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# Probabilities and How many times do I got ta do this?

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This came up while playing Minecraft with a couple mods. In the game there was a machine that offered a trade of your stuff for a specific chance to get some cool loot. The same basic thing occurs in Perfect World, where you can trade some of your gathered resources for a chance to improve some equipment. And again in Tynon, where you can trade stars for a chance to upgrade a player character (although the mechanics are a bit more convoluted in Tynon).

If there is a 10% chance to succeed each time we burn a resource, then 90% of the tries will fail. Burning two resources would be 90% * 90% = 81% both fail. Three tries would be 90%*90%*90%=73% all fail, and so on. So we can express the chance of not getting any success as P(TotalFail) = P(eachTryFail)^numTries.

We know the eachTryFail is 0.90, so we've currently got P(TotalFail) = P(0.9) ^ numTries.

But I want to know how many times I should burn a resource, in order to be resonably certain of getting at least one prize. In order to do so, I have to define what Reasonably Certain means for this problem. It's a willingness to be wrong, a risk factor, and for this problem where the penalty is just a little amount of my time, I'll go with the standard 95% certainty. If the situation were about public safety, or a production line, we'de pick a much higher value. But for this exercise, being wrong 1/20 doesn't hurt too much.

With this decision in place, we now have a definition for TotalFail; we have decided that it will be 1/20.

P(totalFail) = P(eachTryFail)numTries

0.05 = 0.9numTries

log(0.05) = log(0.9) numTries

log(0.05)/log(0.9) = numTries

numTries =~ 28

Edited by AngleWyrm

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Yes, it looks correct. You have to try 29 times so the probability of total fail goes below 0.05.

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I think what you are after is the probable number of trials until the first success, which is given by the geometric distribution.

-Josh

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I think what you are after is the probable number of trials until the first success, which is given by the geometric distribution.

Looks like I've done pretty much this.

Edited by AngleWyrm

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