This would be very hard to analyse for an arbitrary deck; but, your general premise is correct. The variance expected by a deck with an arbitrarily large number of copies of each card (that is, drawing randomly with
replacement) would have higher variance.
The following example can be taken as an analogy for lands and non-lands in a deck.
For the sake of analysis we can consider the following “deck”.
20x Black Lotus
40x Lightning Bolt
This deck will win when it casts 3 Black Loti and 7 Lightning Bolts. This will happen on or after the 10th card is drawn, when that card is either the 3rd Black Lotus or 7th Lightning Bolt.
The probability of winning on ith draw is:
PW(i) = PW_Bolt(i) + PW_Lotus(i)
Without Replacement
For the normal version of the deck, this will take somewhere between 10 and 43 draws inclusive.
Where the chance of drawing the 7th Bolt is:
PW_Bolt(i) = Ways to draw i-1 cards with 6 Bolts * Chance to draw Bolt
PW_Bolt(i) = Choose(40,6)*Choose(20,(i-1)-6)/Choose(60,i-1) * (40-6)/(60-(i-1))
And the chance of drawing the 3rd Lotus is:
PW_Bolt(i) = Ways to draw i-1 cards with 2 Loti * Chance to draw Lotus
PW_Lotus(i) = Choose(20,2)*Choose(40,(i-1)-2)/Choose(60,i-1)*(20-2)/(60-(i-1))
We can validate this probability calculation by checking that it sums to 1.
The expected draws that this deck will need to win is:
E = sum[i*PW(i),{i,10,27}] = 11.97
And the variance is:
Var = sum[(E-i)^2*PW(i),{i,10,27}] = 4.5
With Replacement
There is no bound on how many draws it could take to get the winning combination; but, the basic logic still holds.
The chance of drawing the 7th Bolt is:
PW_Bolt(i) = Ways to draw i-1 cards with 6 Bolts * Chance to draw Bolt
PW_Bolt(i) = Choose(i-1,6)*(2/3)^6*(1/3)^(i-1-6) * (2/3)
And the chance of drawing the 3rd Lotus is:
PW_Bolt(i) = Ways to draw i-1 cards with 2 Loti * Chance to draw Lotus
PW_Lotus(i) = Choose(i-1,2)*(1/3)^2*(2/3)^(i-1-2) * (1/3)
This also adds to 1 in the limit.
The expected draws that this deck will need to win is only slightly higher:
E = sum[i*PW(i),{i,10,27}] = 12.39
But the variance is much higher:
Var = sum[(E-i)^2*PW(i),{i,10,27}] = 7.32
دیدگاهتان را بنویسید