FinsRule
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Joined: Dec 23, 2009
March 31st, 2022 at 8:43:35 AM permalink
Assuming both contestants are using correct strategy, how often should we expect the first two contestants on the “spin the wheel” game to both go over? This allows the third contestant an automatic win.

I root for it every time and I’ve never seen it. Has anyone else seen it?
Joeman

• Posts: 2430
Joined: Feb 21, 2014
March 31st, 2022 at 8:50:19 AM permalink
Quote: FinsRule

Assuming both contestants are using correct strategy, how often should we expect the first two contestants on the “spin the wheel” game to both go over? This allows the third contestant an automatic win.

I root for it every time and I’ve never seen it. Has anyone else seen it?
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I have seen it multiple times in the Bob Barker era. I never liked it because I enjoy seeing the contestants get \$1.00 on the wheel, and when the first two go out, it gives the 3rd contestant less of a chance. They get to spin once for the dollar, but obviously, they can't spin it a second time.
"Dealer has 'rock'... Pay 'paper!'"
ThatDonGuy
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Joined: Jun 22, 2011
March 31st, 2022 at 9:27:21 AM permalink
As Joeman said, it happens quite a bit.

Note that, although the third spinner cannot take a second spin in this case, if at least one of the first two does not go over a dollar but the third player's first spin is enough to win, they can take a second spin, at least when Barker was hosting; he would always ask, "You don't want to spin again, do you?," and as far as I know, no one did.
TigerWu
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Joined: May 23, 2016
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March 31st, 2022 at 10:42:28 AM permalink
This is supposedly the optimal strategy for playing the Wheel, sans bonus payments:

–Contestant 1 spins if she gets 65 or fewer on the first spin
–Contestant 2 spins if she gets 50 or fewer on the first spin, if she gets 55, 60 or 65 and her score ties contestant 1, or if she spins a score less than contestant 1
–Contestant 3 spins if she gets 50 or fewer on the first spin and ties another contestant, if she gets 55, 60 or 65 and ties two contestants, or if she spins a score lower than the other contestants’ totals

Using these strategies, ...contestant 1 wins 30.82 percent of the time, contestant 2 wins 32.96 percent, and contestant 3 wins 36.22 percent. That’s a pretty decent advantage to being the last spinner.

As far as factoring in the bonus payments (the way it is actually played on the show), the strategy is exactly the same except for Contestant 2:

–Contestant 2 spins if she gets 55 or fewer on the first spin, or if she gets 60 or 65 on the first spin and ties contestant 1, or if she has to spin to get more than contestant 1.

Source
FinsRule
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Joined: Dec 23, 2009
March 31st, 2022 at 2:08:05 PM permalink
Well, my first question might just be too hard to answer.

Let’s go with the next question.

How often will someone end up with the max possible amount - \$1.95?

I can’t recall ever seeing that either.
billryan

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Joined: Nov 2, 2009
March 31st, 2022 at 2:22:00 PM permalink
Quote: FinsRule

Well, my first question might just be too hard to answer.

Let’s go with the next question.

How often will someone end up with the max possible amount - \$1.95?

I can’t recall ever seeing that either.
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Why would someone risk .95?
The difference between fiction and reality is that fiction is supposed to make sense.
FinsRule
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Joined: Dec 23, 2009
March 31st, 2022 at 2:26:02 PM permalink
You know the answer…. If the first person got \$1 exactly.
FinsRule
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Joined: Dec 23, 2009
March 31st, 2022 at 2:26:28 PM permalink
Quote: FinsRule

You know the answer…. If the first person got \$1 exactly.
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Or second person.
TigerWu
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Joined: May 23, 2016
March 31st, 2022 at 2:36:00 PM permalink
Quote: FinsRule

Well, my first question might just be too hard to answer.

Let’s go with the next question.

How often will someone end up with the max possible amount - \$1.95?

I can’t recall ever seeing that either.
link to original post

The max possible amount would be \$2.00, if you hit \$1.00 twice in a row, or two spins adding up to \$1 and a bonus spin on top of that hitting \$1.

It happened twice in one day last year, and apparently it happened twice in one day back in 2017 as well. You could probably do the math pretty easily; there are 20 spots on the wheel, so a 1/20 chance of hitting \$1 straight up, and another 1/20 chance of hitting it straight up again. I don't know how to figure spinning \$1 out of two spins....the 20 slots are \$.05 to \$1.00 in increments of \$.05, so you'd have to figure the odds of hitting exactly \$.05 and then exactly \$.95, \$.10 and \$.90, etc.

There are thousands of episodes, each with, what, 6 contestants per episode spinning the wheel? So it probably happens more than you'd think, someone hitting the max.
avianrandy
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Joined: Mar 7, 2010
March 31st, 2022 at 2:40:06 PM permalink
Seems like here lately I have seen some contestants get a total of 10 cents.nickrl on first spin,Nickle on second spin.seems like the odds would be same for maximum.
ThatDonGuy
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Joined: Jun 22, 2011
March 31st, 2022 at 2:56:41 PM permalink
Quote: TigerWu

Quote: FinsRule

Well, my first question might just be too hard to answer.

Let’s go with the next question.

How often will someone end up with the max possible amount - \$1.95?

I can’t recall ever seeing that either.
link to original post

The max possible amount would be \$2.00, if you hit \$1.00 twice in a row, or two spins adding up to \$1 and a bonus spin on top of that hitting \$1.
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The question is, what is the probability that (a) one (or both) of the first two spinners get \$1, either in one spin or a two-spin total, and then (b) someone who spins after them gets 95 cents on their first spin and then, having to spin again, spins \$1 on their second spin? Obviously the probability of getting \$1.95 is 1 / 400, but that doesn't take into account the probability that an earlier player will have spun \$1.
TigerWu
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Joined: May 23, 2016
March 31st, 2022 at 4:02:00 PM permalink
Quote: ThatDonGuy

The question is, what is the probability that (a) one (or both) of the first two spinners get \$1, either in one spin or a two-spin total, and then (b) someone who spins after them gets 95 cents on their first spin and then, having to spin again, spins \$1 on their second spin? Obviously the probability of getting \$1.95 is 1 / 400, but that doesn't take into account the probability that an earlier player will have spun \$1.
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So, assuming the first person gets \$1 on the first spin, that's 1/20...

If they get \$1 on the second spin, the odds of that are 1/400...

Then the second person spins, the odds of their getting \$1 on their first spin is 1/8,000, right? And then 1/160,000 for \$1 on the fourth spin?

You just have to calculate odds for four \$1 spins in a row; the actual spinners are irrelevant.
ThatDonGuy
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Joined: Jun 22, 2011
March 31st, 2022 at 4:49:38 PM permalink
Quote: TigerWu

So, assuming the first person gets \$1 on the first spin, that's 1/20...

If they get \$1 on the second spin, the odds of that are 1/400...

Then the second person spins, the odds of their getting \$1 on their first spin is 1/8,000, right? And then 1/160,000 for \$1 on the fourth spin?

You just have to calculate odds for four \$1 spins in a row; the actual spinners are irrelevant.
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I think you're still misunderstanding it.

What is the probability of either of these two happening:
(a) The first spinner gets \$1, either in one spin or a combination of two spins, using optimum strategy to decide whether or not to take their second spin (so if the first spin is 95, the first spinner will stop with the 95 rather than risk going over with a second spin), and then either, or possibly both, of the other two spin 95 on their first spin and \$1 on their second;
(b) Just the second spinner gets \$1, again using optimum strategy to decide whether or not to take their second spin, and then the third spinner gets 95 on their first spin and \$1 on their second.