What's the chances of flipping ten heads in a row at some point when flipping a coin 100 times?
October 16th, 2011 at 12:01:42 PM
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My friend and I have been playing backgammon and are pretty evenly matched but we have noticed that quite frequently one of us will go on a long winning streak. In our last 100 games I have had 10 wins in a row, 7 wins in a row and 6 wins in a row. Is this likely to have happened by random chance? i.e. If i flipped a coin 100 times what's the most heads/tails i would expect to get in a row?
Thanks Steve
Thanks Steve
October 16th, 2011 at 12:08:06 PM
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In roulette, its not uncommon to see 15 reds or blacks in a row.
Or more. 10 in a row is a common occurrence.
Or more. 10 in a row is a common occurrence.
"It's not called gambling if the math is on your side."
October 16th, 2011 at 12:13:35 PM
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Thanks for the quick reply. Do you know how uncommon? I'm trying to work it out mathematically - ideally I want to work out a graph that says for instance "On average you will get 15 single wins, 10 double, 5 triple etc".
October 16th, 2011 at 12:27:07 PM
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Yes. Here is a run probability table for 100 trials:Quote: thebeardkingMy friend and I have been playing backgammon and are pretty evenly matched but we have noticed that quite frequently one of us will go on a long winning streak. In our last 100 games I have had 10 wins in a row, 7 wins in a row and 6 wins in a row.
Is this likely to have happened by random chance?
For at least 1 run of a certain length or more. Example: at least 7 in a row at least 1 time. 7+ = 31.75%
To answer your thread title question of 10 in a row.
At least 1 run of length 10 or more in 100 trials: 0.044137229 or about 1 in 23
At least 2 runs of length 10 or more in 100 trials: 0.000777775 or about 1 in 1286
| length | prob for N=100 |
|---|---|
| 3+ | 0.999738255 |
| 4+ | 0.972715042 |
| 5+ | 0.810109599 |
| 6+ | 0.546093619 |
| 7+ | 0.317520387 |
| 8+ | 0.170207962 |
| 9+ | 0.087558927 |
| 10+ | 0.044137229 |
| 11+ | 0.022029266 |
| 12+ | 0.010941678 |
| 13+ | 0.005421525 |
a handy streak calculator can be found at A handy Streak Calculator
Quote: thebeardkingi.e. If i flipped a coin 100 times what's the most heads/tails i would expect to get in a row?
Thanks Steve
You are looking for the longest run of either tails or heads for a fair coin for 100 trials.
Data below from a simulation.
The math for me for the longest run in N trials is too messy.
group middle freq freq/100
-----------------------------------------------
2.50 <= x < 3.50 3.00 2775 0.03%
3.50 <= x < 4.50 4.00 279971 2.80%
4.50 <= x < 5.50 5.00 1651160 16.51%
5.50 <= x < 6.50 6.00 2644439 26.44%
6.50 <= x < 7.50 7.00 2274305 22.74%
7.50 <= x < 8.50 8.00 1463360 14.63%
8.50 <= x < 9.50 9.00 817229 8.17%
9.50 <= x < 10.50 10.00 430158 4.30%
10.50 <= x < 11.50 11.00 218683 2.19%
11.50 <= x < 12.50 12.00 109456 1.09%
12.50 <= x < 13.50 13.00 54739 0.55%
13.50 <= x < 14.50 14.00 27242 0.27%
14.50 <= x < 15.50 15.00 13407 0.13%
15.50 <= x < 16.50 16.00 6691 0.07%
16.50 <= x < 17.50 17.00 3165 0.03%
17.50 <= x < 18.50 18.00 1629 0.02%
18.50 <= x < 19.50 19.00 813 0.01%
19.50 <= x < 20.50 20.00 383 0.00%
20.50 <= x < 21.50 21.00 222 0.00%
21.50 <= x < 22.50 22.00 80 0.00%
22.50 <= x < 23.50 23.00 48 0.00%
23.50 <= x < 24.50 24.00 20 0.00%
24.50 <= x < 25.50 25.00 10 0.00%
25.50 <= x < 26.50 26.00 10 0.00%
26.50 <= x < 27.50 27.00 2 0.00%
27.50 <= x < 28.50 28.00 2 0.00%
28.50 <= x < 29.50 29.00 0
29.50 <= x < 30.50 30.00 0
30.50 <= x < 31.50 31.00 1 0.00%
-----------------------------------------------
grouped data
items: 10,000,000
minimum value: 3.00
first quartile: 6.00
median: 7.00
third quartile: 8.00
maximum value: 31.00
mean value: 6.98
midrange: 17.00
range: 28.00
interquartile range: 2.00
mean abs deviation: 1.34
sample variance (n): 3.21
sample variance (n-1): 3.21
sample std dev (n): 1.79
sample std dev (n-1): 1.79
