Back to the first question. When we left, we had just worked out the probability of having 0, 1, 2 or 3 minions after the mulligan when going first, if you throw back any minions you see in the mulligan.
3 minions: about 0.000047 or in scientific notation 4.7155909224874888299440023578e-5 (0.0047%)
After mulligan, the set aside cards are shuffled back into the deck which still has 27 cards remaining. Number of minions left in the deck is 27 minus however many minions you have after the mulligan.
We then use the above probabilities to get the odds of getting a certain number of minions after drawing 5 cards from the deck.
0 minions after mulligan case, 8 minions remain in the deck of 27, we draw 5 so we are using H(27, 8, 5, M). These probabilities will eventually be multiplied by the chance to have 0 minions after the mulligan (about 0.81), so I'll do that as well.
1 minion after mulligan case, 7 minions remain in the deck of 27, we draw 5 so we will be using H(27, 7, 5, M-1) to get the probability of having M after 5 draws. The M-1 is because we already have 1 minion. These probabilities will eventually be multiplied by the chance to have 1 minion after the mulligan, so I'll do that as well.
All other cases involve having 2 or more minions after the mulligan, so we can add the probabilities up from the 0 minions after mulligan case and the 1 minion after mulligan case where we ended up with 1 minion to get the probability of having 1 minion after drawing 5. This is
2 minions after mulligan case, 6 minions remain in the deck of 27, we draw 5 so we will be using H(27, 6, 5, M-2) to get the probability of having M after 5 draws. The M-2 is because we already have 2 minions. These probabilities will eventually be multiplied by the chance to have 2 minions after the mulligan, so I'll do that as well.
Multiplying by 0.00752136752136751325670498084288 = 0.00189585417679063043420038830025
This number is pretty small. The calculations from now on aren't going to make much difference to the final results, so we can do an estimate of them now.
For 2 minions case after 5 draws we add up the conditional (multiplied) probabilities from the 0, 1 and 2 minions after the mulligan case, where we ended up with 2 minions, we get
0.27374390998671124086323672848368 + 0.07474459570565355835995911022496 + 0.00189585417679063043420038830025 + numbers smaller than that from the 3 minion case which we haven't yet done (numbers will be smaller still)
This is 0.35038435986915542965739622700889 or about 35%
For 3 minions after 5 draws we add up the numbers from the 0 and 1 minion after mulligan case to get an estimate. The case where we had 2 minions after the mulligan won't make much difference, estimate is:
0.09661549764236867324584825711189 + 0.05276089108634362545906095823446 which is 0.14937638872871229870490921534635, so about 15%
For 4 minions after 5 draws we estimate again using the figures from the 0 and 1 minion after mulligan case. This is
0.01341881911699567621233107117516 + 0.01465580307953990782213329987306 = 0.02807462219653558403446437104822 which is about 2.8%
For 5 minions, we do the same thing to get the estimate. The numbers we add are small
5.6500291018929120130223927897759e-4 + 0.00154271611363578136265991240384 = 0.00210771902382507256396215168282 which is about 0.2%
Those estimates should be good enough, the 6 minion number was already tiny so the chances of that happening is remote.
Summary (estimated, but should be accurate enough):
When you go first, if you throw away all minions from your hand in the mulligan, and have 8 minions out of 30 cards in your deck, the chances to end up with a certain number of minions after drawing 5 cards after the mulligan during normal gameplay, is
0: 0.11731881856573350530027732134375 (11.7%)
1: 0.34701914087882641766126856945563 (34.7%)
2: 0.35038435986915542965739622700889 (35.0%)
3: 0.14937638872871229870490921534635 (14.9%)
4: 0.02807462219653558403446437104822 (2.8%)
5: 0.00210771902382507256396215168282 (0.2%)
6+: negligible
Adding up the numbers gives us estimated chance for drawing 5 or less minions by turn 5 (since we didn't bother to do the case for 6)
so even with an estimate that leaves only 0.6% of the delicious probability pie left unaccounted for.
