Can somebody please clarify
In a fair coin toss (as an example), mathematically speaking how often should you expect to lose 4 decisions in a row?
I'm guessing it is once per 64 coin flips, as there are 16 possible combinations and only LLLL can exist once. but I think this is way off the mark as you would have overlaps when factoring in prior possible combinations.
So for example,
WWLL
LLWW would return 4 losses in a row if it occurred in that sequence. OR
WLLL
LWWL
There must be an easier way to figure this out, with all the possible combinations.
The reason for the question, I was running through something and after placing 232 bets, I'm seeing 4 losses in a row occur 8 times, and a maximum of 5 losses in a row once. So simply wanted to know if these figures are good, bad or as expected.
Thanks
Quote: unJonFour losses in a row is 1 in 16 not 1 in 64.
That was a typo, corrected...
Quote: CyrusVCan somebody please clarify a rudimental math questions, it would much appreciated.
In a fair coin toss (as an example), mathematically speaking how often should you expect to lose 4 decisions in a row?
I'm guessing it is once per 64 coin flips, as there are 64 possible combinations and only LLLL can exist once. but I think this is way off the mark as you would have overlaps when factoring in prior possible combinations.
So for example,
WWLL
LLWW would return 4 losses in a row if it occurred in that sequence. OR
WLLL
LWWL
There must be an easier way to figure this out, with all the possible combinations.
The reason for the question, I was running through something and after placing 232 bets, I'm seeing 4 losses in a row occur 8 times, and a maximum of 5 losses in a row once. So simply wanted to know if these figures are good, bad or as expected.
Thanks
As noted Your 1 in 64 is squiffy: There are 16 ways that 4 coin tosses could land
First, you need to determine how many sets of potential '4 losses in a row' you are going to encounter.
Let's take your 232 bets: There are 229 sets of 4 in a row that overlap each-other... 1,2,3,4... ,2,3,4,5,,. ... 228,229,230,231... 229,230,231,232
Now, for any set of 4, the probability of 4 losses is 0.5^4 = 1 in 16
So within your 232 wagers, on average you will have 229 x (1/16) = 14.3125 ( call it 14 ) sets of 4 consecutive losses.
Now... If you are also thinking about sets of 5 losses in a row, you need to decide whether you consider 5 losses in a row to include 2 lots of 4 losses in a row.
Looking at the 5 in a row part of the problem:-
For any set of 5, the probability of 5 losses is 0.5^5 = 1 in 32
Within your 232 wagers, on average you will have 228 x (1/32) = 7.125 ( Call it 7 ) sets of 5 consecutive losses. If you only saw 5 losses in a row once, your results were somewhat anomalous.
Enough to go on with?
5 losses in a row, I would not describe as two sets of 4 losses in a row, rather a set of 5 losses in a row.
Just so I'm correct with this, going back to 4 losses in a row, the expectation over 232 outcomes would be 14 and 5 losses in row it would be 7.
I'm getting 4 occurring 8 times and 5 only the once. Sure I understand that it is a very small sample size, yet both extremely promising and exciting.
For your information, I'm also seeing 3 losses in a row occur 17 times and 2 losses in a row 31 times.
232 bets placed, winning 124 losing 108 for a 53.4% win rate, not bad for a negative expectant game,
Damn, I need a excel macro wiz, to test things further....
Quote: CyrusVOneDear, Cheers for that, extremely appreciated, yes it is enough to go on with.
5 losses in a row, I would not describe as two sets of 4 losses in a row, rather a set of 5 losses in a row.
Just so I'm correct with this, going back to 4 losses in a row, the expectation over 232 outcomes would be 14 and 5 losses in row it would be 7.
I'm getting 4 occurring 8 times and 5 only the once. Sure I understand that it is a very small sample size, yet both extremely promising and exciting.
For your information, I'm also seeing 3 losses in a row occur 17 times and 2 losses in a row 31 times.
232 bets placed, winning 124 losing 108 for a 53.4% win rate, not bad for a negative expectant game,
Damn, I need a excel macro wiz, to test things further....
I suspect that when you are seeing 5 losses in a row you are ignoring one or both of the '4 losses in a row' within it. Similarly when you see 4 losses in a row are you not counting any of the 3 losses in a row' within those?