-----------------------------------------------
cumulative
-----------------------------------------------
2.50 <= x < 3.50 3.00 2775 0.03%
3.50 <= x < 4.50 4.00 282746 2.83%
4.50 <= x < 5.50 5.00 1933906 19.34%
5.50 <= x < 6.50 6.00 4578345 45.78%
6.50 <= x < 7.50 7.00 6852650 68.53%
7.50 <= x < 8.50 8.00 8316010 83.16%
8.50 <= x < 9.50 9.00 9133239 91.33%
9.50 <= x < 10.50 10.00 9563397 95.63%
10.50 <= x < 11.50 11.00 9782080 97.82%
11.50 <= x < 12.50 12.00 9891536 98.92%
12.50 <= x < 13.50 13.00 9946275 99.46%
13.50 <= x < 14.50 14.00 9973517 99.74%
14.50 <= x < 15.50 15.00 9986924 99.87%
15.50 <= x < 16.50 16.00 9993615 99.94%
16.50 <= x < 17.50 17.00 9996780 99.97%
17.50 <= x < 18.50 18.00 9998409 99.98%
18.50 <= x < 19.50 19.00 9999222 99.99%
19.50 <= x < 20.50 20.00 9999605 100.00%
20.50 <= x < 21.50 21.00 9999827 100.00%
21.50 <= x < 22.50 22.00 9999907 100.00%
22.50 <= x < 23.50 23.00 9999955 100.00%
23.50 <= x < 24.50 24.00 9999975 100.00%
24.50 <= x < 25.50 25.00 9999985 100.00%
25.50 <= x < 26.50 26.00 9999995 100.00%
26.50 <= x < 27.50 27.00 9999997 100.00%
27.50 <= x < 28.50 28.00 9999999 100.00%
28.50 <= x < 29.50 29.00 9999999 100.00%
29.50 <= x < 30.50 30.00 9999999 100.00%
30.50 <= x < 31.50 31.00 10000000 100.00%
-----------------------------------------------
October 16th, 2011 at 12:33:20 PM
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For either heads or tails where p=0.50. The expected number of streaks (runs) per length per 100 trials.Quote: thebeardkingThanks for the quick reply. Do you know how uncommon? I'm trying to work it out mathematically - ideally I want to work out a graph that says for instance "On average you will get 15 single wins, 10 double, 5 triple etc".
| Length | Average | Or more |
|---|---|---|
| 1 | 12.75 | 25.25 |
| 2 | 6.3125 | 12.5 |
| 3 | 3.125 | 6.1875 |
| 4 | 1.546875 | 3.0625 |
| 5 | 0.765625 | 1.515625 |
| 6 | 0.378906 | 0.75 |
| 7 | 0.1875 | 0.371094 |
| 8 | 0.092773 | 0.183594 |
| 9 | 0.045898 | 0.09082 |
| 10 | 0.022705 | 0.044922 |
| 11 | 0.01123 | 0.022217 |
| 12 | 0.005554 | 0.010986 |
| 13 | 0.002747 | 0.005432 |
| 14 | 0.001358 | 0.002686 |
| 15 | 0.000671 | 0.001328 |
Simulation result data
group middle freq freq/100
------------------------------------------------
0.50 <= x < 1.50 1.00 127471930 50.48%
1.50 <= x < 2.50 2.00 63132663 25.00%
2.50 <= x < 3.50 3.00 31256267 12.38%
3.50 <= x < 4.50 4.00 15482681 6.13%
4.50 <= x < 5.50 5.00 7661983 3.03%
5.50 <= x < 6.50 6.00 3791121 1.50%
6.50 <= x < 7.50 7.00 1873678 0.74%
7.50 <= x < 8.50 8.00 928206 0.37%
8.50 <= x < 9.50 9.00 458358 0.18%
9.50 <= x < 10.50 10.00 227831 0.09%
10.50 <= x < 11.50 11.00 112425 0.04%
11.50 <= x < 12.50 12.00 55271 0.02%
12.50 <= x < 13.50 13.00 27549 0.01%
13.50 <= x < 14.50 14.00 13852 0.01%
14.50 <= x < 15.50 15.00 6758 0.00%
15.50 <= x < 16.50 16.00 3348 0.00%
16.50 <= x < 17.50 17.00 1599 0.00%
17.50 <= x < 18.50 18.00 828 0.00%
18.50 <= x < 19.50 19.00 410 0.00%
19.50 <= x < 20.50 20.00 193 0.00%
20.50 <= x < 21.50 21.00 119 0.00%
21.50 <= x < 22.50 22.00 35 0.00%
22.50 <= x < 23.50 23.00 26 0.00%
23.50 <= x < 24.50 24.00 8 0.00%
24.50 <= x < 25.50 25.00 6 0.00%
25.50 <= x < 26.50 26.00 4 0.00%
26.50 <= x < 27.50 27.00 1 0.00%
27.50 <= x < 28.50 28.00 0
28.50 <= x < 29.50 29.00 0
29.50 <= x < 30.50 30.00 0
30.50 <= x < 31.50 31.00 1 0.00%
------------------------------------------------
grouped data
items: 252,507,151
minimum value: 1.00
first quartile: 1.00
median: 1.00
third quartile: 2.00
maximum value: 31.00
mean value: 1.98
midrange: 16.00
range: 30.00
interquartile range: 1.00
mean abs deviation: 0.99
sample variance (n): 1.94
sample variance (n-1): 1.94
sample std dev (n): 1.39
sample std dev (n-1): 1.39
------------------------------------------------
cumulative
------------------------------------------------
0.50 <= x < 1.50 1.00 127471930 50.48%
1.50 <= x < 2.50 2.00 190604593 75.48%
2.50 <= x < 3.50 3.00 221860860 87.86%
3.50 <= x < 4.50 4.00 237343541 93.99%
4.50 <= x < 5.50 5.00 245005524 97.03%
5.50 <= x < 6.50 6.00 248796645 98.53%
