Not sure how to title this except - Probability of a sequence occuring based on a %

Nukkas

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Jun 14, 2024
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Been way too many years outside of basic arithmetic alone but I am trying to work out the formula for the likelihood of a string of success based on a %

For example - If the odds of a outcome are 50% (coin toss for example) and I need to have 3 in a row, whats the probability.

This needs to be expanded to cover more consecutive events with a changing % (5 x 49%, 4 x 62% etc etc). I know it shouldn't be difficult but my aging Friday brain just ain't functioning.

Where am I going with this - Well I play a game that has upgrade options. Each upgrade requires you to either have a successful outcome or not. When it's not successful you have to start again. Each upgrade tree increases in cost, and the probabilities generally get worse (starting for outright success 2 x 100%, down to just painful levels like 5 x 49%) I am trying to work out the expected cost of each upgrade. Using the previous 2 x 100% at 1000 means it will cost 2000, and that's guarenteed however at the lower levels, the cost per action might be 3000 but the chances are much slimmer. You might get away with it being 15000, or it could be 150000. You can use other resources to boost the odds, or flat out just pay for the upgrade and that's what I am trying to work out, whether some upgrades are just worth it, or spinning the wheel of fun.

Hopefully that ramble provides enough information.

Thanks
 
For example - If the odds of a outcome are 50% (coin toss for example) and I need to have 3 in a row, whats the probability.
If outcomes of coin tosses are independent of each other then the probability are multiplied, i.e. [imath]p = (1/2)^3 = 1/8 = 12.5\%[/imath].
This needs to be expanded to cover more consecutive events with a changing % (5 x 49%, 4 x 62% etc etc).
Same as above: just multiply the probabilities, provided the events are independent.
 
If outcomes of coin tosses are independent of each other then the probability are multiplied, i.e. [imath]p = (1/2)^3 = 1/8 = 12.5\%[/imath].

Same as above: just multiply the probabilities, provided the events are independent.
Thanks, I knew it looked too simple but the brain just wasn't math'in.
 
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