I may go back and improve the estimate by doing the rest of the calculation from where you had 2 minions after the mulligan. But I'm going to have a beer for now ;)
Great work man ! So, if i understand correctly, these numbers make my test numbers look weird. Because after 5 turns, i had :
1 minion : 1 time (out of 30 games, so roughly 3%)
2 minions : 9 times (out of 30 games, so roughly 30%)
3 minions : 14 times (out of 30 games, so roughly 46%)
4 minions : 6 times (out of 30 games, so roughly 20%)
5 minions : 1 time (out of 30 games, so roughly 3%)
Now , i know that i didn't throw away all minions in all 30 games. But the times i kept a minion were too few (and it was only 1 minion), to cause, in my opinion, such a large difference. What do you think ?
You always keep Barnes in the mulligan (I'll go over this next I think). That adds +1 to number of minions kept after the mulligan straight away, when you see Barnes (which is 1/10th of the time if you go first).
Going second may have more of an effect I may look at that next instead.
You give no info as to whether you had Barnes or not and whether you went first or second, therefore you can't add up the number of minions at the end and compare to the result I calculated (which I hope is correct) - it only applies in the no Barnes, mulligan all minions, going first case.
You also ran Skulking Geist in half the games I think, and if you keep that in the mulligan as well against Druids (maybe Shaman too) that would make a big difference as well.
Another factor I've not even talked about is that if you throw away a spell in the mulligan (because it is too high cost) that means you are more likely to get a minion back instead of a spell. I assume only getting rid of minions in the mulligan.
You're right. But even so: In 30 games, since we assume the game is "fairly random", (and since Blizzard says you alternate going 1st and 2nd about the same times to make it fair), we can safely assume i went 1st 15 times. So that means i saw Barnes 1.5 times. I dont know what is the % to see Barnes if we go 2nd. I assume it isnt very different as its only 1 card difference. But let's say it's ....2.5 times for the other 15 games. That makes 4 times i kept Barnes, out of 30 games.
Will 4 games out 30 make up the huge difference in the numbers we see? It doesn't seem so. If i subtract 4 minions from my 30 games stats, my numbers will not change much.
The half Geist thing was with the previous experiment (different deck), so don't try to calculate anything with that.
So 3/35 to see a Primordial Glyph from the first one and 3/35 to see a Cabalist Tome. (You see 3/35 cards from the discover, so this is the chance to see a specific one, since there are no dupes).
Then you have chance to not get molten reflection as one of your 3 spells is 34/35 * 34/35 * 34/35, which is 39304/42875. Chance to see one is therefore 1 - (39304/42875) = 3571/42875.
Multiplying the 2 together is (3*3571)/(35*42875) = 105/153106625 = 6.8579658130404219934963624206333e-7 (small number indeed).
Multiply by 100 to get a % we get around 0.000069% I believe.
EDIT: No, multiply that by 3/35 again (you need 2 discovers to hit not 1), which is 0.0000059% or so
EDIT2: First answer was getting a Cabalist's Tome from just 1 glyph so not needing another one to get it.
Actually, that changes the odds somewhat since the first choice must NOT include Cabalist's Tome, unless they be trolling for lols.
Will work out that in a bit...
EDIT3: Do you want the chance to get exactly glyph into glyph (no cabalist) into cabalist into molten reflection or do you want the chance to end up with a molten reflection at the end of it all so molten from 1st or 2nd glyph or cabalist from 1st into molten as well (and if mana available more glyphs from the cabalist, lol)?
Back to the first question. When we left, we had just worked out the probability of having 0, 1, 2 or 3 minions after the mulligan when going first, if you throw back any minions you see in the mulligan.
