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# Thread: On Average How Many Blackjack Hands Have to be Played to Win 5 in a Row

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## On Average How Many Blackjack Hands Have to be Played to Win 5 in a Row

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?

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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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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?

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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.

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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

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Originally Posted by DSchles
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.
Don
I have been thinking about this part too. The 0.475^5 gives you the probability of winning/pushing five hands in a row. Is this equivalent to winning exactly five hands, excluding pushes? Mathematicians needed.

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Originally Posted by DSchles
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.

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Originally Posted by aceside
I have been thinking about this part too. The 0.475^5 gives you the probability of winning/pushing five hands in a row. Is this equivalent to winning exactly five hands, excluding pushes? Mathematicians needed.
Just forget about the pushes. You don't win 47.5% of hands. The higher number accounts for the pushes. It's for resolved hands.

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Originally Posted by Midwest Player
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?

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.

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Originally Posted by James989
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 !
Let me verify the coin flipping case first. The probability of getting 3 heads in a row is 0.5^3=0.125, that is once in every 1/0.125=8 sets of 3-flips. In theory, if we flip 24 times, we would almost certainly have a 3-head streak.

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