6.50 <= x < 7.50 7.00 250670323 99.27%
7.50 <= x < 8.50 8.00 251598529 99.64%
8.50 <= x < 9.50 9.00 252056887 99.82%
9.50 <= x < 10.50 10.00 252284718 99.91%
10.50 <= x < 11.50 11.00 252397143 99.96%
11.50 <= x < 12.50 12.00 252452414 99.98%
12.50 <= x < 13.50 13.00 252479963 99.99%
13.50 <= x < 14.50 14.00 252493815 99.99%
14.50 <= x < 15.50 15.00 252500573 100.00%
15.50 <= x < 16.50 16.00 252503921 100.00%
16.50 <= x < 17.50 17.00 252505520 100.00%
17.50 <= x < 18.50 18.00 252506348 100.00%
18.50 <= x < 19.50 19.00 252506758 100.00%
19.50 <= x < 20.50 20.00 252506951 100.00%
20.50 <= x < 21.50 21.00 252507070 100.00%
21.50 <= x < 22.50 22.00 252507105 100.00%
22.50 <= x < 23.50 23.00 252507131 100.00%
23.50 <= x < 24.50 24.00 252507139 100.00%
24.50 <= x < 25.50 25.00 252507145 100.00%
25.50 <= x < 26.50 26.00 252507149 100.00%
26.50 <= x < 27.50 27.00 252507150 100.00%
27.50 <= x < 28.50 28.00 252507150 100.00%
28.50 <= x < 29.50 29.00 252507150 100.00%
29.50 <= x < 30.50 30.00 252507150 100.00%
30.50 <= x < 31.50 31.00 252507151 100.00%
Being football day, I am just too good busy to show my work or provide links to other posts that show how to do this yourself. That will be added later.
Hope this can help.
Enjoy
added:
My daughter has a web page for the "probability of runs" and "expected number of runs" and other goodies.
Here is the link below.
The Excel tables are LIVE, meaning you can change the values and see the results, even if you do not have Excel on your computer.
Expected number of runs per N trials
October 18th, 2011 at 8:28:13 PM
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Quote: guido111
length prob for N=100 3+ 0.999738255 4+ 0.972715042 5+ 0.810109599 6+ 0.546093619 7+ 0.317520387 8+ 0.170207962 9+ 0.087558927 10+ 0.044137229 11+ 0.022029266 12+ 0.010941678 13+ 0.005421525
Guido's table above is absolutely correct.
Adding to the table, even small deviations from 50/50 make it even more likely that the person who is slightly better will get a significant streak. Picking a mid length streak of 7 wins in a row, and giving yourself a slight advantage over your opponent leads to the following:
50%-50%: 31.752% of getting 7+ in a row
51%-49%: 33.877% of getting 7+ in a row
52%-48%: 38.546% of getting 7+ in a row
53%-47%: 42.161% of getting 7+ in a row
...
60%-40%: 68.910% of getting 7+ in a row
The solution to this problem can be worked out as a recursive relationship, but it is not possible to express it as an algebraic formula. It takes some higher order mathematics to show that it can't be boiled down to a formula.
The gambler's fallacy is often expressed as the quantitative statement, that a gambler often believes that past events influence future event, and often relies on betting systems. But you can express the "gambler's fallacy" in more quantitative terms, by saying that most gambler's underestimate the length and frequency of streaks.
Most untrained people will not be able to fake a simple random stream including the results of coin tosses. An overwhelmingly large percentage of people don't put in long enough streaks.
July 29th, 2015 at 9:20:50 AM
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ands at a 60/40 advantage there could easily be more than 1 such runQuote: pacomartin60%-40%: 68.910% of getting 7+ in a row
i only know of one that has done this (he, yes, BruceZ did it in Excel and in R... i wonder if he is left-handed?)
others have shown how to do this (in heavy math papers) with inclusion-exclusion (lots of math for those that love lots)
i show a 29.059% chance of at least 2 such 7+ run streaks
wow!
that makes 0.398508652 the probability of exactly one such run in 100 trials
and it goes from there
the 50/50
at least 1 run: 0.3175203874966 <<<< this was an ask the wizard question too
https://wizardofodds.com/ask-the-wizard/253/
Q#3
at least 2 runs: 0.0490016781438
at least 3 runs: 0.0043295727700
yes, BruceZ also mentioned B4 that this can also be easily done in a few columns in Excel
the multiple streak thing
so I be off to do just that
Mully
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