5 Card Double Up Question

I understand that with 1 card being dealt to 5 people the chance of me having the highest card or tying with another player/s is 1/5.
Also, if I were to make a side bet with 1 of the 5 players my chances of winning would be 1/2 between them and I based on our 2 cards if we push on ties. However, sometimes we would both be tied with or beat by the other 3 players.

This makes the double up feature on a video poker machine a 50/50 proposition for a single double up attempt. I am assuming against the same player over and over for a first double attempt.

My question regards doubling multiple times. I find that my results of successfully doubling 1x is 1/2, 2x is very close to 1/4, 3x in a row are slightly more than 1/8, doubling successfully 4x in a row is significantly more than 1/16, 5x in a row has not been seen by me in hundreds of attempts.

Based on this I believe the double up feature containing 5 cards to be a 50/50 for a single wager but I would like to know how to figure out the chances for multiple doubles 2x,3x,4x,5x, etc.

If you've only done a few hundred attempts then it's quite likely to get a result different from the expected answer. Assuming an infinite deck then the chances of a tie are 1 in 13 (smaller with fewer decks to 1 in 17 with 1 deck) and the wins and losses should be the same (by symmetry as they say in mathematical circles).

I used such a test when looking for bias in my shuffling algorithm and found if you used a hashing method (e.g. put the cards in one of N boxes where N>5*number of cards in deck, rather than a proper method) then there was a bias for cards to stay in the same order.

Success of doubling up "n" times is just

1/2^n

So 1, 2, 3, 4... successes in a row are 1/2, 1/4, 1/8, 1/16...and so on. Being well above 1 in 16 on 4 doubles is quite fortunate.

The probable outcomes for me in the game are Win, Lose, Push.

In a coin flip, true 50/50, I should have the typical results. Based on the fact that there are 3 probable results 50% of the time my opponent benefits from all the other possibilities affecting results and 50% of the time I benefit so I used .4165 as my number of possible outcomes.

So my thinking is leading me to believe that my chances rounded on multiple double ups are as follows;
2x= 1/5.75 attempts
3x=1/13.75
4x=1/33
5x=1/79

where it should be if there were only 2 probabilities
2x=1/4
3x=1/8
4x=1/16
5x=1/32

Quote: MaxPen

I understand that with 1 card being dealt to 5 people the chance of me having the highest card or tying with another player/s is 1/5.
Also, if I were to make a side bet with 1 of the 5 players my chances of winning would be 1/2 between them and I based on our 2 cards if we push on ties. However, sometimes we would both be tied with or beat by the other 3 players.

This makes the double up feature on a video poker machine a 50/50 proposition for a single double up attempt. I am assuming against the same player over and over for a first double attempt.

My question regards doubling multiple times. I find that my results of successfully doubling 1x is 1/2, 2x is very close to 1/4, 3x in a row are slightly more than 1/8, doubling successfully 4x in a row is significantly more than 1/16, 5x in a row has not been seen by me in hundreds of attempts.

Based on this I believe the double up feature containing 5 cards to be a 50/50 for a single wager but I would like to know how to figure out the chances for multiple doubles 2x,3x,4x,5x, etc.

I think you've got the gist of it; each time (assuming a non-predetermined result; I personally don't think those double-up features are random) you have a 50-50 shot, so 1 is 50%, 2 is 25%, 3 is 12.5%, etc.

Due to the introduction of more than two probabilities I believe the double up feature is only 50/50 up to and until the completion of the first attempt. After the initiation of second successive attempt you are basically playing a slot machine if there are more than 2 cards dealt out and 2 possible outcomes. I believe the house edge on the 2nd successive attempt and beyond to be equivalent to betting the any seven bet in craps.

Quote: tringlomane

Success of doubling up "n" times is just

1/2^n

So 1, 2, 3, 4... successes in a row are 1/2, 1/4, 1/8, 1/16...and so on. Being well above 1 in 16 on 4 doubles is quite fortunate.

Should have said less than more(corrected). The reason for my question is that from the beginning there is more than 2 possible outcomes. Win, Lose, and Push. If I push the first attempt it is inconsequential. I don't lose money. I get to try again. If I push after the the 2nd successful consecutive attempt I now have to fade the possibility of losing 6 times before I get to my desired result of doubling 5x initial starting wager. Then there is the other 3 cards. It just seems, to me, to be some kind of paradoxical problem. I don't see how this is like flipping a coin with only two possible results. I do see how the first attempt is 50/50 with the house as to whether I lose money or win money with an additional option of being neutral. Also 20% of the time I only have the option to lose, possibly push, and not win at all which is never the case with a coin flip.

