• A person is gambling.
  • Person has an equal chance to roll on 1 through 50.
  • Each roll equals the amount they get (rolling a 45 will get you 45 dollars).

I have no trouble figuring out the average. But I'm having difficulty finding out how many times the person needs to roll in order to "reach" this average or higher.

The actual question is: What are the number of times they must roll to reach a 70%, 80%, and 90% confidence level average?

The average is (1+2+...+49+50)/50 = (n+1)/2 = 25.5 -- I figured, if they roll once they have a 1 in 50 chance of hitting the average. And if they roll twice they have a 1 in 50 chance of hitting average twice?? So 2 in 50? But doing this 50 times does not "guarantee" getting the average. So I'm obviously missing some serious error here.

It's about figuring out how much money is required to hit the "average" amount.

  • $\begingroup$ "To reach" means in sum? I.e. if you roll many times, do you get to keep all previous intake or get whatever the last throw was? $\endgroup$ – gt6989b Jul 17 '13 at 19:24
  • $\begingroup$ I hope you mean "get atleast" when you say "reach", because it's pretty darn difficult to get exactly $25.5$ in one throw. $\endgroup$ – Arthur Jul 17 '13 at 19:27
  • $\begingroup$ You have unlimited money and keep rolling until you are 70% certain, 80% certain, 90% certain of getting about 25-26 dollars. $\endgroup$ – victoroux Jul 17 '13 at 19:32
  • $\begingroup$ @Arthur I meant average lol $\endgroup$ – victoroux Jul 17 '13 at 19:32
  • $\begingroup$ The question is more about figuring out how much capital you need to start out with to be sure about getting the average :/ $\endgroup$ – victoroux Jul 17 '13 at 19:33

So I wrote a little program to spin that wheel 10,000,000 times. And then I ran it multiple times.

Hitting/beating the average (26) on a 1-50 wheel:

1 spin : 50%

2 spins: 38%

3 spins: 10%

4 spins: 2%

5-8 spins: less than 0.2% (that's total, not each)

Since I don't know the cost of a spin, I can't finish your calculations, but I hope this is close to what you were looking for.


You've got a sampling distribution. A sampling distribution is a distribution of the averages of doing something N times. So, all your sampling distributions are centered at 25.5, you're correct there. What I gather you're asking is what the relationship between N and the chance of you getting at least average. This probability remains fixed. You will always have a 50% chance of getting average or higher, as all sampling distributions are normal, symmetric distributions. However, as N increases, variability decreases. So you become less likely to be very far off the mean.

  • $\begingroup$ Yes! So how can I calculate "How likely" I am from the mean based on N? $\endgroup$ – victoroux Jul 17 '13 at 20:26
  • $\begingroup$ Also how would an exponential sampling distribution affect N? $\endgroup$ – victoroux Jul 17 '13 at 20:28
  • $\begingroup$ Sampling distributions are only normal if the population has a finite variance (as is the case here). Some distributions have infinite variance. $\endgroup$ – Dale M Jul 18 '13 at 2:24
  • $\begingroup$ I calculated your STDEV to be 14.577. Standard deviation of a sampling distribution is STDEV/SQRT(N) so plug in that equation and then just use your continues normal probability functions. $\endgroup$ – user86724 Jul 19 '13 at 4:56

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