say, and I throw the hand, and the next four would have made a royal. Happens to much latley for my taste, to seem probable. Or I loose a royal where it gives me two aces, and lets say one of the aves is a spade, and there is a 10 of spades up, I take the two aces, and the rest of the cards come up for a royal. I had this happed about 8 times this month, and it burns me up.

Ama

Iterations:173000000, running for 57957.4 seconds

#Cards held:#RF
0 held:14 (0.35%)
1 held:153 (3.78%)
2 held:826 (20.4%)
3 held:1670 (41.3%)
4 held:1125 (27.8%)
5 held:257 (6.35%)

As you can see the most common source of RF's actually comes from situations where you kept 3 cards.

Zoozie

Yes, I am replying to my own post - and indeed a 4 year old one.
The question was about where your RF 'came from' in the sense of how often it was from holding 0,1,2,3,4 or 5 cards.

I have digged this thread back from the dead, because I saw the Wizard of Odds recently have calculated the exact probablities of this old problem, and this solving it.Overall my simulation gave a quite good approximation though and also gave the result '3 cards' as the biggest contributor to your RF.

CASE CLOSED.

His results:

You guys are into probabilities etc - I cannot get my head round them - could anyone work out a formula for this ;-

Three different machines (all jacks or better with 99.54 expected return) -- If you are playing 1 line \$ (bet per spin \$5) , 10 line - 25c (bet per spin \$12.50) , or 50 line 5c ( bet per spin \$12.50) -- and you have a) a \$500 starting pot b)\$100 c) \$3000 - what is the probability for each combination of running out of money after a)500 spins b) 1000 spins c)3000 spins. I would not expect you to calculate them all, but would expect that one formula or equation would work, you just change the figures.

I would like to be able to extend it further to say (eg) that if I had say a \$1000 start pot and played 1000 hands of a 10 line 25c game (\$12500 bet ), that there is a (say) one in ten cance that I will still have \$1000 left , a one in five chance I would have \$700 left etc - I think that can be done with a probability curve but again do not know how to do it !

If you cannot do it on here , can you point me to a thread or website that might provide that info in simple language !

You guys are into probabilities etc - I cannot get my head round them - could anyone work out a formula for this ;-

Three different machines (all jacks or better with 99.54 expected return) -- If you are playing 1 line \$ (bet per spin \$5) , 10 line - 25c (bet per spin \$12.50) , or 50 line 5c ( bet per spin \$12.50) -- and you have a) a \$500 starting pot b)\$100 c) \$3000 - what is the probability for each combination of running out of money after a)500 spins b) 1000 spins c)3000 spins. I would not expect you to calculate them all, but would expect that one formula or equation would work, you just change the figures.

I would like to be able to extend it further to say (eg) that if I had say a \$1000 start pot and played 1000 hands of a 10 line 25c game (\$12500 bet ), that there is a (say) one in ten cance that I will still have \$1000 left , a one in five chance I would have \$700 left etc - I think that can be done with a probability curve but again do not know how to do it !

If you cannot do it on here , can you point me to a thread or website that might provide that info in simple language !

There is no formula or equation to calculate the probabilities for your question, mostly because VP results don't follow normal distribution so it takes a computer simulation to calculate the precise odds. Luckily for you there is such a simulator at Beating Bonuses, which can simulate 1-hand, 4-hand, 10-hand, 25-hand and 100-hand versions of Jacks or Better (
). You type in starting balance and total wagering and it gives you the probability distribution of the end result. You can also calculate the probabilities to end at certain result after certain amount played so you should get answers to all those questions with the sim.

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