Showcase Showdown: Strategy for the ‘Big Wheel’ on The Price is Right

One of my favorite game shows to watch when I was younger was The Price is Right. To determine who advances to the final round, three contestants spin ‘The Big Wheel,’ which contains 20 possible spots starting with 0.05 in 0.05 increments up to 1.00. The winner is the contestant who spins closest to 1.00 – without going over – in one spin, or a combination of two spins. A friend of mine, a data scientist, recently proposed to me a simple question while on a run: on what value should the first contestant stay, opposed to spinning again?

Since each participant has the option to stay after their first spin, determining the probabilities of victory are complex. Player 1 always has a decision to make: whether they are confident in their first spin value, or if they want to risk busting to spin again and potentially bolster their chances of winning. I thought this would be an interesting problem to simulate, which I did using R. However, the analysis requires several different parallel paths. In addition to the two possible decisions for Player 1 (spin once or spin twice), Player 2 has the same options depending on how their first spin relates to Player 1’s current score. In addition, Player 2 also has to consider that there is a third and final contestant yet to spin. The possible conditions were defined as follows:

Group 1

  • Condition A: Player 1 stays, Player 2 stays if better
  • Condition B: Player 1 stays, Player 2 stays if better or tied
  • Condition C: Player 1 stays, Player 2 spins again – even if better or tied

Group 2

  • Condition D: Player 1 spins twice, Player 2 stays if better
  • Condition E: Player 1 spins twice, Player 2 stays if better or tied
  • Condition F: Player 1 spins twice, Player 2 spins again – even if better or tied

The first (A, B, C) and second (D, E, F) groups only differ by the number of spins for Player 1. Thus, our comparisons of interest are the probabilities of a Player 1 victory comparing Conditions A & D, B & E, and C & F.

Player 3’s decisions are easier to determine because they are last to act. The only judgement call for Player 3 is deciding to spin a second time if their first spin ties them with the current leader(s). I made some assumptions for this scenario:

If Player 3 ties the current leader, they will spin again if the first spin was less than 0.50. This is because the odds of busting (<= 45%) are less than the odds of losing a two-way spin-off (50%). If Player 3 ties both players (i.e., there is a three-way tie), they will spin again if first spin was less than 0.70. Again, this is due to the odds of busting (<= 65%) being less than the odds of losing a three-way spin-off (66.7%).

Once the above rules were established, the simulation was run 1,000,000 times to determine the probability of each player winning based on the potential decision trees (Conditions A-F). By comparing the odds of winning based on Player 1’s first spin (Condition A vs. D and Condition B vs. E), Player 1 should spin again on 0.65 and below (Table 1). However, for the Condition C vs. F comparison, the odds suggest that Player 1 should spin again on 0.60 and below. Therefore, if Player 1 suspects that Player 2 will spin again even if Player 2 ties or leads after one spin, then Player 1 should spin again on 0.60. Unfortunately, there is no way to get inside the mind of this hypothetical Player 2; however, we can explore the odds of Player 2 winning to see when it is advantageous to behave this way. In other words, does it make sense for Player 2 to spin again, even if they are the current leader?

table1
Table 1. Probability of Player 1 winning based on first spin value for six Conditions (A-F). Shaded areas represent, for each possible spin value, which decision (spin once A/B/C or spin twice D/E/F) has greater win probability.

If Player 2 spins 0.60 and elects to spin a second time (even if leading after the first spin), their odds of winning are lower (~26-27%) than if they were to stand (~33-35 %, Table 2). Also, if Player 2 spins 0.65 and elects to spin a second time (even if leading after the first spin), their odds of winning are lower (~23-26%) than if they were to stand (~37-41%, Table 3). Therefore, it does not make sense for Player 2 to spin again if they have Player 1 beat at 0.60 and up. Now that this is established, let’s consider the case of ties (i.e., comparing B/E to C/F). If Player 2 stays when tied with Player 1 at 0.60, Player 2’s odds of winning are 17.43%. If Player 2 spins again when tied at 0.60, his odds of winning are 25.8%. The same trend is evident for ties at 0.65, as well. Therefore, Player 2 has better odds to win by spinning again when tied with Player 1 at 0.60 or 0.65.

table2
Table 2. Probabilities of Player 2 victory, after a first spin score of 0.60. Values on the left correspond to Player 1’s score.
Table3
Table 3. Probabilities of Player 2 victory, after a first spin score of 0.65. Values on the left correspond to Player 1’s score.

Using these probabilities, we should expect Player 2 to spin a second time if tied or losing, but not to spin a second time if leading after Player 1’s score of 0.60 or 0.65. Thus, Condition B/E includes the decisions for Player 2 that maximizes win probability, leading to the final recommendation that Player 1 should spin again on any first spin less than or equal to 0.65.

 

 

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