# Understanding example on VaR

Trying to understand the example below on VaR in Wikipedia. I don't really understand how the 1% VaR is being defined here. Firstly, shouldn't it be 1% Var is 100 since its the amount her looses? And secondly, I dont understand this statement: "the probability that he loses less than zero (which is impossible) is less than 99%.". Given 1% Var = -100 as per their definition shouldnt it be "the probability that he loses less than -100 is less than 99%." and not 0?

More formally, p VaR is defined such that the probability of a loss greater than VaR is less than or equal to p while the probability of a loss less than VaR is less than or equal to 1−p. For instance, assume someone makes a bet that flipping a coin seven times will not give seven heads. The terms are that he gains 100 if it doesn't happen (with probability 127/128) but loses $12,700 if it does (with probability 1/128). The 1% VaR is then -100, because the probability that he loses more than that is less than 1% while the probability that he loses less than zero (which is impossible) is less than 99%. Thanks ## 1 Answer The way I understood it is as follows: • If you DO NOT get 7 heads out of seven flips, then you make 100\$;
• If you DO get 7 heads out of seven flips, then you lose 12 700\$; The probability of NOT getting seven heads is$\frac{127}{128} = 99,21875\%$and$1-p = 0.78125\%$. Therefore, with 99% confidence level VaR you are still making 100\$ (I think that's why there is a minus sign). Since VaR ignores tail events, you are not losing at 99% VaR. Hence the statement:

... the probability that you lose MORE than -100\\$ is less than 1%, while the probability that he loses LESS than zero (which is impossible) is less than 99%.

Maybe taking a look at the original statement by Einhorn may be helpful: