1. ## Roulette question

I need your help with this...and remember i'm pretty stupid.

A roullette wheel (with 37 segments) is spun 5000 times. On average each number (0-36) will be hit 135 times (5000/37)

How do I draw a graph showing the expected distribution of hits after 5000 spins (cos obviously not all numbers will be hit exactly 1 in 37 times)

ta

lee

2. Originally Posted by murless22
I need your help with this...and remember i'm pretty stupid.

A roullette wheel (with 37 segments) is spun 5000 times. On average each number (0-36) will be hit 135 times (5000/37)

How do I draw a graph showing the expected distribution of hits after 5000 spins (cos obviously not all numbers will be hit exactly 1 in 37 times)

ta

lee
While not every number will have the same number of hits in a particular experiment when the wheel is spun $\displaystyle 5000$ times the expected number of hits is the same for all numbers.

Now if you are interested in the distribution of the number of hits on a particular number (lets say the $\displaystyle 0$) then that distribution has a binomial distribution $\displaystyle B(5000, 1/37)$. Now this is difficult to handle because of the large number of spins so we use the normal approximation $\displaystyle N(5000/37, 5000 \times (1/37)\times (36/37))$

CB

3. Captainblack

thanks for the reply, not sure if I made myself clear enough though.

Lets say the segment we are interested in is the Zero.

If we spin the wheel 5000 times the zero might be hit say 140 times
If we spin the wheel again 5000 times , this time the zero might be hit say 132 times.......lets repeat this sequence of 5000 spins say 1000 times.

Zero might have been hit 140, 132, 138, 135, etc etc etc and its these results which I want to graph.

clearly the graph would have 135 hits at its peak and would decline either side of this.

I'll tell you why I'm asking...a casino near me is offering odds of 40/1 on one particular number for a short period each day (instead of the usual 35/1)

clearly this gives the backer a margin of >10%. I've been placing a bet on this number for the last few weeks. Now ive done this 4477 times and hit my number 109 times, which is one spin in 41 (id expect 1 in 37 over time) and I want to know how far from 'normal' 1 hit in 41 is over this relatively short sample

Thanks

4. Originally Posted by murless22
Captainblack

thanks for the reply, not sure if I made myself clear enough though.

Lets say the segment we are interested in is the Zero.

If we spin the wheel 5000 times the zero might be hit say 140 times
If we spin the wheel again 5000 times , this time the zero might be hit say 132 times.......lets repeat this sequence of 5000 spins say 1000 times.

Zero might have been hit 140, 132, 138, 135, etc etc etc and its these results which I want to graph.

clearly the graph would have 135 hits at its peak and would decline either side of this.
I have already answered this in that I have told you what the distribution of the number of times zero (or any other number shows) in 5000 spins. You just need to plot the distribution.

I'll tell you why I'm asking...a casino near me is offering odds of 40/1 on one particular number for a short period each day (instead of the usual 35/1)

clearly this gives the backer a margin of >10%. I've been placing a bet on this number for the last few weeks. Now ive done this 4477 times and hit my number 109 times, which is one spin in 41 (id expect 1 in 37 over time) and I want to know how far from 'normal' 1 hit in 41 is over this relatively short sample

Thanks
Easy. Expected number of hits is $\displaystyle 4477/37=121$, standard deviation of number of hits is $\displaystyle \approx 10.85$, so z-score of the actual number of hits is:

$\displaystyle z=\frac{109-121}{10.85}\approx -1.106$

looking this up on the table of the standard normal distribution we find that there is $\displaystyle \approx 13.5 \%$ of getting $\displaystyle 109$ or fewer hits in that many spins of a fair wheel (in fact the percentage is higher as I have not made a continuity correction the corrected answer is more like $\displaystyle 14.5\%$)

CB