What is the Probability of Winning with Consecutive Spins on a Roulette Wheel?

  • Context: High School 
  • Thread starter Thread starter Ivan91
  • Start date Start date
  • Tags Tags
    Statistics
Click For Summary

Discussion Overview

The discussion centers around calculating the probability of winning with consecutive spins on a roulette wheel, specifically when betting on the numbers 1, 2, 3, and 4. The conversation explores the assumptions behind probability calculations, particularly the independence of spins.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes that the probability of winning at least once in two spins is either 4/38 x 4/38 or 4/38 x 1/38.
  • Another participant agrees with the need for assumptions about independence and explains that if the spins are independent, the probabilities can be multiplied.
  • A different participant asserts that the second spin cannot depend on the first unless the roulette is manipulated, questioning the validity of the calculations.
  • One participant clarifies that (4/38)(1/38) represents the scenario where the same number comes up on both spins, while (4/38)(4/38) represents the scenario where any of the four numbers comes up on both spins.

Areas of Agreement / Disagreement

Participants express differing views on the correct probability calculation, with some supporting the independence assumption while others question its application. The discussion remains unresolved regarding which probability expression is correct.

Contextual Notes

The discussion highlights assumptions about independence and the nature of the roulette spins, which are critical to the probability calculations but remain unexamined in detail.

Ivan91
Messages
4
Reaction score
0
So, I have a very easy question.
Imagine a roulette with 38 pockets (0,00,1,2,3,...36)
Suppose you were to place bets on numbers 1,2,3,4.
What is the probability that at least one of your numbers would be the winner in two consecutive spins of the roulette wheel.

I am pretty sure the answer is either 4/38 X 4/38 , or 4/38 X 1/38.
 
Physics news on Phys.org
That is the correct answer, but you have to realize the assumptions that are made to get the correct answer.

The main assumption you are using is independence. Also due the the nature of the experiment, you are doing a duplicate of the first trial which means each trial has the same probability.

Simple definition of probability is P(Event) = Number of events in Event / Number of events in total.

Independence implies that for two events A and B, P(A and B) = P(A) x P(B).

Your P(A) = 4/38 and since B is the same event as A, P(B) = 4/38

Typically when we talk about pure randomness we typically use independence as a way to emphasize randomness.

If however the second spin of the wheel depended on the first, then you would not be able to use the independence relationship and it would get more complicated.
 
Yeah, but there is no way the second spin depends on the previous one, unless the roulette is illegally manipulated ;)
So, is 4/38 x 4/38 the correct answer?/
 
Givem a fair wheel, (4/38)(1/38) is the probability that one of your four numbers came up on the first spin and then the same number came up on the second spin.

(4/38)(4/38) is the probability that one of your four numbers came up on the first spin and one of your four numbers came up on the second spin.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
7K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K