Quote:joshyeltmanOn average, how many spins does it take to see a previously rolled number? Not necessarily back to back, but to repeat any number that has previously been rolled. A couple of us have made guesses, but we can't figure out an accurate way of calculating the average.

I get a mean of 8.408797212, but invite somebody else to confirm. This is similar to the "birthday problem," which asks, how many people do you need for there to be a 50% or greater chance of a common birthday?

Quote:WizardI get a mean of 7.408797212, but invite somebody else to confirm. This is similar to the "birthday problem," which asks, how many people do you need for there to be a 50% or greater chance of a common birthday?

Funny thing, once I was discussing the birthday problem with someone who turned out to have the same birthday as me, albeit in a different year.

Different year is the key. Make it the same year and the number skyrockets.Quote:NareedFunny thing, once I was discussing the birthday problem with someone who turned out to have the same birthday as me, albeit in a different year.

I remember this being discussed one night on Carson's Tonight Show. I believe the number was in the low 20's.

Johnny "tested" it by proposing to ask the people in one section of the audience. The rows only had 6 or 8 people in them. He was on the third or fourth person in the first row when someone in the second row reacted that they had the same day as the person he had just asked.

Back at my last real job there were about 30 people in the office, and we had two sets of common birthdays, as I recall. There were also a whole bunch of birthdays in June, about 25% of the entire office. If I were still there I'd be enjoying longer breaks and eating lots of cake about this time. As long as I'm on the topic, my two brothers have the same birthday, but four years apart.

Quote:WizardAs long as I'm on the topic, my two brothers have the same birthday, but four years apart.

I also have two brothers with the same birthday, two years apart. I counted back and could never figure out what was so special about the conception date.

Quote:ChuckI also have two brothers with the same birthday, two years apart. I counted back and could never figure out what was so special about the conception date.

I never thought of that before, but both my younger brothers would have been conceived right around my birthday. Hey, I'm the one who is supposed to have fun that day!

Can we please get this thread back on topic?

Quote:joshyeltmanWhat I really want to know is what formula you used to come up with that number. I arbitrarily guessed around 8, and I'm glad I was close, but I'd like to know how to calculate this.

First you need to know the formula for the probability of a repeat repeat number with exactly n spins. That probability is 1-(37/38)*(36/38)*...*((38-n+1)/38). Here are some values by n:

2 0.026316

3 0.077562

4 0.150386

5 0.239819

6 0.339843

7 0.444078

8 0.546485

9 0.641962

10 0.726760

11 0.798666

12 0.856947

13 0.902121

14 0.935606

15 0.959330

16 0.975384

17 0.985749

18 0.992124

19 0.995855

20 0.997927

21 0.999018

22 0.999561

23 0.999815

24 0.999927

25 0.999973

26 0.999991

27 0.999997

28 0.999999

Let f(n) be the probability of a repeat within n spins. Let g(n) be the proability that the first repeat is on exactly the nth spin. Then g(n)=f(n)-f(n-1). Here are some values for g(n)

2 0.026316

3 0.051247

4 0.072824

5 0.089433

6 0.100024

7 0.104235

8 0.102407

9 0.095477

10 0.084799

11 0.071905

12 0.058281

13 0.045175

14 0.033485

15 0.023724

16 0.016054

17 0.010365

18 0.006376

19 0.003731

20 0.002073

21 0.001091

22 0.000543

23 0.000254

24 0.000112

25 0.000046

26 0.000018

27 0.000006

28 0.000002

29 0.000001

Then you take the dot product of the above table to get the expected number of spins, which is...

2*0.026316 + 3*0.051247 + 4*0.072824 + ... = 8.408795574

p.s. In writing this up I realize my original answer was off by exactly 1, I think because I wasn't counting the first spin.

I've seen only one person doing it and it is amusing since most red numbers are covered and the chances of red/black are higher than the straight bets.

Does this increase the chances or reduces the house edge in any way? Is this just as bad as any other "method"?

i.e.: on a single zero roulette, making $5 bets, straight bets on 5-7-9-12-14-16-18-21-23-25 + Split bets on 27/30 and 0/1. (All red numbers and covering 0).