A shout out to my statistician friends

Over the past couple of years, I have been regularly watching “The Price is Right” while eating my lunch. For reasons I cannot logically explain, I find this show highly entertaining. We have set up to record the show each day and I watch whenever I have a chance.

Yesterday I was eating a yummy lunch, a flour tortilla filled with broccoli and a delicious cheddar cheese with beer that Barry had bought, while watching the show. The first unusual thing that happened was that all six of the contestants in the initial rounds won their prizes: this does not seem to happen very often. But the really amazing thing happened when the three in the second round spun the wheel at the end of their round. All three – yes, all three – spun a total of 100 in their two spins of the wheel (the middle one landed on 100 in her first spin). Ordinarily, the prize for spinning 100 is $1000, but because the show is celebrating its 45^th anniversary, the prize was $10,000!

When there is a tie, the players get a second spin of the wheel, just one spin, with the highest number going on to the final round. The first two landed on 100, winning $25,000 each! The third came very close, only 2 numbers away from landing on 100. Those first then each spun once more to determine the player who would enter the final round. I have attached a video clip here.

When Barry got home I had to make him watch it, it was just so unbelievable. So that got me to thinking, what exactly is the probability of spinning 100 on the first spin and then of hitting 100 again in the second spin?

It has been a very long time, not to mention cancer in my brain, since I have thought about such questions, so I am counting on some of my former statistical colleagues who may be reading this to weigh in and correct me, but here goes.

First, in the two spins of the wheel, there are twenty ways to obtain a total of 100:

First Spin	Second Spin
5	95
10	90
15	85
20	80
25	75
30	70
35	65
40	60
45	55
50	50
55	45
60	40
65	35
70	30
75	25
80	20
85	15
90	10
95	5
100

That means that on any given day, there is a 20% chance that a player spinning the wheel will have a total of 100. That sounds about right, that does seem to happen about 20% of the time.

After spinning a total of 100, the player then has one additional spin and must land on 100 to win, which should happen 5% of the time.

So, the probability that one player will have a total of 100 on his or her first one or two spins AND THEN land on 100 on the second spin is 1%, in other words, of every 100 players who spin the wheel, approximately one will win both prizes.

Now the question is, what is the probability that these events will happen for multiple players on the same day? On the first spin of the wheel, when each of the three players have a 20% chance of winning, the probability that all three will do so is 0.8% (0.008).

On the second spin of the wheel, there are 8 possible and equally likely configurations of players who can land on 100:

Number of players who land on 100	Possible configurations	Details of configurations
0	1	No players
1	3	Player 1, player 2, or player 3
2	3	Players 1 & 2, players 2 & 3, or players 1 & 3
3	1	All 3 players

The probability that just two of the three players will land on 100 is based on the binomial probability

In our case, N = 3, x = 2, and π = 0.05, so this comes to 0.007125.

I think we then need to multiply these two numbers, 0.008 (first spin) and 0.007125 (second spin), to get the overall probability of the outcome of the game in question, and that is then 0.000057 (0.0057%). Seems like a very long shot to me!

Help me out here, guys – am I even close to the right answer?

I really do need to get a life!

Love,

Rebecca