5050 A Guide to 100%

I try to make a statistical analysis in order to find the optimal strategy (i.e., the one that on average takes the less button presses) to get all the achievements in the game. In addition, with this guide you may learn how to maximize the profits of an even gamble when the EV is positive.

5050 A Guide to 100%

Hello, my name is Marco, I am a Spanish Bio-statistics Master’s degree student and as of 2023-09-14 I have stumbled upon this short fun game. The thrill of gambling without risking my life savings is one thing that I always enjoy, but the underlying problem of the game to get all the achievements makes it all the more interesting.

After making a quite suboptimal approach to 100% the game, I did some research that I wanted to share with the community. In addition to getting 11 achievements quicker in a simple game, I think some people might gain some cool knowledge about making profit in gambles where you have the advantage.

After reading this guide, please let me know if you miss something, like visual help, deeper explanations or less digressing, and I might try making a better 2nd version of the guide.

Disclaimer: I think it’s worth mentioning that when I mention “gambles where you have the advantage” I’m excluding almost all games that involve real money, like casino games at face value (this wouldn’t include, for example, card counting strategies in Blackjack) or sport bets. I strongly recommend against participating in this type of games if not for the experience and with the aim to make money. No strategy is going to save you out of losing money in the long run.

Objectives

• Figuring out what is the optimal way to get high amounts of points in the shortest time possible.
• Figuring out what the optimal strategy to get all the achievements is.

The first objective is separated from the second because the longest part of the game is to get to 1 million points. In fact, when reading the million point strategy you won’t probably need help getting the other 10 achievements (as some of them involve getting lower than 1 million points and others expect you to spend sums of them). But now that we’re here, solving the toughest challenge, we can go out of our way and try to solve the other.

5050, a game of equal odds?

5050 is a game that on its surface may seem like a game in which the chance to double your bet is 0.5 (50%), and the chance to lose it is also 0.5, hence the title. This was also the case for me during the first swings:

“it’s only fair that the chances are distributed like this, that would make the game fair and, although quite hard, it is thrilling to find out what’s the highest score you can reach”.

But, have we realised that, whenever we play, we win too much? Nah. This must be only product of our imagination. Well, this is not a trivial question. When flipping a coin in real life, we assume the odds of calling the right side to be 50-50, but the developer of a game can set the bar to any other number from 0 to 100 (0 to 1 in probability). With this in mind, is there a way to empirically determine what are the odds? Well, not quite, but I feel we can make a good estimation. In fact, making a series of assumptions that I will talk about later we could even make a right guess. It will all depend on how good our sampling is, and also on if our assumptions are correct.

Sampling

Let’s play a little with the game. I am going to bet all of my points and see the result for, say, a couple times and write down the results:

• N Result
• 1 W
• 2 W

Well, looks like the probability of winning is 1, isn’t it? Obviously not. Let’s do 10 more rolls.

• N Result
• 3 W
• 4 L
• 5 W
• 6 W
• 7 W
• 8 L
• 9 W
• 10 L
• 11 W
• 12 W

This looks more like it. Well, we have won 9 out of 12 times. So, the probability is 0.75 and we can wrap it up, right? The reality is that every time we increase the sample size there is a very high chance that we get a different result, although the result will be more consistent and closer to the truth with higher sample sizes. This is known as the Law of Large Numbers, or Bernoulli’s Golden Theorem as he called it [2].

Then, what is the desired size for estimating the probability ‘p’ of winning in this game? 100 trials? 1,000? 1,000,000?

First we need to decide how good we want our estimation to be. Let’s say we want our estimation to be right 95% of the time. This is a rule of thumb to make good estimations based off empirical evidence.

But we also need to be exigent about the result, so we are going to force it to be not more than 0.05 off (this is, that we want to find p ± 0.05, so if we get the estimation that the probability of winning is 0.75, the true value could lie between 0.7 and 0.8 with a 95% probability).

Now brace up, because I’m going to need to make a big jump that I could only explain with a blackboard and chalk, but I’m going to leave a link [3] of a math stackexchange problem very similar to this one.

