# Probabilities of certain events in a lucky wheel game

• MHB
• emmi2104
In summary, the conversation discusses a game with a wheel of fortune with ten equal sectors, five labelled with 1, three with 2, and two with 3. The game is played with pairs of candidates who spin the wheel, and the winner is the one with the higher number. The minimum number of candidate pairs needed to ensure at least one draw with a probability of 0.95 is 7. The probability of at most 143 out of 376 matches ending in a draw is found using the binomial theorem.

#### emmi2104

Given information:
A wheel of fortune with ten equal sectors is used for a candidate game. Five of these sectors are labelled only with the number 1, three only with the number 2 and two only with the number 3.

The game for a pair of candidates is as follows: The two candidates �K1 and K2, independently of each other, each spin the wheel of fortune once. The winner is the candidate who has "spun" the higher number. If the numbers are the same, the game ends in a draw.

• f) The game described above is played with 𝑛 pairs of candidates. What is the minimum number of candidate pairs with which the game must be played so that with a probability of at least 0.95 at least one game ends in a draw?
and
• h) With what probability 𝑝 do at most 143 of the 376 matches end in a draw?

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I see this is an old unanswered question. Perhaps it was a homework problem at one time. I would start by finding out the probability of (not a tie) in a single match.

That would be 1 - (0.5^2+0.3^2+0.2^2) = 0.62
Now what's the minimum value of n s.t. 0.62^n <= 0.05? Trial and error (or taking the log of .05 base 0.62 and rounding up) says 7

Does h require the binomial theorem?