Remy Gibson
United States
http://boardgametop10s.com/audio/bg10.xml

Oh, would that I were actually submitting a strategy article regarding the calculation of probability! In actuality, though, I'm seeking help. Though I know how to calculate basic probability of certain things being drawn, even slightly complex situations befuddle my ability to think things through.
The most basic probability situation would be the "draw a card" trigger. For example, if there is one particular card that you want to draw, and there are five cards in your deck, then you have a 20% chance of drawing that card.
A slightly more complex situation is drawing 2 cards. In this case, you add the successive probabilities together. To continue the previous example, you have a 45% chance of drawing a particular card from a fivecard deck if you have two draws. The math on this is:
 1 in 5 chance (20%) plus a 1 in 4 chance (25%)
(The second chance is 1 in 4, rather than 1 in 5, because when you actually make the second draw, you're now drawing from a fourcard deck.)
But there are more interesting things that probability can tell you. I can see evidence of this in comments made by other players, along the lines of, "He had a 57% chance to beat me in his next two draws." I can only begin to see the hazy outlines of making such a computation, and so I'm seeking help from those mathsinclined among you in how to figure out different situations.
Here are a couple of examples:
 In an endgame situation, I look at my opponent's 10card deck and see that there are several combinations of cards that can bring my downfall. But as long as he doesn't draw that one, I should be okay. How do I compute the probability in a situation like this? (Actually, I think I do understand this one: I believe the answer is roughly 64%, extending the idea from above.)
 Okay, but what if there are two cards that I don't want him to draw?
... And obviously I just can't think of examples. So maybe a better way to put this is: How do you incorporate probability into your decisionmaking when playing Star Realms?

Dániel Lányi
Hungary Budapest * Not Applicable

Aweberman wrote:  In an endgame situation, I look at my opponent's 10card deck and see that there are several combinations of cards that can bring my downfall. But as long as he doesn't draw that one, I should be okay. How do I compute the probability in a situation like this? (Actually, I think I do understand this one: I believe the answer is roughly 64%, extending the idea from above.)
 Okay, but what if there are two cards that I don't want him to draw?
i think the actual solution would be: all possible five card draws out of a 10 card deck are calculated as combination 5 out of 10 (which is 10!/(5!x5!) = 252) and favorable outcomes are combination 5 out of 9 (which is 9!/(5!x4!) = 126) so favorable/all outcomes = 50% chance which is fairly intuitive, because there is equal chance the card is in each half of the deck
with two cards you don't want it's pretty much the same, except instead of 9 you'd have 8, and then it would be 56/252=25% (which again is fairly intuitive, since you'd have 50% for both cards, and then you multiply 50% by 50% which is again 25%
of course all this relies on your opponent not having any cards that draw cards
btw i use this website for calculating combinations and permutations: www.mathsisfun.com/combinatorics/combinationspermutationsc...

Dániel Lányi
Hungary Budapest * Not Applicable

Aweberman wrote:
A slightly more complex situation is drawing 2 cards. In this case, you add the successive probabilities together. To continue the previous example, you have a 45% chance of drawing a particular card from a fivecard deck if you have two draws. The math on this is:
 1 in 5 chance (20%) plus a 1 in 4 chance (25%)
(The second chance is 1 in 4, rather than 1 in 5, because when you actually make the second draw, you're now drawing from a fourcard deck.)
again, correct me if i'm wrong, but a 5 card deck has 5!=120 possible ways of being ordered if you take the card you want at a fixed position (either top, or second) the rest could still have 4!=24 so number of ways with your card on top + number of ways with your card being secong from top divided by all ways the deck could be = (4!+4!)/120 = (24+24)/120 = 48/120 = 0.4
which, again, is very intuitive since the card would have to be in 2 out of 5 positions which is 2/5=0.4

