Question 1204052: Illinois Lotto In this game, the player chooses six different integers from 1 to 40. If the six match (in any
order) the six different integers drawn by the lottery, the player wins the grand prize jackpot, which starts at
$1 million and grows weekly until won. Multiple winners split the pot equally. For each $1 bet, the player
must pick two (presumably different) sets of six integers.What is the probability of winning the Illinois Lottery
Lotto with a $1 bet?
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
Let's consider a similar example with smaller numbers.
We'll have a game where 10 ping-pong balls are placed into a bin.
The winning ball is painted red, and the other 9 are white.
The player is then allowed to make 1 selection at random.
What is the probability of winning?
Answer: 1/10
Reasoning: There is 1 ball we want out of 10 total.
That seems fairly straight forward. There's not much else to say here.
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Let's change the rules to make things a bit more interesting. The player is now allowed 2 random selections without replacement.
What are his or her chances of winning now?
Answer: 2/10
Reason: Perhaps the logic is obvious, but I'll be methodical here.
Consider three scenarios
Scenario A: win on 1st selection
Scenario B: win on 2nd selection
Scenario C: do not win
P(A) = 1/10 as mentioned earlier
P(B) = (9/10)*(1/9) = 1/10
The 9/10 refers to losing on the 1st selection and 1/9 is the probability of winning on the 2nd selection.
So,
P(A or B) = P(A)+P(B) ... mutually exclusive events
P(A or B) = 1/10 + 1/10
P(A or B) = 2/10
Mutually exclusive events are such that they cannot happen simultaneously.
One or the other could happen, but both cannot happen at the same time.
In terms of notation, events A and B are mutually exclusive if and only if P(A and B) = 0.
It will allow the formula P(A or B) = P(A)+P(B)-P(A and B) to condense to P(A or B) = P(A)+P(B).
Another term for "mutually exclusive" is "disjoint".
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One more example.
The player is now allowed 3 selections without replacement.
What are the chances of winning now?
Answer: 3/10
Reason
Scenario A: win on 1st selection
Scenario B: win on 2nd selection
Scenario C: win on 3rd selection
Scenario D: do not win
P(A) = 1/10
P(B) = (9/10)*(1/9) = 1/10
P(C) = (9/10)*(8/9)*(1/8) = 1/10
P(A) and P(B) were explained earlier.
P(C) has the 9/10 to refer to losing on the 1st selection, 8/9 to mean the player lost on the 2nd selection and 1/8 is them winning on the 3rd selection.
P(A or B or C) = P(A) + P(B) + P(C) ... mutually exclusive events
P(A or B or C) = 1/10 + 1/10 + 1/10
P(A or B or C) = 3/10
Conjecture: There are n items in a set, with 1 of them being the winning item.
A player is allowed k selections where k is an integer between 1 and n.
The probability of winning is k/n.
I'll let the student prove this conjecture.
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Now back to your homework problem.
There are 40 balls labeled 1 through 40 placed into a churning bin. The player (or likely a machine) selects 6 balls at random without replacement. The winner must match all 6 values in any order.
The first thing we need to figure out is: how many combinations are there?
As the key word "combination" implies, we'll use the nCr combination formula.
This is where order doesn't matter. Something like {1,2,3} is the same as {3,1,2}.
We have n = 40 items to pick from and r = 6 selections.
Let's apply those to the nCr formula below.
n C r = (n!)/(r!(n-r)!)
40 C 6 = (40!)/(6!*(40-6)!)
40 C 6 = (40!)/(6!*34!)
40 C 6 = (40*39*38*37*36*35*34!)/(6!*34!)
40 C 6 = (40*39*38*37*36*35)/(6!)
40 C 6 = (40*39*38*37*36*35)/(6*5*4*3*2*1)
40 C 6 = (2,763,633,600)/720
40 C 6 = 3,838,380
There are roughly 3.8 million different combos possible.
The numerator 40*39*38*37*36*35 has us start at 40 and count down by 1 until filling up 6 slots.
That computes the number of permutations.
The 6*5*4*3*2*1 = 720 is the number of ways to arrange any set of 6 items.
We divide by 720 because the permutation value over-counts by a factor of 720.
If the player was allowed 1 selection of 6 values, then the chances of them winning would be 1/(3,838,380) = 0.00000026052658 approximately.
But the player is allowed 2 different selections, so their chances double to get 2/(3,838,380) = 1/(1,919,190) = 0.00000052105319
Sure the probability is still very very small, but it's a slight boost.
I'm now using the conjecture mentioned in the previous sections.
So this is why some people prefer to play the lotto with multiple tickets to boost their chances. Their chances are still incredibly low. Perhaps their money may be better spent elsewhere, invested, or saved.
If some eccentric billionaire or millionaire wanted a guaranteed win, then all they have to do is buy all 3.8 million combos costing them 1.9 million dollars roughly. It would be a bad idea because they will win an amount less than what they paid (assuming the jackpot is 1 million dollars).
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To wrap things up, the probability of the player winning the Illinois Lotto is 0.00000052105319
This decimal is approximate.
Round it however your teacher instructs.
I recommend rounding to at least 7 decimal places to avoid getting a result of 0.
The value is really small but it's not actually 0 itself.
The fraction form of the probability is 1/(1,919,190) which is read as "1 out of 1,919,190", or we can say something like "1 out of 1.9 million" depending how accurate you want to be.
Sometimes you'll find probability expressed in the format "1 out of N" where N is some (really large) positive integer.
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