# Wheel odds

101 posts

The list of what items to expect next is really useful, thanks!

0

This discussion has been closed.

101 posts

The list of what items to expect next is really useful, thanks!

0

This discussion has been closed.

## Replies

That is correct, missed that one when splitting it up

It didn't add up to 100% when spAnser posted the odds, either.

So the way I read it:

Bucket 1:

Clause Co / 275 Gift Cards

156/10041

Bucket 2:

Brown House / Purple House / Pink House / Orange House / Cash

2022/10041

Bucket 3:

Snow Monster / 250 Gift Cards

168/10041

Bucket 4:

Flanders / Van Hauten / White / Muntz / Skinner / Cash

2022/10041

Bucket 5:

Helter Shelter / 225 Gift Cards

225/10041

Bucket 6:

Simpsons / Cletus / Blue / Willie's / Krabapple / Cash

2022/10041

Bucket 7:

Vishnu / 225 Gift Cards

225/10041

Bucket 8:

Marge / Maggie / Homer / Snowball / 50 Gift Cards

1348/10041

Bucket 9:

Mr. Plow / Plow King / 100 Gift Cards

505/10041

Bucket 10:

Lisa / Grandpa / Bart / Dog / 50 Gift Cards

1348/10041

If you already have the item and it's not a Snowman, that item will be removed and the next in the list will be added; Snowmen will always be there because you can always have more than one. Folks who add house lights from the last time around still get the Skinner House Lights because they weren't available back then and I already had the Scorpio Volcano Lights so I don't know which bin those go in. Probably in Bucket 2 but not sure at what level and would only be available if you bought the Volcano.

I built my Monte Carlo simulator in Microsoft Excel. First I built a spreadsheet that created 1000 random numbers using Excel's built-in random number generator. I put these all in a table and they represent the first 1000 spins that a player takes on the wheel.

Each number is multiplied by 10041 and then I use a formula to determine which bucket that "spin" ended up in. Each row in the table has ten columns representing the number of prizes left after that spin. These columns all have formulas that determine whether the number of prizes left should be the same as the row above, or one less than the row above (based on the spin outcome). With a few more formulas I find which row is the first with no more prizes and then capture:

- How many spins it took to win all the prizes

- What was the last prize

So, the spreadsheet quickly simulates a single player making up to 1000 spins. But I don't just want to simulate one player, I want to simulate thousands so I can figure out the odds. Fortunately, Excel comes with a built in programming language so I was able to write a small macro to update this spreadsheet thousands of times and record the results. About an hour or two later I had some data to share. I first put this up on another forum and it received some positive feedback, so I thought I would register here and share it with a wider audience. The odds below are a little different than what I posted on the other site because I have increased the number of simulations to 100,000 simulated players.

Assumptions:

1. The first post in this thread (and the one immediately proceeding this one) accurately capture the odds for each bucket and the number of prizes in the bucket.

2. Each spin is truly independent of the spins that came before

3. EA isn't planning to tweak the odds.

4. The simulated player had no prizes from last Christmas (i.e., the player needs to "win" all 28 prizes to empty the wheel).

Observations:

1. There was a wide range of outcomes, that corresponds with the reports we've seen on this forum of how the wheel actually works. You can get very lucky and empty the wheel without going into donut-range, or you can spin the wheel hundreds of times and still not have all the prizes. Overall 80 percent of players will need between 55 and 199 spins to win all the prizes, with 10 percent getting luckier and 10 percent getting unluckier.

2. Having a lot of prizes already from last Christmas doesn't seem to be too much of an advantage. The simulations show that 82.8% of players starting with no prizes will receive one of the unique items last.

Average number of spins required (out of 100,000 simulations): 117.7

Median number of spins required (out of 100,000 simulations): 101.5

Fewest number of spins required (out of 100,000 simulations): 30, occurred 3 times

Highest number of spins required (out of 100,000 simulations): 805, one of 5 occurrences over 700 spins

Here is a table showing the odds of clearing the wheel after X spins:

Spins - % of players clearing the wheel

55 - 10.0%

67 - 20.5%

78 - 30.4%

90 - 40.7%

102 - 50.2%

117 - 60.5%

134 - 70.1%

158 - 80.0%

199 - 90.0%

239 - 95.0%

293 - 98.0%

334 - 99.0%

370 - 99.5%

455 - 99.9%

Which prize will be the last off the wheel?

Houses 1 (4 prizes): .4%

Houses 2 (5 prizes): .9%

Houses 3 (5 prizes): .9%

Snowmen 1 (4 prizes): 2.6%

Snowmen 2 (4 prizes): 2.6%

Mr. Plow/Plow King (2 prizes): 9.9%

Helter Shelter: 15.3%

Ice God: 15.3%

Yeti: 24.7%

Claus Co: 27.5%

Notes: Because of rounding the above adds up to 100.1%. Each of the pairs that *should* have the same outcome are the same to a single decimal, but they differ in the second decimal. For example Houses 2 occurred 877 times out 100,000 simulations, whiles Houses 3 occurred 893 times. In the end, expressing the odds with more than one decimal point would be misleading. For the sake of completion the actual raw count of last prize (in the order above) are 352, 877, 893, 2553, 2614, 9874, 15268, 15338, 24724, 27505.