The probability that a random card is strictly greater than another random card (from a single deck) is 8/17.

So, the probability for n successes in a row is (8/17)ⁿ

Quote: PeeMcGee

The probability that a random card is strictly greater than another random card (from a single deck) is 8/17.

So, the probability for n successes in a row is (8/17)ⁿ

That appears to be a piece of the puzzle. If I was playing the war sidebet in BJ that would be all the info needed. On a 5 card double up feature there is in essence 3 more spots playing that could affect results. Anyone else have a way to discount that and factor the 20% of the time possibility of only a lose or push scenario?

Quote: MaxPen

Quote: PeeMcGee

The probability that a random card is strictly greater than another random card (from a single deck) is 8/17.

So, the probability for n successes in a row is (8/17)ⁿ

That appears to be a piece of the puzzle. If I was playing the war sidebet in BJ that would be all the info needed. On a 5 card double up feature there is in essence 3 more spots playing that could affect results. Anyone else have a way to discount that and factor the 20% of the time possibility of only a lose or push scenario?

I’m not well versed in video poker, so maybe I’m missing something here.

From my understanding, you are only concerned about if your card is higher than the dealer’s (both random cards). The three other cards have no influence on this probability.

I'm not sure what you mean by 20% probability of lose or push.

There are 5 cards dealt out on the double up feature. The machine has one and the other 4, I can choose 1. 1 out of 5 times I will have the lowest or be tied with the lowest card with no possibility to win against the machine's card.

Quote: PeeMcGee

Quote: MaxPen

Quote: PeeMcGee

The probability that a random card is strictly greater than another random card (from a single deck) is 8/17.

So, the probability for n successes in a row is (8/17)ⁿ

That appears to be a piece of the puzzle. If I was playing the war sidebet in BJ that would be all the info needed. On a 5 card double up feature there is in essence 3 more spots playing that could affect results. Anyone else have a way to discount that and factor the 20% of the time possibility of only a lose or push scenario?

I’m not well versed in video poker, so maybe I’m missing something here.

From my understanding, you are only concerned about if your card is higher than the dealer’s (both random cards). The three other cards have no influence on this probability.

I'm not sure what you mean by 20% probability of lose or push.

If you pick the same rank of card as the dealer has, its a push.

Some people claim that out of the 4 card they have to chose from, once they pick a card, they notice frequently that none of the remaining cards even had a chance to win.

I have heard this many times over the years.

personally i just pick the same card each time no matter what. I don't look at the other cards they are meaningless to me.

Quote: MaxPen

The probable outcomes for me in the game are Win, Lose, Push.

i simulated this a few years ago as i did not believe at that time and even now
that one double-up round with 5 cards was 50/50

here is what i gets and gots (from the simulations and calculations)
when the 1st card is for the Dealer, it is always shown on the machines i have played,

and one can choose one of the other 4 cards

push = 3/51
win = 24/51
lose = 24/51
(this agrees with another poster of 8/17)

so to win the next 5 in a row = (24/51)^5
or abouts 1 in 43.3 (not the 1 in 32 for a true 50/50 game)

of course, not counting the push, it is a true 50/50 game

but i (one) has to count the push as it does happen and
really does not break a win streak
or
start a lose streak


the average number of games (trials) to play (a push counts as a game played) to get 5 in a row
(including a push that does not break the streak)
from me Markov chain = 65.875

a 50/50 bet requires only 62 trials on average for a run of length 5

the chance for a run of 5 in X number of games played:
10: 0.100183717
29: 0.338026235
47: 0.505052623
147: 0.901600306
289: 0.990074074 <<<< 1 in 100 still require more than 289 games played to see the 5 run

hope this helps out
i did not show my work other than me photo of it

Sally

Quote: MaxPen

There are 5 cards dealt out on the double up feature. The machine has one and the other 4, I can choose 1. 1 out of 5 times I will have the lowest or be tied with the lowest card with no possibility to win against the machine's card.

Yea, that can happen. Such a scenario is included in the 9/17 probability that you push or lose.

