Quote:OnceDear

average number of times is 1/Probability which for a streak of 5 is about 36.7

Is this the probability of red 5 streak occuring?

Say im tracking both red and black, would i divide 36.7 by 2?

Quote:DanthemanIs this the probability of red 5 streak occuring?

Say im tracking both red and black, would i divide 36.7 by 2?

I'm not the math guy in this discussion, but it appears the one in apx. 36.7 refers to one(red) or the other(black) occurring five times in a row. Five spins in a row. We were unable to continue an intelligent discussion about it due to personal attacks being launched. A sad day actually....Cheers mate.

Quote:NokTangI'm not the math guy in this discussion, but it appears the one in apx. 36.7 refers to one(red) or the other(black) occurring five times in a row. Five spins in a row. We were unable to continue an intelligent discussion about it due to personal attacks being launched. A sad day actually....Cheers mate.

An intelligent discussion and a discussion involving the emotional control regarding a game of probability are two different things.

Quote:DodsferdAn intelligent discussion and a discussion involving the emotional control regarding a game of probability are two different things.

I'm not clear why you post that?

Having the emotional control to take your $100.usd a day to live on, after laying apx 36 to 1 odds, seems intelligent enough to me. What have I got wrong? I think we all know you "can't" do this hour after hour, it takes "emotional control" to just do it once a day, have your prime rib dinner, do your laundry, watch the news, and sleep in the comfort of AC or heat, with running water etc..

I have a thread here:Quote:DanthemanI am wanting to find out the average number of spins it would take playing European roulette, if betting both black and red, to see a streak length of 2,3,4,5 etc

on average How many trials

easy to use formula for wait time until binomial run (streak) of successes

Now,

data for you

18/37

0 Roulette

(all calculated values here checked with simulations)

run length:average number of spins

a=red

2:6.280864198

3:14.96622085

4:32.81945397

5:69.51776649

b=black

2:6.280864198

3:14.96622085

4:32.81945397

5:69.51776649

c=red or black

2:3.140432099

3:7.483110425

4:16.40972699

5:34.75888325

c=a+b-d

c=a/2 or b/2

d=red and black

2:9.421296297

3:22.449331275

4:49.22918095

5:92.05098036

d=a+b-c

Now lots of partially correct answers in this thread so far.

not picking on any one (or two)

How about this...

5 in a row for red

many say (18/37)^5=about 1 in 37 spins on average

that 37 is not spins

(Unless you count overlapping streaks and that is way too confusing to even attempt)

37 is the number of trials or attempts and each attempt is NOT 1 spin in length.

we need more for the average length of each trial

1+p+p^2+p^3+p^4,p=(18/37) = 3550231/1874161

result: about 1.894304171

(sure we can use 1 plus the sum of a geometric series

but this is different too)

1/(18/37)^5 * 3550231/1874161 = 131358547/1889568 =

69.52

Wolfram Alpha

in agreement with simple simulation results

grouped data

items: 1000000

minimum value: 5.00

first quartile: 23.00

median: 49.00

third quartile: 95.00

maximum value: 925.00

mean value: 69.48 <<<<<

midrange: 465.00

range: 920.00

interquartile range: 72.00

mean abs deviation: 48.35

sample variance (n): 4301.07

sample variance (n-1): 4301.07

sample std dev (n): 65.58

sample std dev (n-1): 65.58

hope this helps out current and future visitors

averages are like apple eating

probability distributions are like a 10 course meal.

I am so hungry.

hope no errors

fun question!

Sally