Price is Right Question

So, the big wheel on Price is Right has 20 numbers on it starting with 5 up to 100 in increments of 5. For those that are somehow not familiar with how it works, three people get to spin the wheel. Whoever has the highest spin, wins. However, each person may spin one time, or two times. If they get over 100, they cannot win.

My question is, what number should the first person stay on and not spin a second time around. I am trying to figure out how to go about calculating this. I want to know what number gives them greater than .5 probability of remaining in first after the other two people try to get higher than the first person without going over 100.

Any help would be appreciated!

My guess is 65. Two-thirds of the first spins will be below, and only one third above. However, anyone 25 or below will have a good chance of catching or passing you on their second spin. I suspect Kelly comes into play too.

What happens if you tie the highest spinner on your first spin? Can you decline to spin again, and force a playoff, or must you try to improve or bust?

If you tie, I believe you are correct, they go to a playoff of one spin...in which case, tying would give you a 50% chance of winning, since it is only one spin.

There are 3 people:

1st person spins once. The person may elect to take a second spin or stand pat. Going over 100 "busts" you and you're out.
2nd person spins once. If the spin is greater than the 1st person's total score, he may elect to take a second spin or stand pat. If the spin is less than the 1st person's total, he must spin a second time. Going over 100 "busts" you and you're out.
3rd person spins once. If the spin is greater than the leader's total score, he may elect to take a second spin or stand pat. If the spin is less than the 1st person's total, he must spin a second time. Going over 100 "busts" you and you're out.

In any case, spinning exactly a 100 (on the first spin or the aggregate of two spins) wins a bonus $1,000 and a single bonus spin with a chance to win an additional $5,000 (2/20) or $10,000 (1/20). All other values yield no further bonus.

This process happens twice a show. The winners of each competition gets to go to the "Showcase Showdown" where they bid on prize packages worth $12,000-25,000 or so.

My completely anecdotal evidence put the winning spin somewhere around 75-80 as the mode winning score. There is a definite disadvantage to going first (same as in blackjack, really) and that person probably must play more conservatively than the others, making a 65 a good guess without doing any math, in my book.

Here is the math I did for the probability of the SECOND PERSON ONLY beating the first person, when the first person stays with a score of 60:

(.05)(.4)(11)+(.05)(8)+.5(.05)=.645

The (.05)(.4) comes from there being a 1/20 chance of spinning a 5, and a 8/20 chance of getting another number that puts the score above 60. This could happen for 5, 10,...,55 which is eleven numbers. That probability is assuming a first spin is not already greater than 60. The probability of scoring above 60 on the first spin is 8/20, which is where the .05*.8 comes from. Then I added the tie at the end. There is a 50% chance that they win the tie-breaker if the first spin is actually 60.

Now I'm thinking I need to recalculate to add in, what if they tie with a score of 60, and the 2nd player spins twice, instead of just landing on it the first time? Thoughts?

Adding to the .645 probability, I decided I should add (.05)(.05) for each of the possible ties on the 2nd spin, which would be 11 total possibilities.

So, the probability should be .645+(.05)(.05)(11)=.65875

This is my final calculated probability that the 2nd player can beat a score of 60.

Wow! I found that the first player must have at least a 90 to have better than 50/50 shot of winning.

I also calculate that the first player should stand on a first spin of 55 or higher. If the first player stands on 55, there is a 71.25% chance that the second player will beat him, but if the player pushes it and rolls for more, there is a 55% chance of busting and and overall probability of 71.97% of busting or losing to the second player.

Wizard has answered this (very last question at bottom):
https://wizardofodds.com/ask-the-wizard/tvgames/

The contestant that has won the most on the show to that point gets the advantage of going last.

Yeah I saw this. I ended up getting the same answers as he did through my calculations. Let me know if anybody wants to see the full calculations as to what I did to get those answers.