Need 27 "wins" at 20% chance with 46 tries

[Soggyshorts]

Hello, I think it's been too many years since I was in school...

I'm trying to figure out what my chances are to win something that has a 20% chance. I get 46 tries, but need at least 27 wins.
I managed to figure out my chances of winning at least once using

1-0.8^46 =0.999965155%

but I'm not sure where to go from there.

Any help would be appreciated, thanks!

 

[ksdhart2]

Well, you're actually nearly to the solution, just a bit more reasoning is needed. You've correctly identified that the probability of getting at least one win is the same as 1 minus the probability of getting zero wins. If we let N be the number of wins, then the notation for that is:

\(\displaystyle P\left(n\ge 1\right)=1-P\left(N=0\right)\)

Extending that principle, the probability of getting at least two wins is 1 minus the probability of getting zero wins minus the probability of getting exactly one win, or:

\(\displaystyle P\left(n\ge 2\right)=1-P\left(N=1\right)-P\left(N=0\right)\)

Then the probability of getting at least three wins is....? The probability of getting at least 27 wins is...?

 

[Soggyshorts]

At least 3 wins is
P(n≥3)= 1-P(N=2)-P(N=1)-P(N=0)
At leat 27 wins is
P(n≥27)= 1-P(N=26)-P(N=25)-P(N=24)-P(N=23)....-P(N=0)


Ok, the gears are slowly starting to turn again

EDIT:
Back to at least 2 wins

P(n≥2)= 1-P(N=1)-P(N=0)
P(n≥2)= 1-0.2^46-0.8^46

ahh **** I thought I had it for a sec. brb

EDIT2: I think I know what I'm missing, I know how to get the probability of at least one win, but for the formula to work I need to know how to get the probability of exactly 1 win.
Not sure how to do that

Edit3:
is it the 46 that needs to change? 45,44,43?
P(n≥2)= 1-0.8^45-0.8^46
=-0.999921599
looked good for a moment before I saw it was negative... I think I'm stuck.

 

[Ishuda]

Hello, I think it's been too many years since I was in school...

I'm trying to figure out what my chances are to win something that has a 20% chance. I get 46 tries, but need at least 27 wins.
I managed to figure out my chances of winning at least once using

1-0.8^46 =0.999965155%

but I'm not sure where to go from there.

Any help would be appreciated, thanks!

How many ways can you win M times. See the binomial distribution
https://en.wikipedia.org/wiki/Binomial_distribution
Now add them up.

 

[Soggyshorts]

How many ways can you win M times. See the binomial distribution
https://en.wikipedia.org/wiki/Binomial_distribution
Now add them up.

I'm going to have to come back tomorrow, that wiki page is little over my head.

 

[ksdhart2]

I appreciate that the Wiki can certainly seem intimidating, but it's really not as bad as it seems. Basically, it says that, given n trials, to get exactly k wins you have to have k wins and n-k losses. For instance, to get 1 win you'd also have 45 losses. So the formula should be:

\(\displaystyle \displaystyle P(N=1)\overset{?}{=}0.2^1 \cdot 0.8^{45}\)

Except, well, not exactly. That only accounts for one of the ways you can have exactly 1 win. Out of 46 trials, you have to choose 1 of them to be the one you win. So there are "46 choose 1" possible outcomes. Thus the complete formula is:

\(\displaystyle \displaystyle P(N=1)={46 \choose 1} \cdot 0.2^1 \cdot 0.8^{45}\)

Then for finding exactly 2 wins, you'd have to pick 2 of 46 elements to be winners, or use "46 choose 2." To calculate these values, you can apply the general formula for "n choose k:"

\(\displaystyle \displaystyle{n \choose k} = \dfrac{n!}{k!(n-k)!}\)

Or, if you're familiar with Pascal's Triangle, "n choose k" is equal to the (k+1)th element of the nth row of Pascal's Triangle.

 

[Ishuda]

I'm going to have to come back tomorrow, that wiki page is little over my head.

