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Ask the Wizard #305
Wizard, I challenge you to a bet. Here are the rules:
 You choose any pattern you like of heads (H) and tails (T). For example, HTT.
 After revealing your pattern, I will choose mine.
 We will flip a coin over and over until one pattern appears in order. The one who chose that pattern wins.
 I will give you 3 to 2 odds.
Do you accept? I will do it as many times as you wish.
odiousgambit
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.
Odiusgambit Game Strategy
Player 1  Player 2  Pr. Player 2 Wins 

HHH  THH  87.50% 
HHT  THH  75.00% 
HTH  HHT  66.67% 
HHT  THH  75.00% 
THH  HTT  75.00% 
THT  TTH  66.67% 
TTH  HTT  75.00% 
TTT  HTT  87.50% 
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.
Mike H. from New Jersey
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 4^{5}=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 4^{3} = 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 fivesymbol 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 foursymbol 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 threesymbol 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 twosymbol win of symbol s equals n(s,1)*n(s,2)*b(s,3)*t(4)*t(5)*v^2
Naranjas1
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 tiebreaker 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, tiebreaker round, if the wheel lands on $0.05 or $0.15, then the player shall win $10,000.
 In the first, and only first, tiebreaker 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.
Showcase Showdown Statistics
Question  Answer 

Expected $1000 winners first round  0.253790 
Probability 2player tie  0.113854 
Probability 3player tie  0.004787 
Expected $10000 winners second round  0.024207 
Expected $25000 winners second round  0.012104 
Expected total prize money  $798.45 
Probability any given player wins $1000  0.084597 
Probability any given player wins $10000  0.008069 
Probability any given player wins $25000  0.004035 
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.