Assuming that the binomial (coded 0 for losses and 1 for wins) distribution of the results in our game can be approximated to a normal distribution [4] by Central Limit Theorem we have an optimistic formula that will let us know what a good enough n will be. This value is optimistic in that it assumes p is equal to 0.5 but the n doesn’t increase greatly unless p approaches 0 or 1, but since we are approximating the binomial to a normal distribution, we already need to make sure p is not too small or too big. As a rule of thumb, the approximation is good if np > 5 and n(1-p) > 5, when n is the sample size. So, the high sample sizes are most probably going to be good enough to approximate p.

The result that we get from the formula is n = 1.96^2/(4*0.05^2). This ends up being 196. 196 multiplied will be higher than 5 as long as p lies between 0.97 and 0.03.

So now we only need to push the button 184 more times:

• N Result
• 13 W
• 14 W
• 15 L
• 16 W
• 17 W
• 18 W
• 19 W
• 20 W
• 21 L
• 22 W
• … …
• 192 L
• 193 W
• 194 W
• 195 W
• 196 W

Which yielded a total number of 152 wins and 44 losses. But first I need to also reveal something: The first time I pressed the button in my 44 runs, I always succeeded. Which leads me to think that 44 of these 152 wins are forced. That being the case, the real number of coinflips that I performed was 152, and thus I need to do 44 other real flips.

• N Result
• 155 W
• 156 L
• 157 L
• 158 W
• 159 L
• 160 W
• 161 W
• 162 L
• … …
• 194 W
• 195 W
• 196 W

This time around you can see some losses in a row and overall a bigger distribution of losses. The real number of wins (not including the first button press) amount to 137 out of 192, or 0.714 as our estimation of probability (vs 0.710 when we didn’t realise there were forced successful results and worked with only 152 trials).

What is the confidence interval of this proportion? If we did the math right before, it shouldn’t be very different to 0.714 ± 0.05.

After doing the math, I got that the result is 0.714 ± 0.06 (because of our optimistic calculation of n=196), not 0.05, but it’s good enough. This means that the value could be between 0.65 at worst and 0.77 at best, 95% of the times. Getting a much more precise measurement would probably require to multiply the number of trials, so we are going to leave it here and assuming the developer put a rounded number, we are going to work with the number 0.7, so 7 out of 10 button presses are wins.

Decision making – Getting to 1 million

Now that we think our chance of winning is 0.7, we can make decisions based off on that. The first thing you may have noticed (and one of the aims of this guide) is that the EV is positive (>0). Specifically, it is EV = 0.7x – 0.3x = 0.4x, where x is the number of points we bet (which can only be positive). This roughly means that for every bet we are going to win on average 40% of what we put on the line.

But that doesn’t mean that every type of bet is going to net us benefits. For example, always going all in is going to do the exact opposite, it’s going to lead us to bankruptcy the fastest. This is because, although our winnings are fast, the probability of having to start all over again is 0.3 each roll, and the probability only goes higher for every round we survive. Take into account that to go from 100 to 1,000,000 we need to multiply our entire bankroll by 2 ⌈log2(1e6) – log2(1e2)⌉ times, which equals to 14, so we would have a probability to lose all before reaching the million equal to 1-0.7^14, or 0.9932. Following this strategy could only be optimal for speedrunning, but would maximize the number of button presses altogether vs any other one.

The opposite type of bet is the most conservative of them all. Betting 1 each time. Well, assuming the button worked every time this way, you would be in to press the button 999900 times (999800 times if you bet 100 in your first go, by the way), and if we pressed the button to see the result every second, we would take 12 days without rest to arrive to 1 million.

Therefore, we are obviously looking for some proportion inbetween. The number must be a proportion and not a fixed amount because, again, the most direct proof of this not being the right strategy is that whatever number you pick at first (from 1 to 100, or 200 with the first roll being confirmed win), the most optimistic number of button presses you are going to take is 5000 (if the button never failed), which again amounts to more than an hour.

This is where I stumbled with the [1] reference, which I put at first because in reality this is the central reference to the whole study. The optimal fraction of your bankroll to bet equals to.