Scott Heise
United States San Jose California

Great topic! I look forward to seeing what folks have to say. I know that I like to do rough estimates of probability at key points in the game. Some situations where
 When I'm second player and I want to know what the probability of my oppponent having $3/4/5 trade for his second buy after having $1/2/3 trade in his first hand.  Estimating how likely I would be to trigger key ally abilities, like Patrol Mech or Blob Carrier, to help me know whether it's worth buying the card or not.  Towards the end of the game, estimating how many more hands my opponent will need to draw enough total combat to kill me (and vise versa) by adding his total combat and dividing by the number of cards in his deck minus card draws.
Aweberman wrote: A slightly more complex situation is drawing 2 cards. In this case, you add the successive probabilities together. To continue the previous example, you have a 45% chance of drawing a particular card from a fivecard deck if you have two draws. The math on this is:
 1 in 5 chance (20%) plus a 1 in 4 chance (25%)
(The second chance is 1 in 4, rather than 1 in 5, because when you actually make the second draw, you're now drawing from a fourcard deck.)
Sorry to be nitpicky, but this isn't quite accurate. The probability of drawing the magic card out of 5 when you draw 2 cards is a bit lower than this because you only get the second draw if the first is unsuccessful.
So the probability is actually: [P_draw1in5] + [P_fail_1in5]*[P_draw1in4] = (1/5) + (4/5)(1/4) = 0.2 + (0.8*0.25) = 0.2 + 0.2 = 40%
Someone on the Star Realms Fancreated Facebook page recently shared this hypergeometric calculator that is very useful for calculating the probability of complex card draw scenarios like this: http://stattrek.com/onlinecalculator/hypergeometric.aspx
For example, for the card draw scenario above:  Population size = # of total cards in deck  Number of successes in population = # of magic cards are in your deck  Sample size = # of cards you're going to draw  Number of successes in sample (X) = how many of the magic cards you want to draw
The probilities it displays below show the chances of drawing exactly the number of magic cards that you want, less than that number, more than that number, etc

Remy Gibson
United States
http://boardgametop10s.com/audio/bg10.xml

See? This is what I mean when I say I don't do probability so well.



Adding on to the posting of the hypergeometric probability calculator above, you can also do multistage calculations using that calculator to get your inputs.
Like, humm, let's see. Let's take a situation that's trickier than just, "she has 10 cards to draw, I'm ok if she doesn't get Dreadnaught" (an easy scenario  just plug "0 successes" into hypergeometric calculator).
Let's build it up slowly and first take a situation where there's a pair of cards you're afraid of. Two Blob Destroyers. You're ok if she gets either one of them + anything else, but not if she gets both. She's drawing 3 cards from a 10 card deck.
Plug that into hypergeometric, still just takes one calculation. Population 10, population successes 2, sample 3, sample successes 1, look at the cumulative probability of 1 (incorporating 0 and 1), you get 0.9333, better known as 14/15ths. Ok, looks pretty safe!
What if there are two pairs you're afraid of? You're ok if she gets one Blob Destroyer, but not the other. You're also ok if she gets either Space Station, or War World, but not both. There's one Space Station and one War World in her card pool too.
The easiest way to do this is to take it for granted that she gets 0 Blob Destroyers, then work out the chance of 0 or 1 yellows given 0 Blob Destroyers. Then do the same for 1 Blob Destroyer.
0 Blob Destroyers: 7/15
Given 0 Blob Destroyers, we know that she's drawn among 8 possible cards. Chance of 01 out of 2 yellows in a pool of 8, drawing 3, is the same style of calculation as above: 25/28 (you'll get in decimal, I'm just simplifying to fractions for ease).
1 Blob Destroyer: 7/15 (one of those funny things where it's the same) Given 1 Blob Destroyer, we know that she's drawn 2 other cards, from a pool of 8 possibilities again. That works out as: 27/28  makes sense, she's really unlikely to get both of them out of only 2 cards.
So our two routes to safety are as follows:
0 Blob Destroyers AND 01 Yellows OR 1 Blob Destroyer AND 01 Yellows
In probability, AND is multiplication, OR is addition.
So that's (7/15 * 25/28) + (7/15 * 27/28), which works out as 13/15 chance that you're not gonna die, which is about 87%.
It's not always that involved! But that's one way to thread your way through those calculations
(and I'm sure I'll have made some vague error above but the general process should be sound!)


