For fun Suppose you roll a die, if you get 1 and 2, i give you 200$, but if you get 3,4,5,6 you give me 100$. whats the chance that you would lose any money if u rolled the die 10 times
I never lose.
What are the chances of me rolling a die to lose? How many possibilities out of 6?
im asking :P
No you asked a question. I am asking you a part of your question. You should know at least this.
i know how to solve it :O just for fun
-.-
xD
The answer should be 20/3 but i might be wrong.
you are wrong for one reason 20/3 >1
I told you I was wrong The chances of me rolling a die to lose is 4/6 == 2/3 I don't know how to account for it 10 times
hmm well think again 2/3 is high value, also its binomial distribution .... try it :)
I honestly don't know. Show me yourself.
i wont reveal lets see if others can do it ^_^
remember my question is asking about losing money, (not guessing wrong)
haha, well the cute thing is the chance is pretty law so if it was gambling i'll play this game :D
expected value = 0
hmmm interesting ...
but no xD
out of 10 rolls, you need to loose at least 7 times to make negative money
yessss exactly xD
simply find P(x >= 7)
yesss !
OHHHH
You can do a probability tree to find out the possibilities of losing
:D
that doesn't look right, one sec
hehe see its small value
binomial distribution : \[P(X\ge 7) = \sum\limits_{k=7}^{10}\binom{10}{k} (4/6)^{k} (2/6)^{10-k} = \dfrac{11008}{19683} \approx 0.559 \]
it should be 1- ur value right ?
\(0.559 = 55.9\%\) chance for loosing money
thats not small, i wouldn't be willing to play this game wid u lol
hehe ok i made same mistake u did id first reply :P
would u play tossing a coin ?
it is wise not to play any casino games wid statisticians :P
lol why so
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