vpFREE2 Forums

Tonopah Nevada

I stayed at the Tonopah Station last night, and they have a unique
promotion going on. When checking into the hotel, you get to roll three
dice. If all three dice come up with the same number, your room is
free. Of course, I did not win and had to pay for my room. That got me
thinking about the odds of winning a free room. My college statistics
book was tossed out two moves ago, so I cannot check on how to
calculate the odds. Anyone help me out with the calculation? Oh, the
check in gal said they give away at least one free room every night.
There is a total of 78 rooms, but I doubt they were even half full last
night.

No need to rush to Tonopah for great VP, or really anything else. I did
not check every machine in town (not a hard job with only two small
casinos), but the best I found was a quarter bank of 8/5 BP with three
way progressive that was quite ripe at the Banc Club. No table game
action at all in town.

hamstockman wrote:
...you get to roll three dice. If all three dice come up with the same
number, your room is free...Anyone help me out with the calculation?

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There are 6 X 6 X 6 = 216 possibilities. Of these, there are 6 that
will get you a free room, so the probability of getting a free room is
6/216 = 1/36 or about 2.8%.

Thanks for the calculation. It was really quite an easy one. With the
room rates about $65/night with tax, the EV of this promotion is about
$1.82. They advertise this promotion on billboards as you drive into
town.

"I guess the promotion might have been in a Chuk-O-Luck cage they
borrowed from the Moose Lodge?"
Yes, it is one of those Chuk-O-Luck cages, but I think they own it, and
it is getting in bad shape.

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--- In vpFREE@yahoogroups.com, "jeffcole2003oct" <jeff-cole@...> wrote:

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There are 6 X 6 X 6 = 216 possibilities. Of these, there are 6 that
will get you a free room, so the probability of getting a free room is
6/216 = 1/36 or about 2.8%.