FYI I'm an Excel And Excel Macro whizz and if you need assistance, I will be happy to help (by PM)
the formula for the 'expected number of runs' has been around for a long time.Quote: CyrusVThere must be an easier way to figure this out, with all the possible combinations.
The reason for the question, I was running through something and after placing 232 bets, I'm seeing 4 losses in a row occur 8 times, and a maximum of 5 losses in a row once. So simply wanted to know if these figures are good, bad or as expected.
Thanks
Mine (found - with proof- from the internet) dates back to 2011
a=(p^run)*(1+((trials-run)*q));
this is for at least X
parameters:
p=1/2;
q=1-p;
run=4;
trials=232;
code
a=(p^run)*(1+((trials-run)*q));
aDec=1.*a;
a1in=1/aDec;
a
aDec
a1in
in pari/gp
(10:00) gp > p=1/2;
(10:00) gp > q=1-p;
(10:00) gp > run=4;
(10:00) gp > trials=232;
(10:00) gp > a=(p^run)*(1+((trials-run)*q));
(10:00) gp > aDec=1.*a;
(10:00) gp > a1in=1/aDec;
(10:00) gp > a
%8 = 115/16
(10:00) gp > aDec
%9 = 7.1875000000000000000000000000000000000
(10:00) gp > a1in
%10 = 0.13913043478260869565217391304347826087
mean # of runs of length 4 or higher = 115/16 (7.18750)
do the same for length 5 and higher and subtract that from 4 to get expected number of runs of length 4
I get 3.609375
in Excel
and a simple table from Excel
average number of runs
| run length X | X | X or more |
|---|---|---|
| 1 | 29.25 | 58.25 |
| 2 | 14.5625 | 29 |
| 3 | 7.25 | 14.4375 |
| 4 | 3.609375 | 7.1875 |
| 5 | 1.796875 | 3.578125 |
| 6 | 0.89453125 | 1.78125 |
| 7 | 0.4453125 | 0.88671875 |
| 8 | 0.221679688 | 0.44140625 |
| 9 | 0.110351563 | 0.219726563 |
| 10 | 0.054931641 | 0.109375 |
simulation data to see
at least 4 in a row
232 trials, 1 million times
grouped data
items: 1000000
minimum value: 0.00
first quartile: 6.00
median: 7.00
third quartile: 9.00
maximum value: 19.00
mean value: 7.19<<<<<
midrange: 9.50
range: 19.00
interquartile range: 3.00
mean abs deviation: 1.82
sample variance (n): 5.18
sample variance (n-1): 5.18
sample std dev (n): 2.28
sample std dev (n-1): 2.28
simulation data to see
exactly 4 in a row
232 trials, 1 million times
grouped data
items: 1000000
minimum value: 0.00
first quartile: 2.00
median: 3.00
third quartile: 5.00
maximum value: 14.00
mean value: 3.61
midrange: 7.00
range: 14.00
interquartile range: 3.00
mean abs deviation: 1.45
sample variance (n): 3.22
sample variance (n-1): 3.22
sample std dev (n): 1.79
sample std dev (n-1): 1.79
hope this helps out some
lots of math
It was basically a fair coin toss question, I deliberately poised the question this way, as there exists some negativity as well as misconception towards the game I am referring to on this board., so I avoided mentioning it.
All I wanted to know, was the expected frequency of 4 losses in a row, excluding 5 losses in a row, this was answered above.
Cheers
They estimated how often a team would win three in a row to start the series, then assumed each game after that would be a 50-50 tossup. They came up with three numbers. How often a team can expect to go up 3-0, how often that team will sweep in four and how often that team will lose in seven games. The mathematical odds almost exactly fit the past results.
The same guy predicted how often someone should pitch a perfect game. My numbers might be slightly off but he predicted one every X amount of games and that there should have been 27 perfect games in history. At the time there were 28.
You got the wrong answer above and 7crapd answered it correctly. The answer you refer to above is wrong in that it solved for “at least” four losses in a row and not “exactly” for losses in a row.Quote: CyrusVI wouldn't know where to start with those codes unless I could paste them into Excel. Not sure if this is relevant, but my initial question has nothing to do with expected streak lengths.
