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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.

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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 1

up vote 2 down vote accepted

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.

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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

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