How many times do I have to try for winning something if my chances are 20%?
In: 1
1 time: 1/5
2 time: 9/25
3time: 64/125
So your chances are over 50% if you try 3 times. If that is not enough, it is very likely that you will get it in 5 tries. However if you are unlucky, you might not get it no matter how much you try. Lets not forget that your chance is always %20 for each try. So if you didn’t get it even after 10 tries, your chance is still %20 for your next try.
On average, you should win once every 5 tries.
In practice, it could take you any number of tries.
If you try that thing enough it’ll eventually work itself out to 20% Odds are very good you’ll get around 20 successes out of 100, though it’s all luck so you might get 15-25. If you’re astronomically lucky you might get a lot more. But like the others said it’s all a 1-5 chance.
It depends how the probability is applied. ex: You play backgammon and have red and black places to land only. You see black has hit 15 times in a row and know that there’s a 50/50 chance of red hitting. You think to yourself “what are the odds of black hitting 16 times in a row and place all your money on red. It ends up black and you lose. The casino shows past plays bc they know the ignorant play on these false probabilities. You have a 50% chance of hitting red every spin regardless of what happened in all the past rolls (assuming no green places to land).
So if your game is anything like backgammon or flipping a coin, you have a 1-5 chance every time the game starts anew. Perhaps you can shine some more light on the game of chance you are playing?
Your chance of winning at least once is 1-(0.8)^n, where n is the number of tries you’re willing to take.
So if you get 10 tries, you have a ~89% chance of winning.
How many times do you need to flip a coin before it comes up heads? Probability-wise, only a few times — but it could *never* come up heads, no matter how many times you flip it.
Same for any percentage. The odds of winning on any particular try remain 20% (or whatever) independent of how many times you’ve tried before. However the odds of losing N times in a row are 0.80^N, and so increasingly *improbable*.
But not *impossible*.
you’ll never reach 100% certainty of winning.
the way to calculate it is actually to calculate the odds of *not* winning.
for each try, the odds of not winning is 80%. or 0.8.
then just raise it to the power of the number of attempts.
so after 1 try, the odds of not winning is 0.8. so the odds of winning are 20%
after 2 tries the odds o not winning is 0.8^(2 =) 0.64 = 64%. So the odds of winning are 36%
after 3 tries the odds of not winning are 0.8^(3) = 0.512 = 51%. So the odds of winning are 49%
after a large number of tries, the odds of not winning at least one get very small, and the odds of winning at least once are very high, but never 100%
It depends on if you’re talking dependent or independent probabilities.
If you have a 5-sided die, each time you roll there is a 20% chance it will land on a ‘1’. No matter how many times you roll it, the chance is still the same for the trial. To calculate the chance of getting some exact sequence of rolls involves some multiplication. A trivial example of this would be “what is the chance I don’t see a ‘1’ after n rolls?” – which is .8^n. This means you are never guaranteed to see a ‘1’ no matter how many times you roll.
On the other hand, if I have a jar with 4 red pebbles and 1 black pebble, the chance of drawing a black pebble is 20%. But if I’ve already drawn a red pebble, the chance of drawing a black pebble is now 25% (since there are only 3 red pebbles left out of 4 total pebbles). The probability of each subsequent trial is dependent on the trials that preceded it. If I draw 5 pebbles from that jar, it should be obvious that I’m guaranteed to get a black pebble.
On average 5.
But if you mean “how many times for the victory to be guaranteed”, the victory is never guaranteed no matter the number of tries or the chances (unless it’s 100%).
A lot of other answers are giving the odds, but to apply those answers to your original question “how many times do I have to try”, the answer is related to how “sure” you want to be. Most people mean, “how many times to I have to try before I can win ‘consistently’?” But “consistent” could mean a number of things.
From the answer of /u/chillbo-swaggins:
>so after 1 try, the odds of not winning is 0.8. so the odds of winning are 20%
>
>after 2 tries the odds o not winning is 0.82 = 0.64 = 64%. So the odds of winning are 36%
>
>after 3 tries the odds of not winning are 0.83 = 0.512 = 51%. So the odds of winning are 49%
In other words, if you took 100 people and had them each try 3 times, around 49 people would win at least once, while around 51 people wouldn’t win anything. But a lot of people probably wouldn’t count a 49% overall chance as good enough to be a consistent win and wouldn’t accept “You have to try 3 times” as an answer.
So maybe a more than half chance is good enough for you? In this case, 4 tries will give you a 59% chance of at least one success. If you prefer 75%, 6 tries will get you odds of about 75% for one success. Maybe you want around 90% chance? You’d have to go up to 10 or 11 tries. As you can see, the price of increasing our confidence is that we have to prepare for trying at lot more (to account for “bad luck” cases). The extreme version of this is, as many stated, that 100% confidence is literally impossible.
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