Do Past Results Affect Future Results?

Do Past Results Affect Future Results? (In BJ, Consider An Infinite Deck)

  • Yes, Past Results Do Affect Future Results (If An Infinite Deck Was Used In BJ)

    Votes: 5 27.8%
  • No, Past Results Have No Affect On Future Draws

    Votes: 13 72.2%

  • Total voters
    18
#1
(For BlackJack and other card games, please imagine if those games used an infinite deck. An infinite deck would have an infinite number of each type of card).

This question is of the utmost importance. Most people give an answer without even spending a moment studying the past results. However, I have spent many - many, many, many - moments studying the past results.

Some of the "alternative" lottery experts out there believe that the past results - spins, draws, hands, etc., - have an affect on future results. An example they give is this: If the past results do not affect future results, then how come a certain result will not occur endlessly? After-all, it's 50% heads or tails. Why wouldn't tails come out hundreds, even thousands of times in a row if there was no memory of the past? Yet it never will come out so biased. If a real coin was flipped as a test, there probably wouldn't ever be a string of more than 25-30 consecutive flips either way.

In the past I have leaned toward the belief of past draws affecting the outcome of future draws. But now after studying the Pick3 lottery for quite some time, I have begun to change my opinion on this question. I am really beginning to believe that the past results DO NOT have any affect on future results.

What are your opinions? What do you think? Please vote and comment. Thanks!

Licentia
 
#2
John Grochowski

I found this article online at this website: http://grochowski.casinocitytimes.com/article/past-results-have-no-effect-on-future-outcomes-37083

Past results have no effect on future outcomes
11 September 2007

By John Grochowski

For about a lifetime now, I've been telling gamblers that in most games, past results have no effect on future outcomes. (Blackjack is an exception, since each card dealt out changes the composition of the remaining deck, altering the odds of the game.)

The roulette ball has landed on eight black numbers in a row? The chances of the next one being black are still 18 in 38. The craps shooter has gone a dozen rolls without a 7? The chances of the next roll being a 7 are 1 in 6, same as always. You've just hit a big slot jackpot? The odds against the big jackpot combination turning up on the next spin are the same as the spin before.

From time to time, I'm asked how that possibly can be. If in the long run, 18 of every 38 spins of the roulette wheel will be red, and there have been eight black numbers in a row, shouldn't you be betting on red? If there have been a dozen craps rolls without a 7, isn't 7 "due"? After all, in the long run, one of every six rolls will be a 7.

The answer is that there doesn't have to be any makeup time. In any game, there will be unusual streaks that seem to defy the odds. Casinos and their customers count on that. Without such streaks, there would be no winners --- we'd all just be handing over our money at a prescribed percentage.

Eventually, such streaks just fade into statistical insignificance, overwhelmed by the sheer number of trials that go on in casinos.

For a simple example of how all this works, I like to use another chance event: flipping a coin. Just as in rolling the dice, spinning the roulette wheel or pulling the slot handle, past coin flips have no affect on future outcomes. There's a 50-50 chance of heads or tails on every flip, regardless of what has gone on before.

This came up recently on an Internet message board in which I participate in the odd discussion --- the odder, the better. The manager of a softball team said a coin flip determined who was the home team, and that he'd lost 13 consecutive flips. What were the odds?

On a 50-50 event like flipping a coin, calculating the odds is easy. Your chances of losing a single flip are 1 in 2. Your chances of losing 13 in a row are 1 in 2 to the 13th power --- start with 2, then multiply by 2 12 times. The answer is 1 in 8,192. In 8,192 sets of 13 coin flips, on the average you'll lose all 13 of them once.

Another poster on this message board suggested you could improve your odds with some judicious selections. "If it hits seven heads in a row, isn't it more likely to come up tails the next time, because of the law of averages?"

No, the odds never change. If heads come up seven times in a row, the odds on the next flip are still even. There never has to be a makeup period --- in the long run, any unusual streaks just fade into statistical insignificance.

"But in the long run, things even out," the poster said. "That leads me to believe that, since you know things will even out, you're more likely to hit the other side if there's been a run on one side."

Recognize that train of thought? That's the same thing as a roulette player who thinks that after eight black numbers, red is due, or a craps player who thinks that after a dozen rolls without a 7, the shooter must be due to roll a 7.

But any evening out to be done is purely statistical, whether we're talking roulette, craps, slots or coin flips.

Let's say the coin flipper starts with 7 heads in a row. At that point, 100 percent of flips have been heads. The expectation is that 50 percent should be heads, so it appears there's evening up to do.

Now let's say future flips come up exactly 50 percent heads and 50 percent tails. (Things rarely come out quite so neatly, of course.) After another 100 flips, 57 have been heads, and 50 tails. Now only 53.3 percent of the flips have been heads, only 3.3 percent above expected average. On thousand flips after the first seven, we have 507 heads and 500 tails. 50.3 percent have been heads. One million flips after the first 7, 500,007, or 50.000035 percent, have been heads.