These were:
0 minions: 0.8145122310639552018940956871994 (81%)
1 minion: 0.1779192455054522743746929953825 (18%)
2 minions: 0.00752136752136751325670498084288 (0.8%)
3 minions: about 0.000047 or in scientific notation 4.7155909224874888299440023578e-5 (0.0047%)
After mulligan, the set aside cards are shuffled back into the deck which still has 27 cards remaining. Number of minions left in the deck is 27 minus however many minions you have after the mulligan.
We then use the above probabilities to get the odds of getting a certain number of minions after drawing 5 cards from the deck.
0 minions after mulligan case, 8 minions remain in the deck of 27, we draw 5 so we are using H(27, 8, 5, M). These probabilities will eventually be multiplied by the chance to have 0 minions after the mulligan (about 0.81), so I'll do that as well.
p(M=0 | (M=0 after mulligan)) = H(27, 8, 5, 0) = 0.144035674470457.
Multiplying by 0.8145122310639552018940956871994 = 0.11731881856573350530027732134375
p(M=1 | (M=0 after mulligan)) = H(27, 8, 5, 1) = 0.384095131921219
Multiplying by 0.8145122310639552018940956871994 = 0.31285018284195628563814987823507
p(M=2 | (M=0 after mulligan)) = H(27, 8, 5, 2) = 0.336083240431066
Multiplying by 0.8145122310639552018940956871994 = 0.27374390998671124086323672848368
p(M=3 | (M=0 after mulligan)) = H(27, 8, 5, 3) = 0.118617614269788
Multiplying by 0.8145122310639552018940956871994 = 0.09661549764236867324584825711189
p(M=4 | (M=0 after mulligan)) = H(27, 8, 5, 4) = 0.0164746686485817
Multiplying by 0.8145122310639552018940956871994 = 0.01341881911699567621233107117516
p(M=5 | (M=0 after mulligan)) = H(27, 8, 5, 5) = 0.00069367025888765
Multiplying by 0.8145122310639552018940956871994 = 5.6500291018929120130223927897759e-4 (small number alert)
The multiplied probabilities should add up to the probability of having 0 cards after mulligan, let's check
0.11731881856573350530027732134375 + 0.31285018284195628563814987823507 + 0.27374390998671124086323672848368 + 0.09661549764236867324584825711189 + 0.01341881911699567621233107117516 + 5.6500291018929120130223927897759e-4
= 0.81451223106395467246114549562853
So that is matches to a large amount of significant figures.
Since all the other cases have at least 1 minion after mulligan, we now know the probability to have 0 minions after drawing 5, it is
p(M=0 | (M=0 after mulligan)) = H(27, 8, 5, 0) = 0.144035674470457 * 0.8145122310639552018940956871994 = 0.11731881856573350530027732134375 (11.7%)
1 minion after mulligan case, 7 minions remain in the deck of 27, we draw 5 so we will be using H(27, 7, 5, M-1) to get the probability of having M after 5 draws. The M-1 is because we already have 1 minion. These probabilities will eventually be multiplied by the chance to have 1 minion after the mulligan, so I'll do that as well.
