For all you math geniuses out there... Help please

[savage6]

In roulette if you are betting on one of the 12's(1st 12, 2nd 12, or 3rd 12) and you stay on the same 12 until a win, how many times in 10,000 spins will you encounter a losing streak of 17 or more (on average)?

Thanks!!

 

[GNICETRY]

You would have a 3% chance (.03) of hitting "A" number on a 38 number wheel. So in 100 spins, theoretically, you should hit the number 3 times. I don't know of any formula to figure out 17x without hitting. But I am not a genius either. :biggrin:

 

[Junior Samples]

This picture should clear some of it up anyway

 

[salvageV6]

Ignoring the 0 and 00 because they're ghey, I assume you are betting a square of 12 numbers out of the 36 available?

That gives you 1 in 3 chances of winning per spin or 2 in 3 chances of losing.

Odds are good that you will lose 17 in a row betting the same 12 square.

In 10,000 spins I'd say about 10 times.

 

[rmar]

32
+17*5/18.2(10)-r45/4-9320(20)^=2

**** I forgot to carry the 2.........

Really dedends on how many beers you have had........





.

 

[JOHNDEEREGN]

Depende if the dealer decides to hit the "magnet" button or not.

 

[EVIL]

Roulette is a game of chance, in theory. So, every spin is completely independent from the previous spin and has a little under a third chance of hitting the 12 number square since the 0 and 00 do add to the odds.

At the casino one time a friend and i were playing black jack when he noticed that the roulette display screen showed all red hits. Red had been hit or 15 times in a row. He says "WOW black is due" i told him not to bother but he went over anyway. After betting on black and losing 7 times in a row he came back all upset.

Every spin is independent from every other spin.