There has been no "making up," period, but we're right on 50 percent heads and 50 percent tails. The original streak of seven heads has just faded against the statistical background.

Streaks happen, and if you're on the winning side, streaks are meant to savor. But there doesn't have to be any equal but opposite streak to make the odds come out right. Given enough trials, normal results from the streak onward are enough to keep the games on track.
 

Katweezel

Well-Known Member
#3
28.20% is a very high figure, right?

I voted yes in direct insolence to the Law of Independent Events. I voted yes because of my own research, particularly with the dealer hand results. The dh has a long-term bust average of 28.20% V player's average of a little over 15%. The difference is because the player must 'go first' and the dealer is forced to hit stiffs. So, just say in a 6-deck shoe, with 5 players and 75% pen, and the dealer had NIL busts for the entire 16-round shoe.

At 28.20%, she should have had around 5 busts, average. I noticed many times that when that happens, the next hand-shuffled shoe - or the one after - may well be the one where there could be an 'adjustment' where perhaps 7,8, or more dealer busts happen. (And if this happens - and you expected it - wow! it could be bonanza time!
 
#4
Licentia said:
I am really beginning to believe that the past results DO NOT have any affect on future results.

This is probably one of the best things I have ever seen you say here on the forums. Granted I have been out of it for a good while and have not seen any of your recent posts but please believe it!


It is possible that in a coin flip it can turn out to be heads or tails hundreds or even thousants of times in a row however the likelyhood of that occuring is damn near -1% ;)

edit: PS sorry for the sig...
 
#7
Katweezel said:
At 28.20%, she should have had around 5 busts, average. I noticed many times that when that happens, the next hand-shuffled shoe - or the one after - may well be the one where there could be an 'adjustment' where perhaps 7,8, or more dealer busts happen. (And if this happens - and you expected it - wow! it could be bonanza time!
Yeah I have noticed shoes like that also, where one shoe is at one extreme, and the next shoe at another extreme. Sitting with a gentleman in the casino who was very friendly and had somewhat of a mind for strategies I would judge. We kept complaining about the first shoe, how awful it was. The second shoe made up for everything however. At least for me it did. Not for everyone else.

Now I am remembering the words I used to speak here. How my strategy would only work if I stayed at the table playing through the entire streak, but if I left the table and returned another day it wouldn't work. But why is it so difficult then in Pick 3 to find any correlation between past draws and future draws? Everything continually works out to 50%, 50%, 50%! It makes me ill!

Licentia
 
#8
standard toaster said:
Originally Posted by Licentia
The odds of picking 1 number correct out of 6 is: '1 in 8.17.
I was reading some of Ion Saliu's pages last night, and I saw this on one of them:

"One lotto number is combined with the rest of the numbers (N-1) taken 5 at a time. C(48, 5) = 1712304 combinations. That's how many times one particular 6/49 lotto number appears in the total amount of combinations. P = 1712304 / 13983816 = '1 in 8.17'."

Now I did a websearch and I find this old thread of mine showing the website where I got that number from:
http://www.blackjackinfo.com/bb/showthread.php?p=94196
http://www.lottery-student.co.uk/guide/calculating-odds.asp (Archive copy)
 
#9
Katweezel said:
At 28.20%, she should have had around 5 busts, average. I noticed many times that when that happens, the next hand-shuffled shoe - or the one after - may well be the one where there could be an 'adjustment' where perhaps 7,8, or more dealer busts happen. (And if this happens - and you expected it - wow! it could be bonanza time!
I decided to pm this post to you Katweezel.

Licentia
 
#10
Sonny, Please Help!

I am determined to better my algebra skills as I think doing so will pay great dividends.

Here is the equation I don't understand:

Code:
 1 – (1 – 1/1000)500 = 1 – (0.999) 500 = 0.3936 = 39.36%
I do not understand how the author gets the result 0.3936

I would do the math this way:

Code:
 1 - (1 - 1/1000)500 ->
 1 - (1 - 0.001)500 ->
 1 - (0.999)500 ->
 1 - 499.5 ->
 498.5
Licentia
 

Sonny

Well-Known Member
#11
Licentia said:
Here is the equation I don't understand:

Code:
 1 – (1 – 1/1000)500 = 1 – (0.999) 500 = 0.3936 = 39.36%
I do not understand how the author gets the result 0.3936

I would do the math this way:

Code:
 1 - (1 - 1/1000)500 ->
 1 - (1 - 0.001)500 ->
 1 - (0.999)500 ->
 1 - 499.5 ->
 498.5
Licentia
Technically you’re both wrong, but I’ll give you partial credit. Double-check the last two lines of your equation and I think you’ll see the error. It’s a tiny error but it could make a huge difference in your results.

