Ask The Wizard #305

No. Nice try though. The person acting second has a huge positional advantage. Here is the strategy for the person acting second and his probability of winning.

As the table above shows, my best chance to win, or your least, is if I pick either THT or HTH, where my chance of winning is still only 1 in 3. I should get 2 to 1 for it to be a fair bet, so getting only 3 to 2, you have a 16.67% edge.

Here is a way to remember the player two strategy. Let P(x) be the player 1's pick for position x. Let O(x) be the opposite of player 1's pick for position x. Player 2 should always pick: O(2) - P(1) - P(2).

This question is asked and discussed in my forum at Wizard of Vegas.

First, let me explain the question for the benefit of other readers. A popular new way for slot machines to pay is according to every "way" on the screen. A "way" is every unique set of positions that go through a paying combination, counting only reels that are part of the win. Let's look at the following picture as an example from a Buffalo slot:

This game as 4⁵=1,024 "ways" to win, because there are four positions shown on each reel. However, in this case, only two ways pay, for two buffalo each. Both include the one buffalo on reel 1. Then 1 way for the buffalo on row 1 of reel 2 and a second way for the buffalo on row 2 or reel 2. Although there are 4³ = 64 ways the game could go through positions on reels 3 to 5, it doesn't matter. "Ways" only count the ways to go through symbols that contribute to the win.

With that explanation out of the way, let me introduce some functions for my answer to your question:

• Let v = number of visible rows on the machine.
• Let n(s,r) = number of times symbol s or a wild appears on reel r.
• Let b(s,r) = number of sequences on reel r with no appearances of symbol s or a wild symbol that may substitute for s. In other words, sequences blocking any wins for a given symbol on the reel because it doesn't appear.
• Let t(r) = Total length of reel r

That said, here are the number of winning combinations according to the number of winning symbols in the win:

• The number of combinations for a five-symbol win of symbol s equals n(s,1)*n(s,2)*n(s,3)*n(s,4)*n(s,5)*v^5.
• The number of combinations for a four-symbol win of symbol s equals n(s,1)*n(s,2)*n(s,3)*n(s,4)*b(s,5)*v^4
• The number of combinations for a three-symbol win of symbol s equals n(s,1)*n(s,2)*n(s,3)*b(s,4)*t(5)*v^3
• The number of combinations for a two-symbol win of symbol s equals n(s,1)*n(s,2)*b(s,3)*t(4)*t(5)*v^2

As a word of explanation to other readers, let me explain what you're talking about. The Showcase Showdown is a game played on the game show The Price is Right. In the Showcase Showdown, each player takes his turn spinning a wheel which has an equal probability of stopping on every amount evenly divisible by .05 from .05 to 1.00 . If the player does not like their first spin they may spin again, adding the second spin to their first, however if they go over 1.00 they are immediately disqualified. In the event of a tie, each player will get one spin in a tie-breaker round, the highest spin wins. In the event of another tie, this process will repeat until the tie is broken.

The main purpose of the Showcase Showdown is to advance to the Showcase. However, there are also immediate cash prizes too, as follows:

• In the first round, if any player get a total of $1.00, whether in one sum or the sum of two spins, he shall win $1,000.
• In the first, and only first, tie-breaker round, if the wheel lands on $0.05 or $0.15, then the player shall win $10,000.
• In the first, and only first, tie-breaker round, if the wheel lands on $1.00, then the player shall win $25,000.

I explain the optimal strategy to the Showcase Showdown in column #101. Assuming that strategy is followed, the following table answers your questions and various other.

The bottom row of the table shows that if you make the Showcase Showdown, without considering your order to spin, your chances of winning $25,000 is 0.004035, or 1 in 248.

This question is asked and discussed in my forum at Wizard of Vegas.