Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a question about probability.

The game is like this:

Every $1000$th submission will win, but the players don't know how many submissions were made before.

Is it better for a player to throw all of his $100$ credits in one time or is it better to throw one then wait then submit the next... and so on?

share|improve this question
    
baf, would you look at my edit and see if it accurately represents what you were trying to write? –  The Chaz 2.0 Apr 11 '12 at 20:49
    
yes it is, thank you. my english is not so good –  baf Apr 11 '12 at 20:57

2 Answers 2

Suppose that you submit, and submit again, and again, with a random and very large number of other people playing between your submissions. Although the lottery increments its number by $1$ each time a single bet is made, the numbers on the tickets that you buy electronically vary in totally unpredictable ways. Then the probability you win on any one play is $\frac{1}{1000}$. So the probability you do not win on that play is $1-\frac{1}{1000}$. Thus, by the assumption of independence, the probability that you lose $100$ times in a row is $$\left(1-\frac{1}{1000}\right)^{100}.$$ This is about $0.9047921$. So the probability you win at least once is about $1-0.9047921$, which is about $0.0952079$.

If you "throw in" all at once, meaning I assume that you are getting $100$ consecutively numbered tickets, the probability you win is $\frac{100}{1000}$, that is, $0.10$, somewhat higher than $0.0952079$. But playing in scattered fashion gives the possibility of winning more than once, while with the "throw in" this is not possible.

If "winning" means winning say $400$ dollars, then the expected (mean) amount of money that you win will be the same with either strategy. Roughly speaking the small lowering of the probability of winning with the scattered play strategy is exactly compensated for, in the sense of expected winnings, by the possibility of winning twice, or even more often.

share|improve this answer

if you buy $n$ lots at random times then each will have $1$ in $1000$ chance of winning

but if you buy one and you lose then the next lot will have $1$ in $999$ chance to be the winning lot if you lose again then it's 1 in 998 etc.

but if you buy them in bulk (i.e. buy the next lot before you know if you won with the last) then it's back to $1$ in $1000$ for each

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.