On a demandé à un Options Trader...21 juillet 2010

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You can't assume there are two white socks and two black socks. This problem is an easy enumeration problem actually. From the problem, you have at least 2 white socks, otherwise it would be impossible to take out 2 in a row. So you have either 2 white socks, 3 or 4. Just compute the three prob of getting two white socks in a row for all case and see which is 50%. P(two w in a row| 2 white socks) = (2/4) * (1/3) = 1/6 => not two socks. P(two w in a row| 3 white socks) = (3/4) * (2/3) = 1/2 => this is our answer. Obviously, if there were 4 socks, we would have a 100% chance of getting two whites in a row. So as calculated, there must be 3 white socks. Moins

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The probability is 0 as there is only 1 black sock. Solution: Assume there are k white socks and 4-k black ones in the drawer. => 0.5 = P(2 white socks in a row) = k/4 * (k-1)/3 => k^2 - k - 6 = 0 => k = 3 is the number of white socks. qed. Moins

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"50% probability of pulling out two white socks in a row" 0.5 = 3/4 * 1/3 (3/4 of a circle has 3 slices, so take 1 slice out and you have half a circle :p) "containing 4 socks , either black or white" So we can deduce that we have 3 white socks and 1 black sock. There is no chance we can pull out more than 1 black sock. Moins

On a demandé à un Trader, Options...16 janvier 2011

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The probability that you don't see a star in an hour is 1 - 0.84 = 0.16 = 16%. This means that the probability you don't see a star in 30 min is 0.4 (since 0.4 * 0.4 = 0.16). This makes sense if you think of one hour as two subsequent 30 min intervals. So if the probability you don't see a star in 30 min is 40%, then the probability that you see at least one star in 30 min is 60%. Moins

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I'm assuming that the chances of seeing at least one shooting star outside within one hour is 84 percent. I have also assumed that the number of shooting stars you see within a one hour period follows a Poisson distribution with parameter L, which will be determined from the given probability statement. We know that the probability of seeing at least one shooting star in a one hour period is 84 percent, so the probability of not seeing a shooting star in a one hour period is 16 percent. This gives us a formula which will allow us to solve for L: .16=exp(-L). This implies that the Poisson distribution of interest has parameter L=-Ln(.16)~1.83. Now, we can define a new random variable counting the number of shooting stars we see in a half hour period and it follows a Poisson distribution with parameter L/2~.92. So the probability of seeing at least one shooting star with a half hour period is approximately 60 percent. Moins

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Suppose the event meets Poisson distribution. lambda is the rate P(event happens in one hour) = 1-e^(-lambda) = 0.84 e^(-lambda) = 0.16 P(event happens in .5 hour) = 1-e^(-lambda/2) = 1-sqrt(0.16)=0.6 Moins

On a demandé à un Option Trader Intern...14 mai 2014

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Same expected payoff, but second one has much less uncertainty (variance)

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the EV is the same: 10,000 dollars

On a demandé à un Options Trader...7 avril 2021

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Think or Swim by TD Ameritrade

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These are the two best options

On a demandé à un Trader, Options...28 décembre 2009

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(50 - 3) * (20 + 3) = 50*20 + (50 - 20) * 3 - 3 * 3 = 1,081

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(35-12) * (35+12) = (35)(35) - (12)(12) 35 squared is....1,225 (quick trick...take all non-units digits and multiple it by its consecutive integer (in this case 3 * 4 = 12), then place the product in front of 25. e.g. 105 squared = 10 * 11 = 110, place it in front of 25 and get 11,025. 12 squared is...144 225-144 = 22-14 = 8 and 5 - 4 = 1, therefore 81 1,081 Moins

On a demandé à un Options Trader...7 avril 2021

On a demandé à un Options Trader...7 avril 2021

On a demandé à un Options Trader...27 octobre 2011

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Provided statements.

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There are various trading strategies. But some one simply follow disciplined trading and stick to that strategy can get success Moins

On a demandé à un Junior Exotic Option Trader...21 mars 2021