Floating Watcher + Spreading Madness probabilities?

In arena, I got a Warlock draft with both Floating Watcher and Spreading Madness. I haven't been able to play them together yet, but the concept of it got me wondering what the expected probabilities look like.

I mean, max gain from this combo is a 22/22 Floating Watcher but chances of that are slim to none. But what are the odds that I kill the Watcher? And what would my average result look like?

I would think of this as a potential comeback strategy, so I wouldn't try this if I had minions on my side of the field. If I did, I'd trade them off first. If necessary for the comparision, assume there is nothing else on the board (best case scenario).

Solution 1:

I'm not sure if Floating Watcher keeps getting buffed if it takes lethal damage during resolution of Spreading madness (similar to avenge triggering on "dead" minions, saving them), or if it just dies if it gets to 0 health during resolution. I wrote a program looking through all the possibilities and the results were as follows, assuming only the Watcher is on the board and at 4 health:

Results if it keeps getting buffed:

Dead: 3.6%  
4:    0.7%  
6:    10%  
8:    23.4%  
10:   27.3%  
12:   20.5%  
14:   10.2%  
16:   3.4%  
18:   0.7%  
20:   0.1%  
22:   0.005%  

Results if it stops getting buffed:

Dead: 7.2%  
4:    0.7%  
6:    8.5%  
8:    21.8%  
10:   26.9%  
12:   20.4%  
14:   10.2%  
16:   3.4%  
18:   0.7%  
20:   0.1%  
22:   0.005%

Solution 2:

Assume besides Floating Watcher there are X other minions on the battlefield, so X+1 minions in total. If one damage is dealt randomly between all X+3 characters (the 2 players in addition), the probability for you being hit is Y:=1/(X+3) (Y defined as 1/(X+3)). The probability for you not being hit is (X+2)/(X+3)=1-(1/(X+3))=1-Y.

Now 9 damage are being dealt randomly between all characters. Since for every damage the target is chosen independently from the other damage targets the probability that you are hit by exactly A of the 9 damage is

(9 choose A) (Y^A)(1-Y)^(9-A)

which is the binomial distribution, where (9 choose A) is the binomial coefficient (Please look it up, calculator e.g. here).

Examples:

The probability for you being hit 9 times is (9 choose 9) Y^9. Assuming there is no other minion on the battlefield except the Watcher, this is 0.00508 % (times 100 to get %).

The probability for being hit 0 times is (9 choose 0) (1-Y)^9. Assuming there is no other minion on the battlefield except the Watcher, this is 2.60122 % .

The probability for being hit exactly 4 times is (9 choose 4) Y^4 (1-Y)^5 = 126 * 0.16257 = 20.48468 % with no other minion on the battlefield.

If you want to know the probability for you being hit more than or equal to 4 times, you have to sum the probabilities for being hit exactly 4, 5, 6, 7, 8 and 9 times. This would be the same as the probability for the Watcher being dead if it does not get buffed during the resolution of Spreading Madness (Which I think is the case), since the probability for you or the Watcher being hit is the same. Please calculate yourself. :-)