Pushes don't break the streak. They are just ignored.
Online casino is offering 10 times $25 bet (250 free credits) for getting a 5 hand win streak in a day. Does this sound like a positive expected value promotion?
Pushes don't break the streak. They are just ignored.
Online casino is offering 10 times $25 bet (250 free credits) for getting a 5 hand win streak in a day. Does this sound like a positive expected value promotion?
Email: [email protected]
If you win or push roughly 54% of the time then it's (0.54^5) to give percent chance of it happening. If the push doesn't advance the steak then the odds are longer. Either way 10-1 payoff ain't close to true odds of a 5 win streak. If pushes are ignored you win roughly 43% of the time so the odds are far longer.
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Last edited by bigplayer; 11-19-2021 at 12:31 AM.
We calculated this probability earlier. Let me quote 21forme:
“As pushes don't count, you are left with 91.2& of hands. Within that, the ratio of wins to losses is 47.5% to 52.5%.”
So, the probability of winning 5 hands in a row is
0.475^5=2.418%, that is, once in every 41 hands of continuously playing.
If the table minimum is $10, you expect to lose 41x10x0.005=$2.05 before you get a free bet of 10x$25. This sounds too sweet to be true?
Last edited by aceside; 11-19-2021 at 09:02 AM.
Aceside, forum etiquette calls for using the quote button (or highlighting a section of a post and then clicking quote), so the source post can be verified.
I considered responding to MWP's question, but as he doesn't respond to my follow-up questions, I decided not to help him any longer with his endless barrage of "is this promo beatable?" posts.
If he bothered to do any of his own research (teach a man to fish...), he could have easily found the answer in the BJ sidebet section of WoO.
No one is calculating this correctly. Streak probabilities are very complex to calculate. Just saying 0.475^5 gives you only the probability of playing five hands and winning them all. It has nothing to do with playing x number of hands and wondering if a winning streak of, say, 5 will occur.
Some examples: 50 hands: 47.6% probability
100 hands: 74.5%
200 hands: 95.3%
300 or more hands: virtual certainty
So, giving you $250 in free bets seems impossible to me. It's impossible to play more than a couple of hours without winning five in a row--unless, of course, you're cheated. And, since it's online, I'd say the chances of that are extremely high.
Don
This indeed sounds very complicated. Often when the question is "how is the probability of something to happen at least one time in n trials", it is easier to compute the contrary (probability of this not to happen) and subtract it from 100%. But the probability of no streak of 5 consecutive wins in e.g. 50 trials is not easy to calculate either.
Thanks Don and everybody else who contributed to this thread. I did get a couple of 5 win streaks in a row in less than 2 hours. The first one was on Spanish 21 which I found out didn't count because the game was not blackjack. I then got another on blackjack. Interesting twist in my 5 hand winning streak. One of the hands in the streak was actually a split and then a double down. I lost one hand of the split but won the double down hand. There is always a message that flashes up on the screen if you have 3 or more wins in a row. Yeah, I got my 250 free credits.
The other day when I was playing to get my 5 hand winning steak, it was quite difficult. Today when playing on a different promotion I must of got 6 or 7 in a couple of hours with the first one being less than 5 minutes.
Just a note for Aceside. The 250 free credits is based on on your lowest wager for the day with a max of 250 free credits. If you only play $10 game you will only get $100 free credits. Even if you play $25 min and then drop to $10 min you free credits are based on your lowest wager.
I sure hope there is no cheating. The game is regulated by the State of MI and is not some off shore casino. If there is any doubt, they always seem to give the player the best deal. However, things have not been going well for me at this casino for the last couple of months. I actually lost around $12,000 from my high a couple of months ago. However, on the other hand I'm still having my best year ever. I'm presently Diamond status and only need less than 20,000 points out of 1,000,000 to reach the top tier of Onyx by the end of the year. Maybe I will even get better promotions then.
Well back to play my latest promotion. I'm only half way thru. I get $1 free credit for every $100 bet up to $1,000 in free credits. I think last time they gave $2., but I'm also chalking up points for the leaderboard at the same time. No free sports bets this week.
Last edited by Midwest Player; 11-21-2021 at 12:17 PM.
Email: [email protected]
Yeah, I just made Onyx tier status. And to top off the weekend I just finished my last promotion over a $1,000 ahead.
WELCOME TO
ONYX STATUS
xxxxx, you've achieved Onyx Status.
HERE'S $500 DK DOLLARS FOR REACHING YOUR NEW STATUS TIER.
You've done it - you've earned Onyx Status for the remainder of this year and for the next year. Click 'Claim Reward' below to get your $500 DK Dollars for free.
Email: [email protected]
Here is how to find rounds required before a 3 hand winning streak(instead of 5 hand winning streak).
Suppose Rn = Rounds required after you get n winning streak. p= win probability, q = loss probability and r = tie probability and z = 1- r
Therefore Rn = p * Rn+1 + q * R0 + r * Rn + 1
A) After you get a 3 hand winning streak, rounds required to play further is zero, so R3 = 0 -----------Eq1
B)
R2 = p * R3 + q * R0 + r * R2 + 1
(1-r)R2 = p * R3 + q * R0 + 1
z*R2 = p * R3 + q * R0 + 1
z*R2 = q * R0 + 1-----------Eq2
C)
R1 = p * R2 + q * R0 + r * R1 + 1
(1-r)R1 = p * R2 + q * R0 + 1
z*R1 = p * R2 + q * R0 + 1-----------Eq3
D)
R0 = p * R1 + q * R0 + r * R0 + 1
(1- q - r)R0 = p * R1 + 1
p*R0 = p * R1 + 1-----------Eq4
From Eq2, Eq3 and Eq4, you can simplify it to :-
R0 = (p^2 + pz + z^2)/(pz^2 - p^2q - pqz)
Suppose p = 0.4332, q = 0.4788, r = 0.088, z = 1 - r = 0.912
R0 = (p^2 + pz + z^2)/(pz^2 - p^2q - pqz) = 17.39934
Average rounds required before you get a 3 hand winning streak is 17.39934. If pushes break streak(p=0.4332, q = 0.5668, r = 0, z=1), average rounds required is 19.93796.
Please note that for a specific game where, p = q = 0.5, r = 0, (coin flipping), Average flips required before you get a 3 winning streak is exactly 14 !
So you can use the same approach to find rounds required before get a 5 hand winning streak. This is a tedious process !
updated :-
Just an estimate for your actual case, average rounds required before you get a 5 hand winning streak(pushes don't break the streak) is 84, so your net profit = 250 - 84 * 25 *0.005 = 239.5.
Last edited by James989; 11-20-2021 at 07:32 AM.
Last edited by aceside; 11-20-2021 at 01:07 PM. Reason: I revised the unit here and number here.
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