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Popular US lotteries have odds ranging from the astronomical 1 in ~300 million in Mega Millions (a double-pick lottery) to the fairly good 1 in ~170 thousand in the Wisconsin Lottery Badger 5 (a pick-5, 31 number lottery). Wheeling systems are usually intended to provide a minimum guaranteed number of wins if some of the drawn numbers are ...
Odds are calculated by the total number of tickets in a scratch ticket game divided by the total number of prizes in that game. Study the odds of winning and the prize structure for the games you ...
[4] Intuitively, guessing any number higher than 2 / 3 of what you expect others to guess on average cannot be part of a Nash equilibrium. The highest possible average that would occur if everyone guessed 100 is 66 + 2 / 3 . Therefore, choosing a number that lies above 66 + 2 / 3 is strictly dominated for every player ...
The lists do not include "4+1" games, such as Florida's Lucky Money, where all five numbers must be matched to win the top prize, but are drawn from two number fields(A similar game, Montana's "Big Sky Bonus", is actually a "four-number" game; the double matrix is 4/31 + 1/16(previously was 4/28 + 1/17). Matching all four "regular" numbers wins ...
Midday: 4 - 8 - 5 - 0; Fireball: 1. Evening: 3 - 7 - 3 - 0; Fireball: 9. Check Pick-4 payouts and previous drawings here. NJ lottery: Where does all the billions in ticket sales money go? Jersey ...
If the car is behind door 1, the host can open either door 2 or door 3, so the probability that the car is behind door 1 and the host opens door 3 is 1 / 3 × 1 / 2 = 1 / 6 . If the car is behind door 2 – with the player having picked door 1 – the host must open door 3, such the probability that the car is behind door ...
This means that the probability of correctly predicting 2 numbers drawn from 49 in the correct order is calculated as 1 in 49 × 48. On drawing the third number there are only 47 ways of choosing the number; but we could have arrived at this point in any of 49 × 48 ways, so the chances of correctly predicting 3 numbers drawn from 49, again in ...
In that case the other envelope contains a/2 with probability 3/5 and 2a with probability 2/5. So either the first envelope contains 1, in which case the conditional expected amount in the other envelope is 2, or the first envelope contains a > 1, and though the second envelope is more likely to be smaller than larger, its conditionally ...