Questions d'entretien

Entretien pour Trading

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Flip a coin. if head, I win 1 point. If tail, u win 1 point. The person who first wins 2 points wins the game. And the loser must pay winner \$1. However, if I have a option which I can increase the stake of the game to \$2/game, what is the value of such option.

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8 réponse(s)

3

Anonymous is right but the option is going to be used for 2/3 times. expected gain with option is 2/3(1) + 1/3(-1) = 1/3. expected gain without option is 0. the value of the option is then 1/3.

Utilisateur anonyme le

4

dont believe above answer is correct: as correctly mentioned, this option would only be used if you are 1-0 up. Of the next 2 coin flips for (assuming i want heads), after the 1-0 scenario is HH, HT, TH, TT. Of which, all but 1 will give me \$2. So \$2 x 0.75 = \$1.5 in. compared to \$-2 x 0.25 = -\$0.5 averaging out the above gives you +\$1 expected value. Going further (i.e. what is this option worth to you before the game starts), you start off 0-0, you wouldnt use it. you go 1 down, you wouldnt use it, you go 1 up you do use it. It will only be used in 50% of scenarios, therefore you pay no more than \$0.5 for it.

Utilisateur anonyme le

1

option is worth 0.25 at time 0. at time 1 (after one flip) the option is worth zero or 0.5 (as mentioned above) depending on whether it will be exercised. draw out a tree and this can be clearly seen

Chainsaw le

0

I'm actually posting this answer partially to ask for the community's input. The way I read the question, it doesn't say when the option can be exercised, so why not exercise it once you've won? Then you can view this as an option that pays 1 if you get 2 heads in 3 tosses or pays 0 if you get 2 tails in 3 tosses. That means if the score is 1-1, the value of the option is worth 0.5, which means that at 1-0 it is worth 0.75 and if the score is 0-1 then it would be worth 0.25, which means that when the score is 0-0 the option is worth \$0.50. This agrees with Anonymous's answer, but seems to be for fairly different reasons. For instance, I don't see why you would exercise it when up 1-0, as opposed to when the game is over (I dunno, much like exercising a European digital option, which is what I thought of when I saw this question).

wlfgngpck le

0

I think if we draw the tree (my score vs. opponent's score): / 2-0 1-0 \ / 2-1 0-0 / 1-1 \ 0-1 / \ 1-2 \ 0-2 (The up-up-up branch and the down-down-down branch are pruned because there's no need for the 3 toss) Then backward induction seems to give time zero value = \$0.5. (boundary condition is \$1 for 2-1 and 2-0, \$0 otherwise.

?H? le

0

Consider the following outcomes and their respective probabilities and payoffs: (1-0) -> (2-0) : 1/4 : +£2 (1-0) -> (1-1) -> (2-1) : 1/8 : +£2 (1-0) -> (1-1) -> (1-2) : 1/8 : -£2 (0-1) -> (1-1) -> (2-1) : 1/8 : +£1 (0-1) -> (1-1) -> (1-2) : 1/8 : -£1 (0-1) -> (0-2) : -£1 The expectation is 1/4 (£2 - £1) = £0.25

John le

0

There is a lot of confusion up here. This is the right solution. If the option is not used, the value of the game is 0. If the option is used at any time (i.e. without an optimal strategy), the value of the game is again 0. So when does it make sense to use the option? Well, it makes sense when I am 1-0 up because in that case I am "closer" to winning the game. What is the value of the game playing optimally then? Before starting the game: E = 1/4*(-1) + 1/4*(+2) = 0.25\$ (because in case of a draw the value of the game is zero). After starting the game, if I am 1-0 up and want to use the option, then the value of the game changes, because I have more information from the game (like when you play poker and odds change when cards are flipped). Therefore, when I am 1-0 up, the value of the game is: E = 1/2*(+2) + 1/4*(+2) + 1/4(-2) = 1\$ After starting the game, if I am 1-0 down (i.e. I lost the first round), I am closer to lose, hence I do not want to use the option! Therefore, when I am 1-0 down, the value of the game by not using the option is: E = 1/2*(-1) + 1/4*(-1) + 1/4(+1) = -0.50\$ If instead I had used the option when being 1-0 down, the value of the game would be -1\$. This is why I should not use the option when I am 1-0 down.

Utilisateur anonyme le

0

Expected value of game w/o option is 0. Expected value with option is conditional on you exercising it which rationally should be only when u are 1-0 up when ur probability of winning is 3/4. U win an extra pound 3/4 of time and pay an extra pound 1/4 of the time so expected value of option is 1/2. Open to suggestions

Utilisateur anonyme le

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