Quote: AxelWolf

Some people claim that out of the 4 card they have to chose from, once they pick a card, they notice frequently that none of the remaining cards even had a chance to win.

This should occur roughly 23.1% of the time. Is the argument that this happens more frequently than it should?

Quote: PeeMcGee

Yea, that can happen. Such a scenario is included in the 9/17 probability that you push or lose.


This should occur roughly 23.1% of the time. Is the argument that this happens more frequently than it should?

It's not my argument however people tend to notice when it happens. But they don't notice when all 4 card actually beat the dealers card. It's probably selective memory.

BBB is a moderator here and knows the Wizard (I'm fairly sure he's in the random camp) She's a smart educated person, someone with table game design experience and a long time gambler, Yet she said this

Quote: beachbumbabs

(assuming a non-predetermined result; I personally don't think those double-up features are random)

IMO that's scary. Usually you hear that stuff from the crazy people.

---------------------------------------------------------------------------------------

Here's one aspect that MANY people have experienced. Once you get to lets say double up # 4 or 5 it should be 50/50 (not including ties). There's many complaints it's significantly lower than that.

In my experience the double ups have been fair. I have doubled up Royals and won, I have doubled up 3 of a kinds into hand pays. If double ups were gaffed, I would assume it would be a dominate feature in all casinos.

So, basically I thought my chances of doubling 5 times, successfully, was 1in79 attempts. Sally is saying 1in65. If we were flipping coins it would be 1in32 attempts.

Axel, no one is saying that the double is not fair. It just does not represent and play as a true 50/50 after the first attempt.

Am I interpreting this wrong?

Mustang Sally I really want to thank you.

Quote: MaxPen

So, basically I thought my chances of doubling 5 times, successfully, was 1in79 attempts. Sally is saying 1in65. If we were flipping coins it would be 1in32 attempts.

Axel, no one is saying that the double is not fair. It just does not represent and play as a true 50/50 after the first attempt.

Am I interpreting this wrong?

Mustang Sally I really want to thank you.

I didn't suggest YOU didn't think they were fair. Others do not think they are fair. BBB even said, she didn't think that they were.

I think BBB said that because, in order for you to have even remotely close results on successfully doubling up "multiple times as per attempts" using the feature, you would have to gaff the machine and have it not be random.

Why the casinos do not have the feature activated all the time is beyond me. I have a couple of thoughts on that though.

One assurance (if you can call it that) is many casinos don't even have this option, and when they do doubling up doesn't earn any points.

Quote: djatc

One assurance (if you can call it that) is many casinos don't even have this option, and when they do doubling up doesn't earn any points.

The reason is that on each individual attempt there is no advantage for them. However, there is an inherent advantage against a player going for multiple successive attempts. I have some thoughts, that I would rather not post about, as to why I would not want it activated, especially on larger denomination machines, if I owned the casino.

Quote: MaxPen

The reason is that on each individual attempt there is no advantage for them. However, there is an inherent advantage against a player going for multiple successive attempts. I have some thoughts, that I would rather not post about, as to why I would not want it activated, especially on larger denomination machines, if I owned the casino.

casinos don't want it because it is fair. it slows down the game that is normally -EV in trade for a break even situation.

Also there was a player profitable glitch that involved machines with double ups, that scared casinos. They went around and disabled them because that glitch only showed up if this feature was enabled.

Quote: AxelWolf

I didn't suggest YOU didn't think they were fair. Others do not think they are fair. BBB even said, she didn't think that they were.

I said that I don't think they are random. "Fair" would be an entirely different question. A lot depends on how it's structured, whether you're offered 5 new picks each round, for example, or you're picking from the remaining cards (which may or may not be locked in anyway). Others have a hi/lo structure. Some don't display what you didn't pick, others do. If, for example, you're trying to pick a card that will beat the dealer, and it's a game that displays results out of 5, sometimes it shows there was only one winner that round, sometimes it shows you had 3 shots of 5, sometimes it shows you could only tie or lose, and there are other possible results.

I guess it's possible your results might be random, but based on the instant you pushed the "double-up" button, not on which cards you picked on the bonus screen. Similarly to many slots, where the instant you touch "spin" or "go" or whatever, your result has been determined, and the rest is entertainment. And, since you don't know what the result of that button-push was (whether it's going to let you be successful 1x this time, 3x the next, 2x the next), sometimes you're only going to get so far before you bust, and other times you will have stopped short of the best result offered.