• f* = p – (1 – p)
• for even bets. In our case, that is
• f* = 0.7 – (1 – 0.7)
• f* = 0.4.

The expected growth rate of our bankroll with this strategy is.

• r = (1 + f*)^p * (1 – f*)^(1-p);
• r = (1 + 0.4)^0.7 * (1 – 0.4)^0.3;
• r = 1.086

Which takes us to 1 million points in ⌈log1.086(1e6) – log1.086(1e2)⌉, or 112 button presses on average. Quite impressive, if you ask me.

Decision making – 100% the game.

So, to achieve 100% we need to fulfill 11 conditions, some of which are redundant with others. I will go over the 11 conditions right now and strike the ones that logically don’t involve more button presses.

• Congratulations! – Succeed once. Redundant.
• Beekeeper – Lose your entire point balance once, and after that click on the bee. Redundant, because after “Going out with a bang” you are going to directly see it, so there is no need to lose on purpose to achieve this one.
• Expecting a loss – Insert 2% or less of your points (with 100 or more points total).
• Oof – Lose at least 60% of your point balance with a point balance of at least 1000. I guess this one is redundant because “Going out” is a 100% loss of 1mill points, but this needs to be tested.
• All or nothing – Go down to the last point.
• Good progress – Reach a point balance of 50,000
• 6 figures – Reach a point balance of 100,000
• Maths challenge – Reach a point balance of exactly 1234.
• Half way there – Reach a point balance of 500,000
• You win! – Reach a point balance of 1,000,000
• Going out with a bang. – Insert all your points with a point balance of at least 500,000. It is not exactly redundant but you will get this achievement as long as you make a button press with all your bankroll and are over 500k, therefore this achievement should be the next button press after getting to 1mill.

And listed now are the conditions that we need to go out of our way to fulfill.

• 3. Expecting a loss – Insert 2% or less of your points (with 100 or more points total).
• 5. All or nothing – Go down to the last point.
• 8. Maths challenge – Reach a point balance of exactly 1234.
• 10. You win! – Reach a point balance of 1,000,000

Finding out what the most optimal route is complicated, so rather than that, I’m going to propose tips for getting the non million achievements in an order that looks very acceptable.

5. All or nothing. This can’t be done in the end of the million point run, because we need “Going out with a bang”. When going down we also need to end the game as quickly as possible (going all in as much as possible) Thus, this achievement and the million need to be done at different runs. However, we don’t need to decide which of the runs we need to do at first. So here’s what I recommend:

At the beginning of the game, bet 100 points (the first try is an assured win anyway).
In the second round, bet 199 (this can either drop you to 1 with a 30% chance or put you in 299).
Third round, bet 298. At this point, the chance to drop to 1 combined with the one before is 51% (1- chance of two successes), so we either get it or don’t. If we don’t, then it’s more convenient to run for the 1234/million and hope for a shorter drop in the second run. If we do, then as mentioned before, it’s time to all in until we lose and thus get back to 100 as quickly as possible or, if you are particularly lucky, after reaching 128 stick to the original strategy of betting 40% of the bankroll.

3. Expecting a loss and 8. Maths challenge. These two achievements are probably easier to do together since when going for the 1234 we need to be precise. First thing is that we don’t want to overshoot. This is because the chance of winning and going further from the 1234 is going to imply not only more tries to get to 1234 but also more tries spent recovering that money. Another important thing is that we don’t want to lose time recovering money to aim back again for the 1234. Therefore, the easiest route is probably to aim for 1233 and then just bet one to get both achievements at once. To aim for 1233, just bet from 40% to 40% until you reach 881, then bet as many points as you need to reach 1233 (just subtract your amount of points from 1233, and bet that amount of points).

After doing this you can keep building up to the million, the billion or virtually any amount that you want. Getting high amounts is trivial if your betting strategy, whichever it is, has a positive rate of growth r. And in our example as long as 0 < (1 + f)^0.7 * (1 – f)^0.3; 0 < f < 0.717, the betting strategy will have a positive rate of growth. The only thing that will vary is the speed at which you get the points.