# How many rounds does it take to be 99% sure of reaching Expected Value?

I have a feeling this might be a common question but I was unable to find the right way of asking, and I'm just a hobbyist at stats/math.

Say I have a bet that costs six dollars. If I lose I get nothing, and if I win I get my six dollars back, plus $\$7$more. I have a$74$% chance of winning. 1. I intend to stop betting once I reach EV. How many times must I make the bet before I can be$99$% sure of first reaching cumulative Expected Value? In other words, assuming$n > 0$, how many rounds until we are$99$% sure of reaching$3.62n$the first time? 2. I intend to stop betting once I reach break-even. How many times must I make the bet before I can be$99$% sure of breaking even for the series of bets the first time? This is not homework, I'm seeking a way to communicate a point on a discussion board about how making decisions according to EV is not always wise if you don't have access to many rounds of a +EV scenario. I'm trying to figure out how to answer these questions for a variety of certainty levels (e.g.$99$%), probabilities (e.g.$74$%), and odds. - What do you mean by "reaching expected value"? – Qiaochu Yuan Dec 18 '12 at 10:12 EV of the above scenario is 3.62. So, earning 3.62 * number of rounds played. The first time that happens. The second question is about reaching 0 the first time. – tunesmith Dec 18 '12 at 10:22 This is not a good way to make the point you want. You want to have a certainty level of being close to the expected value for a given value of close. There are various probabilistic theorems about this although I am not sure which one would make the point best. – Qiaochu Yuan Dec 18 '12 at 10:23 To reach or exceed. – tunesmith Dec 18 '12 at 10:26 ## 1 Answer From your comment it seems I had misunderstood the question. I'm leaving the old answer below and answering what I now understand to be the question: After how many bets is the probability at least$0.99$that at some point after the first bet you had at least once 1) won at least the expected value or 2) broken even? I doubt that you'll find a closed form for either question, since the answer depends discontinuously and in a complicated way on the amounts won and lost. Here's code that calculates the desired probabilities, and below are tables with the results up to$36$bets for 1) and 2), respectively. The probability first exceeds$0.99$after the eighth bet for 2); for 1) the increase is much slower and the probability only exceeds$0.99$after$1520$bets. Probabilities to have won at least the expected value at least once: 1 : 0.74 2 : 0.74 3 : 0.74 4 : 0.8453582399999999 5 : 0.8453582399999999 6 : 0.8453582399999999 7 : 0.8453582399999999 8 : 0.8786593162076928 9 : 0.8786593162076928 10 : 0.8786593162076928 11 : 0.8786593162076928 12 : 0.8962020301044347 13 : 0.8962020301044347 14 : 0.8962020301044347 15 : 0.8962020301044347 16 : 0.9074148648344059 17 : 0.9074148648344059 18 : 0.9074148648344059 19 : 0.9074148648344059 20 : 0.915359865866572 21 : 0.915359865866572 22 : 0.915359865866572 23 : 0.915359865866572 24 : 0.9213629803238053 25 : 0.9213629803238053 26 : 0.9213629803238053 27 : 0.9277678707439793 28 : 0.9277678707439793 29 : 0.9277678707439793 30 : 0.9277678707439793 31 : 0.9323141561409174 32 : 0.9323141561409174 33 : 0.9323141561409174 34 : 0.9323141561409174 35 : 0.9359115934382276 36 : 0.9359115934382276  Probabilities to have broken even at least once: 1 : 0.74 2 : 0.9324 3 : 0.9324 4 : 0.96941776 5 : 0.96941776 6 : 0.9836621940479999 7 : 0.9836621940479999 8 : 0.9905137668250881 9 : 0.9905137668250881 10 : 0.9942048461115609 11 : 0.9942048461115609 12 : 0.996335337075713 13 : 0.9980762525492772 14 : 0.9980762525492772 15 : 0.9988298948577833 16 : 0.9988298948577833 17 : 0.9992541563997228 18 : 0.9992541563997228 19 : 0.9995114393395534 20 : 0.9995114393395534 21 : 0.9996738409902597 22 : 0.9996738409902597 23 : 0.9997791130800898 24 : 0.9997791130800898 25 : 0.999848694382642 26 : 0.9999117901988117 27 : 0.9999117901988117 28 : 0.9999425077210228 29 : 0.9999425077210228 30 : 0.9999613308931266 31 : 0.9999613308931266 32 : 0.9999735347566812 33 : 0.9999735347566812 34 : 0.9999816703500977 35 : 0.9999816703500977 36 : 0.9999871903297415  This is the answer to the question as I had originally understood it: After how many bets is the probability at least$0.99$that you 1) won at least the expected value or 2) broke even after all the bets? Your best chance of getting at least the expected value is not to bet at all; then you get the expected value$0$with probability$1$. If you bet once, you get at least the expected value if you win, that is, with probability$0.74$. By the central limit theorem, the form of the distribution tends to a Gaussian as the number of bets increases, so the probability of getting at least the expected value decreases towards$1/2$. Thus, you never have a probability of$0.99$of getting at least the expected value, unless you don't bet at all. I doubt you'll get a closed form for the second question, but you can find the answer by trial and error: With$12$bets you break even with probability$0.982246$(computation), and with$13$bets you break even with probability$0.992692$(computation). With$14$bets you break even with probability$0.986808$(computation), and with$15$bets you break even with probability$0.994428$(computation). The probability to break even is at least$0.99$for$13$bets and for$15$or more bets. - The first answer really surprises me. The definition of EV means that in the long run, the return on your bets will average the EV. So how could it be that you will never even approach 50% likelihood of reaching the EV? – tunesmith Dec 18 '12 at 18:17 @tunesmith: That seems to be a misunderstanding. I wrote exactly the opposite: You start out with a$100\%$chance of reaching the expected value, then if you bet once it drops to$74\%$, and as you keep betting it gradually decreases towards$50\%$. For a symmetric distribution without a discrete probability at the expected value, such as a Gaussian, the probability of reaching the expected value is$50\%$. – joriki Dec 18 '12 at 18:33 Assume at least one bet, and that betting stops when EV is reached the first time. From what I understand, EV means what you'd gain or lose on average from participating in multiple rounds of a bet. The EV of the above example is (0.26 * -6) + (0.74 * 7) = 3.62 . On average, you will earn 3.62 on every bet. If on average you will earn 3.62 on every bet, then after n bets you'd expect 3.62n . Due to volatility you'd be below 3.62n sometimes, and above 3.62n sometimes. It sounds like you are saying that as n increases, you approach 50% likelihood that you will have never been above 3.62n. – tunesmith Dec 18 '12 at 20:10 @tunesmith: Now I see how you meant the question. "for the series of bets (the first time)" is somewhat ambiguous; I thought you wanted to increase the number of bets and find the least such number for which you have a$99\%$chance to break even in the end. I think the probability that you want, the one for breaking even at some point during the series, not necessarily at the end, does go to$1\$. I doubt that there's a closed form for calculating it, though, since it depends discontinuously on what you win or lose per bet. –  joriki Dec 18 '12 at 23:03
Ok, thanks - well, at least it's a hard question. :-) Hopefully there will be some more suggestions on how to calculate it. I'm still trying to follow the break-even computations, btw - do they both fully take into account the 7/6 odds? I don't see both 7 and 6 in any of them. –  tunesmith Dec 19 '12 at 0:53