Probability distribution of coin flips

Hello,

I have a poker related question of the probability distribution of coin flips when playing poker.
If you run bad for a long period of time at the poker tables and you loosing way more coin flips then you should loose, what is the probability of winning only
1 out of 30?
5 out of 30?

The average is 15, but how can I calculate this? The distribution should be around 15. So it can not be a linear path, right?


Thanks,

The probability of winning exactly K out of N coin flips is (N)C(K) / 2^N

For 1 out of 50, that is 50 / 2⁵⁰, or about 1 in 22.5 trillion.
For 3 out of 50, this is 19,600 / 2⁵⁰, or about 1 in 57.4 billion.

Quote: masterj

I have a poker related question of the probability distribution of coin flips when playing poker.

there are really not that many hands that are exactly 50/50 in a "coin flip"
(so poker players really are talking about a biased coin flip)

I can think of 1 after the flop and 1 after the turn that are exactly 50/50.

poker players say something like 51/49 is a coin flip.
and it is.
they just do not say what kind it is.

but with all these biased coin flips, poker players still think they "should win" 50% of them.
30, should win 15 of them, when the only thing the math says about that is the ratio of wins to attempts will approach 50%
the actual number of wins is meaningless

Quote: ThatDonGuy

The probability of winning exactly K out of N coin flips is (N)C(K) / 2^N

For 1 out of 50, that is 50 / 2⁵⁰, or about 1 in 22.5 trillion.
For 3 out of 50, this is 19,600 / 2⁵⁰, or about 1 in 57.4 billion.

1 out of 30?

for a 50/50 event
about 1 in 35,791,394
using pari/gp calculator
(08:54) gp > 1/(binomial(30,1)/2^30.)
%6 = 35791394.133333333333333333333333333332

5 out of 30?
about 1 in 7,535
(08:55) gp > 1/(binomial(30,5)/2^30.)
%7 = 7534.71309278205829929

Quote: 7craps

1 out of 30?

My bad - I though it said "out of 50" for some reason

Quote: ThatDonGuy

The probability of winning exactly K out of N coin flips is (N)C(K) / 2^N

For 1 out of 50, that is 50 / 2⁵⁰, or about 1 in 22.5 trillion.
For 3 out of 50, this is 19,600 / 2⁵⁰, or about 1 in 57.4 billion.

Is the 2 the same as (1/p)?

Could you do say

(N)C(K) / (1/p)^N ?

Quote: ThatDonGuy

I thought it said "out of 50" for some reason

at first I thought so too.
it seemed like a weird question

a spreadsheet (Excel) can give the total distribution of 30 trials at 50%
quite easily.
many still do not understand the binomial probability distribution
(it is just a special case of the multinomial probability distribution. we all should have leared that in early math classes)

for the OP, one would have to know the hands to continue down that road

successes	prob	1 in
0	9.31323E-10	1,073,741,824.00
1	2.79397E-08	35,791,394.13
2	4.05125E-07	2,468,372.01
3	3.78117E-06	264,468.43
4	2.55229E-05	39,180.51
5	0.000132719	7,534.71
6	0.000552996	1,808.33
7	0.001895986	527.43
8	0.005450961	183.45
9	0.013324572	75.05
10	0.027981601	35.74
11	0.050875638	19.66
12	0.080553093	12.41
13	0.111535052	8.97
14	0.13543542	7.38
15	0.144464448	6.92
16	0.13543542	7.38
17	0.111535052	8.97
18	0.080553093	12.41
19	0.050875638	19.66
20	0.027981601	35.74
21	0.013324572	75.05
22	0.005450961	183.45
23	0.001895986	527.43
24	0.000552996	1,808.33
25	0.000132719	7,534.71
26	2.55229E-05	39,180.51
27	3.78117E-06	264,468.43
28	4.05125E-07	2,468,372.01
29	2.79397E-08	35,791,394.13
30	9.31323E-10	1,073,741,824.00

Quote: ThatDonGuy

The probability of winning exactly K out of N coin flips is (N)C(K) / 2^N

that should say 'fair coin flips'

Quote: RS

Is the 2 the same as (1/p)?

Could you do say

(N)C(K) / (1/p)^N ?

one could do say when p = 1/2
for those that want to know more
(N)C(K) is just the binomial coefficient found in Pascal's Triangle, for example
the number of ways an event could happen where order doesn't matter
it is A in A/B
B = the total ways an event could and could not happen (in the example 2^30 = 1,073,741,824)