l*******1
发帖数: 113 | 1 You flip a coin. Head, +2, tails -1. You have $50 to start with.
What's the prob that you lose all your 50 bucks before 100 tosses.
Can someone provide a closed form solution for the general case?
Thanks | c**********e
发帖数: 2007 | | d*******r
发帖数: 7 | 3 You need at least 84 tails in 100 trials (-x+2(100-x)=-50).
It's binomial so the prob is sum(100,x).5^x.5^(100-x), x from 84 to 100.
> x=84:100
> y=dbinom(x,100,.5)
> sum(y) | l*******1
发帖数: 113 | 4
there is 1/2^50 chance of having all tails in the first 50 tosses, also 50/2
^53 to have 1 head and 52 tails in 53 tosses... so x goes from 50 to 100.
any ideas?
【在 d*******r 的大作中提到】
: You need at least 84 tails in 100 trials (-x+2(100-x)=-50).
: It's binomial so the prob is sum(100,x).5^x.5^(100-x), x from 84 to 100.
: > x=84:100
: > y=dbinom(x,100,.5)
: > sum(y)
| d*******r
发帖数: 7 | 5 Good point. So we should also include a series of hypergeometric
probabilities. Suppose the nth trial is a tail, we need to lose 49 in the
previous n-1 trials. so n has to be 50,53,56,59...,98(every head counteract
2 tails). Just an idea, but it's doable.The other part was discussed. |
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