Some card games (including MTG) include rules for a mulligan that permit the option to redraw your hand in hopes of getting better cards. The more times you draw a hand, the more likely it is that you will eventually draw a hand that has your desired card combination. Can we quantify this?
The probability of two independent events occurring is the product of the separate probability of the two events. By replacing the initial draw and reshuffling before drawing again we make these two draws independent events.
Probability when redrawing hands
Instead of looking at the probability to draw your combination in the first hand or the redraw, we will look at the probability to not draw your combination twice in a row.
- Assume the probability to draw a hand with your desired card combination is P
- That means that the probability of not drawing your combination is 1-P
- The chance of not drawing your combination after one mulligan is (1-P) * (1-P) = (1-P)^2
- The probability of drawing your combination with one mulligan is 1-( (1-P)^2 )
If we are trying to draw a 2-card combination in a 60 card deck where we have 4 of each, our chance after drawing 7 cards is P=14.5%. Being willing to mulligan twice (N=3) increases our chance to 37.5%
1-( (1-P)^N )
= 1-( (1-0.145)^3 )
= 1-( 0.855^3 )
= 1-( 0.625 )
= 0.375
Hello, I am working on a similar probability calculator and have been thinking about the topic of this post deeply for some time. I wanted to discuss some logic with you. in the turn 4 calculation that gets the outcome of 0.307, that appears to assume the same deck size and the prior mulligan draws. To be more accurate, would the turn 4 calculation subtract the cards in your hand from the deck size? So going with 53 cards in deck vs. 60.
ReplyDeleteNot sure here, wondering what your thoughts are!
I think I've answered my own question. you WOULD NOT lower the deck size, because the hypergeometric function is taking this into account with the 7+4 hard size? so then I think you're correct...
ReplyDelete