p(M=1 | (M=1 after mulligan)) = H(27, 7, 5, 0) = 0.192047565960609
Multiplying by 0.1779192455054522743746929953825 = 0.03416895803687013202311869122056
p(M=2 | (M=1 after mulligan)) = H(27, 7, 5, 1) = 0.420104050538833
Multiplying by 0.1779192455054522743746929953825 = 0.07474459570565355835995911022496
p(M=3 | (M=1 after mulligan)) = H(27, 7, 5, 2) = 0.29654403567447
Multiplying by 0.1779192455054522743746929953825 = 0.05276089108634362545906095823446
p(M=4 | (M=1 after mulligan)) = H(27, 7, 5, 3) = 0.0823733432429084
Multiplying by 0.1779192455054522743746929953825 = 0.01465580307953990782213329987306
p(M=5 | (M=1 after mulligan)) = H(27, 7, 5, 4) = 0.00867087823609563
Multiplying by 0.1779192455054522743746929953825 = 0.00154271611363578136265991240384
p(M=6 | (M=1 after mulligan)) = H(27, 7, 5, 5) = 0.000260126347082869
Multiplying by 0.1779192455054522743746929953825 = 4.6281483409073458671721922660494e-5 (teeny number alert)
Sanity check the multiplied values add up to about 0.1779192455054522743746929953825:
0.03416895803687013202311869122056 + 0.07474459570565355835995911022496 + 0.05276089108634362545906095823446 + 0.01465580307953990782213329987306 + 0.00154271611363578136265991240384 + 4.6281483409073458671721922660494e-5
= 0.17791924550545207848560369387954
Good enough for me ;)
All other cases involve having 2 or more minions after the mulligan, so we can add the probabilities up from the 0 minions after mulligan case and the 1 minion after mulligan case where we ended up with 1 minion to get the probability of having 1 minion after drawing 5. This is
0.31285018284195628563814987823507 + 0.03416895803687013202311869122056 = 0.34701914087882641766126856945563
Which is about 34.7%
2 minions after mulligan case, 6 minions remain in the deck of 27, we draw 5 so we will be using H(27, 6, 5, M-2) to get the probability of having M after 5 draws. The M-2 is because we already have 2 minions. These probabilities will eventually be multiplied by the chance to have 2 minions after the mulligan, so I'll do that as well.
p(M=2 | (M=2 after mulligan)) = H(27, 6, 5, 0) = 0.2520624303233
Multiplying by 0.00752136752136751325670498084288 = 0.00189585417679063043420038830025
This number is pretty small. The calculations from now on aren't going to make much difference to the final results, so we can do an estimate of them now.
For 2 minions case after 5 draws we add up the conditional (multiplied) probabilities from the 0, 1 and 2 minions after the mulligan case, where we ended up with 2 minions, we get
0.27374390998671124086323672848368 + 0.07474459570565355835995911022496 + 0.00189585417679063043420038830025 + numbers smaller than that from the 3 minion case which we haven't yet done (numbers will be smaller still)
This is 0.35038435986915542965739622700889 or about 35%
For 3 minions after 5 draws we add up the numbers from the 0 and 1 minion after mulligan case to get an estimate. The case where we had 2 minions after the mulligan won't make much difference, estimate is:
0.09661549764236867324584825711189 + 0.05276089108634362545906095823446 which is 0.14937638872871229870490921534635, so about 15%
For 4 minions after 5 draws we estimate again using the figures from the 0 and 1 minion after mulligan case. This is
0.01341881911699567621233107117516 + 0.01465580307953990782213329987306 = 0.02807462219653558403446437104822 which is about 2.8%
For 5 minions, we do the same thing to get the estimate. The numbers we add are small
5.6500291018929120130223927897759e-4 + 0.00154271611363578136265991240384 = 0.00210771902382507256396215168282 which is about 0.2%
Those estimates should be good enough, the 6 minion number was already tiny so the chances of that happening is remote.
Summary (estimated, but should be accurate enough):
When you go first, if you throw away all minions from your hand in the mulligan, and have 8 minions out of 30 cards in your deck, the chances to end up with a certain number of minions after drawing 5 cards after the mulligan during normal gameplay, is
0: 0.11731881856573350530027732134375 (11.7%)
1: 0.34701914087882641766126856945563 (34.7%)
2: 0.35038435986915542965739622700889 (35.0%)
3: 0.14937638872871229870490921534635 (14.9%)
4: 0.02807462219653558403446437104822 (2.8%)
5: 0.00210771902382507256396215168282 (0.2%)
6+: negligible
Adding up the numbers gives us estimated chance for drawing 5 or less minions by turn 5 (since we didn't bother to do the case for 6)
0.11731881856573350530027732134375 + 0.34701914087882641766126856945563 + 0.35038435986915542965739622700889 + 0.14937638872871229870490921534635 + 0.02807462219653558403446437104822 + 0.00210771902382507256396215168282
= 0.99428104926278830792227785588566
so even with an estimate that leaves only 0.6% of the delicious probability pie left unaccounted for.