-Sonny-
 

UK-21

Well-Known Member
#12
Licentia said:
(For BlackJack and other card games, please imagine if those games used an infinite deck. An infinite deck would have an infinite number of each type of card).

. . . This question is of the utmost importance. Most people give an answer without even spending a moment studying the past results. However, I have spent many - many, many, many - moments studying the past results. . ..
What are your opinions? What do you think? Please vote and comment. Thanks!

Licentia
On my first visit to a casino ever I was sitting with Mrs N at a roulette table, and the number 33 came up three times in succession. I thought "Wow! what are the odds of that happening?"

So: (36 x 36 x 36)/1 or 46,656/1

Stick a quid on any number and let it ride for three spins and that's two years earnings in your pocket (based loosely on the UK's average wage).

What about a fourth time? What are the odds of it happening four times in succession? (36 x 36 x 36 x 36)/1, or 1,679,616/1. Over one and a half million to one. So should I have put a quid on 33 so I never have to work again?

Of course not. The odds of it coming up again are the same as for any other number on the wheel, 36-1.

I found a good page to explain this on the WoO site.
 

Katweezel

Well-Known Member
#13
According to script

newb99 said:
On my first visit to a casino ever I was sitting with Mrs N at a roulette table, and the number 33 came up three times in succession. I thought "Wow! what are the odds of that happening?"

So: (36 x 36 x 36)/1 or 46,656/1

Stick a quid on any number and let it ride for three spins and that's two years earnings in your pocket (based loosely on the UK's average wage).

What about a fourth time? What are the odds of it happening four times in succession? (36 x 36 x 36 x 36)/1, or 1,679,616/1. Over one and a half million to one. So should I have put a quid on 33 so I never have to work again?

Of course not. The odds of it coming up again are the same as for any other number on the wheel, 36-1.

I found a good page to explain this on the WoO site.
Interesting Newb. Reminded me of the ole movie called The Sting. Robert Redford and his two pals conned a numbers' runner out of twelve grand cash. With Redford's 4g share in his clutches, he puts the lot on #7 at roulette, due to an intuitive message. The croupier reaches under the table and pushes a button as the ball is spinning. Of course, the ball lands in 7 but mysteriously jumps out, into a neighbour. Redford is stunned of course because he KNEW it was landing in 7. How did he know? Because he read the script. :grin:
 

Kasi

Well-Known Member
#14
standard toaster said:
It is possible that in a coin flip it can turn out to be heads or tails hundreds or even thousants of times in a row however the likelyhood of that occuring is damn near -1% ;)
Not really, if it's a fair coin :grin:

At what point might you draw the conclusion the coin is not random or fair vs the conclusion "cr*p happens" with a fair coin? :grin:
 

Kasi

Well-Known Member
#15
newb99 said:
The odds of it coming up again are the same as for any other number on the wheel, 36-1....
Wow, a roulette table with no zeroes.

PM me where this game is available.

Talk about voodoo heaven.
 
#17
Code:
 1 - (1 - 1/1000)500 ->
 1 - (1 - 0.001)500 ->
 1 - (0.999)500 ->
 1 - 499.5 ->
 498.5
Are you suggesting that 498.5 is wrong because it's supposed to be a negative?

-498.5

???

That equation is Ion Saliu's equation by the way. The 500 is supposed to be a "power" I believe, not necessarily a .999*500. But I don't know how to write the 500 as a power in a post. I am not sure if that makes any difference though.

I will post the page below where the equation is found, but I will not make it a link to prevent him following it back here:

saliu.com/formula.htm

Your help Sonny is much appreciated!

Licentia
 
#18
newb99 said:
37 slots on the wheel. 1 chance in 37 of any number coming up. 36-1 against ????
He's used to the 37-1 Roulette Wheel with 38 numbers. Kasi mistook 36-1 for 1 in 36 chance. After reading Kasi's post I fell victim to reading it wrong also!

Licentia
 

aslan

Well-Known Member
#20
Licentia said:
Code:
 1 - (1 - 1/1000)500 ->
 1 - (1 - 0.001)500 ->
 1 - (0.999)500 ->
 1 - 499.5 ->
 498.5
Are you suggesting that 498.5 is wrong because it's supposed to be a negative?

-498.5

???

That equation is Ion Saliu's equation by the way. The 500 is supposed to be a "power" I believe, not necessarily a .999*500. But I don't know how to write the 500 as a power in a post. I am not sure if that makes any difference though.

I will post the page below where the equation is found, but I will not make it a link to prevent him following it back here:

saliu.com/formula.htm

Your help Sonny is much appreciated!

Licentia
I got 0.3936210551388153
 
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