Now I understand your result. Starting with the initial draws of 1, 2, 3 and 4, it’s identical to your top distribution, 16 possible results with one 2, two 3’s, three 4’s, four 5’s, three 6’s, two 7’s and one 8. This is the set of outcomes we get by adding the Duplicate card after the first draw, it matches the dice roll, which makes sense.
Adding in the Duplicate card during the first draw as well adds 4 new outcomes 2, 4, 6, & 8 to the normal 16, so the distribution is now two each of 2, 3, 7, 8 and four each of 4, 5, 6.
I think you can do this by “ignoring” the duplicate drawn first with a single shuffle.
Draw 2 cards. If the duplicate is the first card drawn, ignore it, draw one more card and use the two non-duplicates. If the duplicate is the 2nd card drawn, then use the first card and you got doubles.
You need to draw the 20 first, which occurs with 1/21 probability. Then the duplicate is drawn with 1/20 probability, i.e. the overall probability is 1/420, not 1/400. Granted, this is closer but still not the same.
Is that in response to my second message? I claim that it’s 1/20 that you draw the 20 first. Because there are only 20 cards you can draw first. If you draw the duplicate first, you reshuffle and start over.
After you’ve drawn a non-duplicate, there are 20 cards, each with equal probability, left to draw.
I see. That’s yet another different game, then. The point was to have 1 shuffle only. If you allow for potentially infinite re-shuffles, it defeats the purpose. Otherwise, for any finite number of shuffles, the probabilities never quite get there.
Might as well have 20 cards and shuffle twice each time.
For double 20s:
1/420 chance after 1 shuffle
(1/21)(1/420) after 2 shuffles
(1/21)(1/21)(1/420) after 3 shuffles
(1/21)(1/21)(1/21)(1/420) after 4 shuffles
and so on
It’s a geometric series with the limiting value of (1/420)*(21/20) = 1/400 as the number of shuffles approaches infinity.
This (IIUC) is the method that @Folly proposed and @Johnny_Bravo rejected because it potentially involved more than one shuffle.
It would, indeed, give the same probabilities as rolling two dice (assuming random shuffling). The average number of shuffles would be slightly more than one (1.05?), but the maximum number of shuffles required is, as you point out, infinite.
Maximum is infinite… but unless you’re using one deck of cards to roll dice for 50 RPGers a night every night, it’s practically bounded at like 4. When was the last time you rolled 5 natural 20s in a row (and it’s a bit harder than that)?
The expected value? Yes, it’s a weighted sum where you need to multiply the probability of that many shuffles by the number of shuffles as well, i.e. sum{0->inf}(20*n/21^n)
I appreciate the input, folks! As promised, I am having trouble grokking the explanations (let alone the absolute gibberish in the two posts immediately above this one ) but I trust the consensus of you big-brains.
What the @$%& is so hard about shuffling that you absolutely REFUSE any suggestion that includes a second shuffle?
How about this: Take a standard deck of cards; remove all face cards. Shuffle. Pick a card. Pick another card. If it’s the same color, pick a third; if it’s the same color, pick a 4th; etc. Bottom line, you get one black card to represent the first die and a red card to represent the second die. Hearts and Spades represent 1 - 10; Diamonds and Clubs represent 11-20. One shuffle; multiple draws; simulates 2d20.
That does not make sense to me. You always shuffle once. 1/21 of the time you shuffle again. (1/21)^2 of the time you shuffle again.
The expected number of shuffles you do is 1 + SUM{0->inf}((1/21)^n).
Expected value should be used when you have weighted probabilities. But there’s no weight to the shuffles that changes. Each shuffle is worth one shuffle.
(I am open to being wrong about this. It’s been a few decades since I took combinatorics. But please explain :))
It looks to me like the expected number of shuffles is (1)(\frac{20}{21}) + (2)(\frac{1}{21})(\frac{20}{21}) + (3)(\frac{1}{21})^2(\frac{20}{21}) + (4)(\frac{1}{21})^3(\frac{20}{21}) + ...
There may be a clever way to get an exact value of this series, but I just calculated the first 15 or so terms and got 1.05.
Because I don’t @&%&ing want to? Because this is FQ and I had a very specific question? Because I owe you exactly zero explanation why, though you can be assured that it’s a rich tapestry full of harrowing intrigue and featuring an emotional, heartbreaking conclusion?
You know what…this is right. I’m not sure what I punched in wrong to my calculator, but this converges to 1.05 - the formula is right but my output was wrong.
) but I trust the consensus of you big-brains.