ksdjdj
Joined: Oct 20, 2013
• Posts: 909
January 13th, 2019 at 11:52:08 PM permalink
There is a game that is similar to Predictor, see link below:

https://wizardofodds.com/games/predictor/

The key points or differences for my game are:

There is no help file that tells you the maximum payout for the game.

The minimum bet is \$1.00

You have determined that the maximum payout is at least \$6133.

You can draw a maximum of 8 carts per game.

If you lose, you then start a new game back at the 1st cart.

In 5000 games, you have never seen a cart's number repeat in the same game.

From a method similar to card counting, you have worked out the RTP for the best choice on each cart is as follows:

1st Cart is 98%
2nd Cart is 98%
3rd Cart is 100%
4th Cart is 102.5%
5th Cart is 105%
6th Cart is 107.5%
7th Cart is 110%
8th Cart is 113%

Your chance of advancing from one cart to the next cart is 35%.

Your chance of winning on all 8 carts is about 1 in 4,440.74... (0.35 to the power of 8)

1. If your goal was, "always play this game to the 8th cart (or bust)", would the EV be about 38.1...% of your initial bet?
(If not, tell me what the EV is and why my figure is wrong)

2. If the EV mentioned in the 1st question is correct and you bet \$1.00 on the first bet, then what would the minimum bank roll need to be for an RoR of 1% or lower? (if solvable from the information provided in this post)
Last edited by: ksdjdj on Jan 14, 2019
GaryJKoehler
Joined: Oct 22, 2015
• Posts: 143
January 16th, 2019 at 3:16:34 AM permalink
In Predictor, "6.The odds offered are expressed on a "for one" basis and are 96% of fair value, rounded to the nearest penny. " What about your game?
charliepatrick
Joined: Jun 17, 2011
• Posts: 2321
January 16th, 2019 at 4:25:13 AM permalink
I'm confused as I agree that if you keep going you have a 1 in 4440,75 chance of winning \$6133.39; so it is obvious you keep going.

Since in essence it is a simple bet where p=.000225188 and your chance of losing is (1=p)^N. This means that there is about a 36.7% chance of losing \$4441 without winning and just over 25% of losing 6133 times in a row.
ksdjdj
Joined: Oct 20, 2013
• Posts: 909
January 19th, 2019 at 8:38:57 PM permalink
Quote: charliepatrick

I'm confused as I agree that if you keep going you have a 1 in 4440,75 chance of winning \$6133.39; so it is obvious you keep going.

Since in essence it is a simple bet where p=.000225188 and your chance of losing is (1=p)^N. This means that there is about a 36.7% chance of losing \$4441 without winning and just over 25% of losing 6133 times in a row.

Thanks for the post.

Yeah it seemed obvious to keep going to me too.
I just wanted to make sure that I had the EV right using the figures in the OP.

If you know the answer to the second question in the OP, that is the one that I really want answered if possible.
ksdjdj
Joined: Oct 20, 2013
• Posts: 909
January 19th, 2019 at 8:46:49 PM permalink
Quote: GaryJKoehler

In Predictor, "6.The odds offered are expressed on a "for one" basis and are 96% of fair value, rounded to the nearest penny. " What about your game?

Thanks for the post.

The odds in my game are expressed "for one" and the fair odds for a person picking at random would average out to about 96%***.

***: But remember, in my game you are always picking the best choice based on RTP.
RubyPetersen
Joined: Feb 2, 2019