I just can't equate those "double-up" bonus games with a genuine 50/50 shot, and as MS parsed it out, it's not one anyway. 8/17, right?

Quote: beachbumbabs

I said that I don't think they are random. "Fair" would be an entirely different question. A lot depends on how it's structured, whether you're offered 5 new picks each round, for example, or you're picking from the remaining cards (which may or may not be locked in anyway). Others have a hi/lo structure. Some don't display what you didn't pick, others do. If, for example, you're trying to pick a card that will beat the dealer, and it's a game that displays results out of 5, sometimes it shows there was only one winner that round, sometimes it shows you had 3 shots of 5, sometimes it shows you could only tie or lose, and there are other possible results.

I guess it's possible your results might be random, but based on the instant you pushed the "double-up" button, not on which cards you picked on the bonus screen. Similarly to many slots, where the instant you touch "spin" or "go" or whatever, your result has been determined, and the rest is entertainment. And, since you don't know what the result of that button-push was (whether it's going to let you be successful 1x this time, 3x the next, 2x the next), sometimes you're only going to get so far before you bust, and other times you will have stopped short of the best result offered.

I just can't equate those "double-up" bonus games with a genuine 50/50 shot, and as MS parsed it out, it's not one anyway. 8/17, right?

OK Sally JR.

Lets talk about IGT normal game king double up. That's what he is obviously talking about.

So you don't think its random? but you think its fair? Why make it nonrandom? Gaming approves a non random VP?

The push is really not an issue, that's just making it take longer to achieve his goal. it shouldn't affect anything.

When does it show you you could only tie? IGT double up feature on standard machines you never have the possibility of only a tie. Example: dealer has x card, even if 3 other cards are the same, the remaining card must be higher or lower.

Quote: AxelWolf

OK Sally JR.

Lets talk about IGT normal game king double up. That's what he is obviously talking about.

So you don't think its random? but you think its fair? Why make it nonrandom? Gaming approves a non random VP?

The push is really not an issue, that's just making it take longer to achieve his goal. it shouldn't affect anything.

When does it show you you could only tie? IGT double up feature on standard machines you never have the possibility of only a tie. Example: dealer has x card, even if 3 other cards are the same, the remaining card must be higher or lower.

The push is a huge issue in multiple consecutive double up attempts.

As far as tie goes; machine has 8 and I selected 8 out of the four cards and the other 3 cards are 7,6,3. I had 1 way out of 4 to tie and 3 ways to lose. Another, machine has 7 and I have selected 7 out of the 4 cards while the other 3 are 10,j,k leaving me 1 way to tie and 3 ways that could have won but didn't. The tie puts me at an extreme disadvantage when attempting "multiple" double ups as compared to flipping a coin "true 50/50". For a single wager it is a 50/50 proposition with a 23% chance of resulting in a push.

This problem is similar to the 1/6 and 1/11 dice problem. It matters how you get there.

Quote: MaxPen

The push is a huge issue in multiple consecutive double up attempts.

As far as tie goes; machine has 8 and I selected 8 out of the four cards and the other 3 cards are 7,6,3. I had 1 way out of 4 to tie and 3 ways to lose. Another, machine has 7 and I have selected 7 out of the 4 cards while the other 3 are 10,j,k leaving me 1 way to tie and 3 ways that could have won but didn't. The tie puts me at an extreme disadvantage when attempting "multiple" double ups as compared to flipping a coin "true 50/50". For a single wager it is a 50/50 proposition with a 23% chance of resulting in a push.

This problem is similar to the 1/6 and 1/11 dice problem. It matters how you get there.

Unless its causing you to play extra - EV hands, then it does not affect what your doing. You lose or gain nothing when you tie. You simply try again with new cards. Assuming it does not stop (max number of times you can double, limiting you prior to your goal.)

So far it has been determined that the offer is not a true 50/50 at all and negatively affects the player based on just the tie factor alone.
This results in changing the outcome of multiple consecutive attempts from 1/2^n to 8/17^n. However as MS pointed out and to which I agree, although not to the detrimental level of expectation that I initially thought, the other factors make it even worse.

