simple EV calculation, but I don't know how to set it up?

[benneh]

apologize if this isn't this place. wasn't sure where else to put this.

this should be really simple, but i just dont know how to set it up. no need to solve it for me, i dont want you guys doin my homework (finals today)

Dealt a 5 card hand from standard 52 card deck

if one king shows up, you win $10. otherwise, you lose $1

what is your EV?

clarification: At least one king. having a pair or trips does not affect the outcome

[LetYouDown]

Well, the probability of 5 random cards containing at least one king should be roughly:

1 - C(48,5)/C(52,5) = 34.1% or about 1.93 to 1. I come up with roughly $2.76 profit per "hand".

[probman]

Let p=prob(no king shows up in five cards). It is easy to see that

p = (48/52)(47/51)(46/50)(45/49)(44/48) = C(48,5)/C(52,5).

We then have

EV = pE[V|no kings] + (1-p)E[V|at least one king]
EV = (-1)p+10(1-p)
EV = 10-11p

[phoenixbets]

Fastest way is to calculate odds of not getting any K at all, i.e., do it "backwards." That's
52 cards, 4 are Kings, 48 aren't, so it's
48/52*47/51*46/50*45/49*44/48 or .66 or you will not get one 66% of the time. That means you will get one 34% of the time. In other words, it's awfully close to lose twice, then win once.

I think I'm right, but if not I'll find out soon enough and learn something along with you.

You want to take it from there, based on what you said?

[benneh]

LetYouDown had it right the first time, though I didn't pick up on it til later on in the day when I realized his C stood for combination. Duh! I'll do it out with variables sine i dont have a calculator on me to speed it up. should be easy to follow along in case anyone else wanted to know how to figure somethin out like this (though i doubt it unless you're playin weird gamblin games)



x = C48,5/C52,5

EV = (1-x)(10)+(x)(-1) for some answer like $2.75 or so