I may go back and improve the estimate by doing the rest of the calculation from where you had 2 minions after the mulligan. But I'm going to have a beer for now ;)
Great work man ! So, if i understand correctly, these numbers make my test numbers look weird. Because after 5 turns, i had :
1 minion : 1 time (out of 30 games, so roughly 3%)
2 minions : 9 times (out of 30 games, so roughly 30%)
3 minions : 14 times (out of 30 games, so roughly 46%)
4 minions : 6 times (out of 30 games, so roughly 20%)
5 minions : 1 time (out of 30 games, so roughly 3%)
Now , i know that i didn't throw away all minions in all 30 games. But the times i kept a minion were too few (and it was only 1 minion), to cause, in my opinion, such a large difference. What do you think ?
You've not recorded the data correctly.
You always keep Barnes in the mulligan (I'll go over this next I think). That adds +1 to number of minions kept after the mulligan straight away, when you see Barnes (which is 1/10th of the time if you go first).
Going second may have more of an effect I may look at that next instead.
You give no info as to whether you had Barnes or not and whether you went first or second, therefore you can't add up the number of minions at the end and compare to the result I calculated (which I hope is correct) - it only applies in the no Barnes, mulligan all minions, going first case.
You also ran Skulking Geist in half the games I think, and if you keep that in the mulligan as well against Druids (maybe Shaman too) that would make a big difference as well.
Another factor I've not even talked about is that if you throw away a spell in the mulligan (because it is too high cost) that means you are more likely to get a minion back instead of a spell. I assume only getting rid of minions in the mulligan.
You're right. But even so: In 30 games, since we assume the game is "fairly random", (and since Blizzard says you alternate going 1st and 2nd about the same times to make it fair), we can safely assume i went 1st 15 times. So that means i saw Barnes 1.5 times. I dont know what is the % to see Barnes if we go 2nd. I assume it isnt very different as its only 1 card difference. But let's say it's ....2.5 times for the other 15 games. That makes 4 times i kept Barnes, out of 30 games.
Will 4 games out 30 make up the huge difference in the numbers we see? It doesn't seem so. If i subtract 4 minions from my 30 games stats, my numbers will not change much.
The half Geist thing was with the previous experiment (different deck), so don't try to calculate anything with that.
What are the odds of getting primordial glyph into primordial glyph into cabalist tome into molten reflection?
I've been unwell but feeling alot better today, which format (wild/standard), how many mage spells are there in total?
standard, there are 35
So 3/35 to see a Primordial Glyph from the first one and 3/35 to see a Cabalist Tome. (You see 3/35 cards from the discover, so this is the chance to see a specific one, since there are no dupes).
Then you have chance to not get molten reflection as one of your 3 spells is 34/35 * 34/35 * 34/35, which is 39304/42875. Chance to see one is therefore 1 - (39304/42875) = 3571/42875.
Multiplying the 2 together is (3*3571)/(35*42875) = 105/153106625 = 6.8579658130404219934963624206333e-7 (small number indeed).
Multiply by 100 to get a % we get around 0.000069% I believe.
EDIT: No, multiply that by 3/35 again (you need 2 discovers to hit not 1), which is 0.0000059% or so
EDIT2: First answer was getting a Cabalist's Tome from just 1 glyph so not needing another one to get it.
Actually, that changes the odds somewhat since the first choice must NOT include Cabalist's Tome, unless they be trolling for lols.
Will work out that in a bit...
EDIT3: Do you want the chance to get exactly glyph into glyph (no cabalist) into cabalist into molten reflection or do you want the chance to end up with a molten reflection at the end of it all so molten from 1st or 2nd glyph or cabalist from 1st into molten as well (and if mana available more glyphs from the cabalist, lol)?