It was basically a fair coin toss question, I deliberately poised the question this way, as there exists some negativity as well as misconception towards the game I am referring to on this board., so I avoided mentioning it.
All I wanted to know, was the expected frequency of 4 losses in a row, excluding 5 losses in a row, this was answered above.
Cheers
the formula I showed was for at least 4 in a rowQuote: unJonYou got the wrong answer above and 7crapd answered it correctly. The answer you refer to above is wrong in that it solved for “at least” four losses in a row and not “exactly” for losses in a row.
and to get 'exactly 4 in a row' one subtracts the value from 'at least 5 in a row' from 'at least 4 in a row'.
Here is Excel in Google sheets. this is an old sheet. I had to find it first.
it should be easy to follow the formula in each cell.
https://docs.google.com/spreadsheets/d/1kHPMGTFkM-L4HDoyweCXxhyMJijFUtbP59Pcq43Eyzs/edit?usp=sharing
That figures seems way to low, in the real world (at the tables) I can encounter 4 losses in a row quite a few times within 70 trials, it is more or less guaranteed. Because no 4 column bet option survives any baccarat shoe without encountering an entire column loss (4 bets), no matter what bet selection is deployed.
That seems a ridiculously low figure to me, no matter what game you are playing. I was originally referring to losing a series of 50-50 bets, are you sure that Google doc is not referring to the appearance of 4 similar outcomes, which is not exactly the same thing, although I'm pretty certain to the maths people will think otherwise.
For sure, odds are reflective on the actual bets placed, but I would contend a streak of 4, such as HHHH which should occur 6.25% of the time, is not the same as, picking H, then five spins later, picking T, then 8 flips later picking H, then H again somewhere else. I know a lot will disagree, I accept that, that doesn't matter. It is just the figure of 3.6 is hard to swallow, because that is hardly ever going to happen while your sat at the tables.
Quote: CyrusVAre you saying that in 232 trial, one would expect to see, an exact 4 losses in a row 3.6 times?...
...
That seems a ridiculously low figure to me, no matter what game you are playing.
7Craps spreadsheet is spot on. You'll expect that streak of exactly 4 losses about 3 to 4 times in your sequence of 232 trials. If you check out the actual random results in the spreadsheet that i sent you you'll see it's perfectly reasonable to see between 0 and 8 times.
Show your working out for that assertion, so that we can show you where you went wrong. You did go wrong!Quote: Cyrus5I would contend a streak of 4, such as HHHH which should occur 6.25% of the time,
Show either your random streams of results* and we will count them for you, or show some maths. Remember that your HHHH must not be part of a sequence HHHHH if it's to be counted.
*The spreadsheet that I sent you can generate them for you by the boatload.
I think that has been mathematically proven over the decades from many sources. Testing of the Zumma books demonstrated this, I'm sure number crunching on the wiz's Baccarat shoes produced the same conclusion. I would have thought that was a mathematical given.
I just want to add regarding this expectation of 3 or 4 times in 232 decisions, which I would assume to equate to approx once per single Baccarat shoe. If you were merely flipping a coin, betting Bank only, applying any methodology you care to invent, only losing 4 decisions in a row once, simply isn't going to happen, despite what the maths states, it is going to more frequent and just because let's say for example, you lose 4 in a row, twice, trice within 70 hands, doesn't mean that you shouldn't expect a clean run in the next shoe.
The maths might indeed be correct and won't dispute that, but it will never transpire like that at the gaming tables, been there and tried it too many times.
Quote: CyrusVLong term number crunching of 50-50 outcomes covering 1000's of decisions, has shown that singles HTH consists of 50% of all outcomes (same applies to coin flips), doubles that end at two 25% of the time, triples that stay at three 12.5% of the time, quads 6.25%, repeats of 5 and stopping at five 3.1% and the streaks (repeats) greater than 5 for the remaining balance.
I think that has been mathematically proven over the decades from many sources.
Nope. Those statements are just plain wrong or out of context. To be absolutely clear. In 232, 50:50 (win : lose) decisions, you will see approximately 3.5 streaks of EXACTLY 4 losses. That's about 1.5%.
Your 6.25% is a different answer to a different question.