You don't need the complete article. Perhaps a different example might help a little more: Suppose it were 3 chances for at least 1 win. For 1 win and 2 losses, we would have
p¹ (1-p)²
or, in this case
0.2¹ 0.8²
but you can arrange those three [=\(\displaystyle \displaystyle {3 \choose 1}\)] ways, that is WLL, LWL, or LLW. Thus your chances of winning exactly 1 time and losing twice is
3 0.2¹ 0.8² = 0.384

Since \(\displaystyle \displaystyle {3 \choose 2}\) = 3, your chances of winning exactly 2 times is
3 0.2² 0.8¹ = 0.096

Since \(\displaystyle \displaystyle {3 \choose 3}\) = 1, your chances of winning exactly 3 times is
1 0.2³ 0.8⁰ = 0.008

Thus your chances of winning at least once is
0.384 + 0.096 + 0.008 = 0.488
which agrees with one minus losing all three times
1 - 0.8³ = 0.488

The notation \(\displaystyle \displaystyle {n \choose k}\) is the binomial coefficient
\(\displaystyle \displaystyle {n \choose k}\, =\, \dfrac{n\, (n-1)\, (n-2)\, (n-3)\, ...\, (n-(k-1))}{k!} = \dfrac{n\, (n-1)\, (n-2)\, (n-3)\, ...\, (n-(k-1))}{k\, (k-1)\, (k-2)\, (k-3)\, ...\, 1}\)

 

[pka]

I'm trying to figure out what my chances are to win something that has a 20% chance. I get 46 tries, but need at least 27 wins.
I managed to figure out my chances of winning at least once using

This is the sum you need. It calculates the probability of winning from 27 to 46 times.
\(\displaystyle \displaystyle\sum\limits_{k = 27}^{46} {\dbinom{46}{k}{{\left( {.2} \right)}^k}{{(.8)}^{46 - k}}} \). See the calculation here.

This is straightforward binomial probability.

 

[Soggyshorts]

This is the sum you need. It calculates the probability of winning from 27 to 46 times.
\(\displaystyle \displaystyle\sum\limits_{k = 27}^{46} {\dbinom{46}{k}{{\left( {.2} \right)}^k}{{(.8)}^{46 - k}}} \). See the calculation here.

This is straightforward binomial probability.

Thank you very much!

9.64450776902122191578750317733347328×10^(-9)



How do I take the last step and convert it to a percent chance?

 

[ksdhart2]

Thank you very much!

9.64450776902122191578750317733347328×10^(-9)



How do I take the last step and convert it to a percent chance?

You'd convert that the same way you can convert any other number to a percent. Just as 5 is 500%, 3.76 is 376% and 0.67 is 67%. Is it the scientific notation that's tripping you up? If so, perhaps it will help to realize that multiplying by 10^-9 is the same as moving the decimal point 9 places to the left. The answer's going to be very small, far far less than 1% (actually not even 1% of 1%) but that's to be expected of a really rare event.

9.64450776902122191578750317733347328 * 10^-9 = 0.00000000964450776902122191578750317733347328

 

[Soggyshorts]

You'd convert that the same way you can convert any other number to a percent. Just as 5 is 500%, 3.76 is 376% and 0.67 is 67%. Is it the scientific notation that's tripping you up? If so, perhaps it will help to realize that multiplying by 10^-9 is the same as moving the decimal point 9 places to the left. The answer's going to be very small, far far less than 1% (actually not even 1% of 1%) but that's to be expected of a really rare event.

9.64450776902122191578750317733347328 * 10^-9 = 0.00000000964450776902122191578750317733347328

That's what I feared. I thought it was too small a chance so I must be wrong. At a glance I figured I "should" get 9 wins, so getting 27 wouldn't be impossible.
I know it's a math forum and I probably am not supposed to use words like "should" or "impossible" but for my purposes anything under 1 in thousand(let alone 1 in 100mil) might as well be.

Thank you again!