Virgin Galactic trip to Mars progressive jackpot probability - expected value?

On a long, slow trip to Mars, Virgin Galactic is entertaining its passengers with a prize contest.

Every hour there is a game with 40% odds each time. If you don't win that hour, then you have the same odds to try again and again (independent trials) with increasing prize amounts per the following sequence: $1, $1, $2, $2, $5, $5, $100, $100.

The sequence restarts after not winning 8 straight or if the passenger wins before then.

The ship's computer automatically enters you in the contest, passengers need not be present or awake to play, nor is there a risk for them to play.

What is Virgin's expected cost per hour, per passenger? That is, what is the expected value of each hour of travel for a passenger?



...sounds like fun. Any comments on the math? Good luck guys! Lucy.

no risk for them to play????????

it wouldn't be odds, then, it would just be a 40% chance to win.

this sounds like a homework assignment

Right. This is like Roulette, or dice. No matter how many times you do it, it's still gonna be 40%. (Not sure if that's in the house's favor or the passengers', but it doesn't really matter.)

This is not a progressive jackpot as I understand the concept.

If I were Virgin Galactic, I'd just run a keno game.

Wow.

This must be a doozy. I'm getting a lot of views. Can anybody attempt the math?

Quote: SteamPnkLucy

Wow.

This must be a doozy. I'm getting a lot of views. Can anybody attempt the math?

The math isn't too hard. It just sounds like a homework assignment.

Quote: SteamPnkLucy

This must be a doozy. I'm getting a lot of views. Can anybody attempt the math?

I think several people have already worked out the math, but, since this smells like a homework assignment, we'd like to see your own attemp before we provide our results.

For what it's worth, it's not that complicated.

BTW: A real gambling application, not homework.

Quote: SteamPnkLucy

Wow.

This must be a doozy. I'm getting a lot of views. Can anybody attempt the math?

What a sell!!!

You're trying to work Psychology on a Forum full of Advantage Players, Dealers, Casino Higher-Ups, Game Designers and just straight-up hardcore lovers of gambling such as myself? I see the wool, but it ain't going over my eyes!!!

You'll never get anywhere by having people do your work for you, unless it is extremely complex. I'm going to help you, and I'll even do it for you if I have to, but you have to take a stab at it first. If I just do it, you'll have learned nothing.

I will get you started:

What did you say, eight passengers?

The first hour gives all passengers the opportunity to each win $1.00. They are 40% to win this dollar (and are risking nothing) so 1 *.40 = $0.40/passenger

There are eight passengers, so 8 * .40 = $3.20 paid out on all first hour wins, on average.

SECOND RUN

I will get you started, you are expected to finish, but you can finish by applying the same principles.

In order for there to BE a second run, the first run must have failed. You need to account for the 60% chance of first run failure, followed by the 40% chance of second run success.

6/10 * 4/10 = $0.24 expected to be paid to each passenger on the second run.

$0.24 * 8 = $1.92

You have to divide that by two, however, because it has now been two hours.

$1.92/2 = $0.96

ON YOU:

The hard part is done, figuring out what you need to figure out, so to speak. If you do the rest by yourself, you will at least learn the procedure. There's also a final step after you calculate the returns for each individual item, but you should really be able to figure that one out.

I will check your work when you are done. If you make a minor mistake, I will correct it for you and explain what the mistake was. If you are not even close, I will ask that you refer back to the procedure in this post.

These kind of problems are all about procedure.

It's per passenger, S***, now I have actually done everything for you except the calculations.

This is where I got stuck before I posted:

p = probability of winning

n = the hour (trial, drawing, etc.)

k_n = prize amount at the n^th trial

For n = 1 to 8 and k_n is 1,1,2,2,5,5,100,100 we have SIGMA ((((1-p)^(n-1))*p)*k_n)/n

Divide by n each time to get EV for each trial, correct? Hmmm...

Any new eyes on the math here?

Thanks.

Quote: SteamPnkLucy

This is where I got stuck before I posted:

p = probability of winning

n = the hour (trial, drawing, etc.)

k_n = prize amount at the n^th trial

For n = 1 to 8 and k_n is 1,1,2,2,5,5,100,100 we have SIGMA ((((1-p)^(n-1))*p)*k_n)/n

Divide by n each time to get EV for each trial, correct? Hmmm...

Any new eyes on the math here?

Thanks.

Don't even mess with that crap!

I'm going to pick the Fourth trial semi-random and show you how you do it:

In order for there to BE a fourth trial, the first three trials must lose, which is 60% of the time or 6/10:

6/10 * 6/10 * 6/10 * 4/10

.6 * .6 * .6 * .4 = x

x = .0864 (Liklihood of winning on the fourth trial)

Now, you get paid $2.00 on a fourth trial win so: 2 * .0864 = $0.1728 is the exxpected value of a win on trial four.

However, you need to know the cost of trial four (and actually the total) per hour. Trial four wins on the fourth hour resulting in an hourly cost of:

$0.1728/4 = $0.0432 on Trial Four Wins Alone.

I hope everyone can pick out the equation there.

Sorry, there's no math script here. That tiny subscript is for the index and sigma is the big 'E' for summation.
Maybe you can see through all of the parenthesis.

Seperately, one person put together an EV for the restart time of the sequence. Very clever. Then they made it the denominator of the jackpot EV. Result is EV per hour.

Maybe that's the same as the equation I posted.

Thanks again for whoever wants to put their mind to work.

here it is again

p = probability of winning

n = the hour (trial, drawing, etc.)

k_n = prize amount at the n^th trial

For n = 1 to 8 and k_n is 1,1,2,2,5,5,100,100 we have SIGMA ((((1-p)^(n-1))*p)*k_n)/n

Quote: SteamPnkLucy

BTW: A real gambling application, not homework.

Okay, so you're planning a real trip to Mars. (I was giving you the benefit of the doubt and figuring that it was for a book you were writing.) Since the purpose is to entertain the passengers during the long trip, wouldn't keno be more interesting for them than a game that you can play without even being aware that you're playing it? As a bonus, the house edge is huge. I read somewhere that keno was invented by a Chinese warlord trying to fund an army. Armies are expensive things.

Or, if you're wanting something rigged in the passengers' favor, just offer Baccarat without the Banker commission (and not that no-commission EZ-Bacc, either). Or Roulette that pays out at true odds (assuming that you've worked out some kind of artificial gravity system that allows the wheel and ball to function as on Earth).