Quote: AxelWolf

You lose or gain nothing when you tie. You simply try again with new cards. y

You pay opportunity costs by having to attempt again. Additionally, your bankroll is jeopardized "-ev situation" when the machine has a 10 and you have 4 choices of say 8,7,6,5, which is definitely not possible in a 50/50 situation. On the first attempt that situation is irrelevant to your bankroll but for every multiple attempt it is highly relevant.

Quote: MaxPen

You pay opportunity costs by having to attempt again. Additionally, your bankroll is jeopardized "-ev situation" when the machine has a 10 and you have 4 choices of say 8,7,6,5, which is definitely not possible in a 50/50 situation. On the first attempt that situation is irrelevant to your bankroll but for every multiple attempt it is highly relevant.

What about when the dealer has a 4 and the cards are 8,7,6,5? Now you are guaranteed a win. Asian unless you are limited to the number of times you can double up it shouldn't have any affect on reaching your goal.

As far as bankroll management that's a different story.

I remember seeing some ploppy at a big strip casino getting a handpay on vp, but no paying hand on display. I asked him what he hit and he told me he kept doubling up. It was a $25 game and i am assuming he doubled a middling hand like a straight or flush a couple of times.

A 5 unit win per credit bet would win you $750. Only need to double 3 times to win $6k.

This casino still offers double up, but has a very tight reputation. I wonder why.....

Quote: djatc

I remember seeing some ploppy at a big strip casino getting a handpay on vp, but no paying hand on display. I asked him what he hit and he told me he kept doubling up. It was a $25 game and i am assuming he doubled a middling hand like a straight or flush a couple of times.

A 5 unit win per credit bet would win you $750. Only need to double 3 times to win $6k.

This casino still offers double up, but has a very tight reputation. I wonder why.....

Yes, and if he tries that flipping a coin he has a 1in8 chance of success costing him 8 tries at 750 per for a total of 6000. If you do what the ploppy did using the double up feature you would be successful rounded down 1in 9.5 attempts costing you at 750 per for a total of 7125.
Hopefully, more casino's will activate the double up feature as a result of this thread. I tried high and low to find info about this before posting the question. Casino's should activate all available double up features.

This discussion started to spill in to another thread. I had some responses from this thread to spark discussion:

If you're going to do combinational analysis, where you say "what are my odds of doubling 4 times in a row?" then yes the answer would be .5^4... with each round being .5^2, .5^3, etc, etc. Basically your odds get cut in half every time you want to parlay another double.

However... Couldn't, and shouldn't, these be looked at as independent trials? Supposedly the deck is re-shuffled and there are fair odds of any card being dealt to any of the 5 spots. This would erase any "state" of the game and, for my understanding, create independent trials. So on each trial you do still have a 50/50. If you don't have a set number of double ups in mind, wouldn't each attempt be it's own 50/50 still?

I don't agree with a couple of statements. One being:

Quote: MaxPen

There are 5 cards dealt out on the double up feature. The machine has one and the other 4, I can choose 1. 1 out of 5 times I will have the lowest or be tied with the lowest card with no possibility to win against the machine's card.

The cards are all completely random. Thus you may have a round where the dealer has an ace, and your cards to choose from are 2-3-4-5. Then again you are just as likely to have a round where the dealer has a 2 and your cards are A-K-Q-J. This is supposedly completely random and shouldn't affect the odds of the independent trials.

Everyone seems to be stuck on the fact that there is 5 cards which is really confusing to me. The other cards are essentially discards. They are worthless to the outcome. Just your card and the dealers card matter and they are both "random." Showing you the other unchosen cards is the same as just dealing more cards off the deck and showing you more and more cards. Who cares? These cards didn't effect the outcome at all. If this were blackjack and card counting, they're just unseen cards and have no effect on the RC.

Also, pushes should be ignored as nothing happens. No wins, no losses, just another re-do of a "random" draw. Sally even states ignoring ties it's 50/50. I don't believe her math on page 2 is correct (big statement, I know) because she's assuming if you go W-W-W-W-P-W that's not 5 wins in a row. She states a push could be the start of a losing streak. Yeah, that has no mathematical barring at all though. You're just as likely to win your next hand after a push as lose. That's like saying the shooter through the dice out of the tub, the next role could be the 7 now!! The dice leaving the tub is the same as a push, nothing happens and you do the game again, which is an independent trial from the previous games.