I don't encourage my students to gamble, but as the field of probability of was begun as a study of games of chance, it seems silly to ignore gambling. When I do use casino games for educational purposes, I always make clear how the house always has the advantage.
My statistics students will soon be exploring the field of probability, so I've taken one for the team and signed up for a (free) casino web site: www.videopoker.com . When I go to Nevada I often enjoy some time playing Hundred Play Draw Poker, which turns out to be an excellent way to demonstrate the difference between theoretical probability and experimental probability. Needing some screenshots for explanation, I took another one for the team and came home from work today and played some video poker.
For example, what's the probability of getting a full house if you're dealt 2 pair? Theoretical probability says there are 4 cards remaining out of 47 that would give you the full house, so the theoretical/mathematical probability of getting a full house would be 4/47, or approximately 8.5%. I played until I was dealt 2 pair:
To get a full house, I could get either a 7 or a 4. There are 2 of each remaining in the deck, so those are the 4 cards I could get, out of 47 remaining in the deck, that would give me a full house. In Hundred Play Draw Poker, you hold the same cards on up to 100 hands (as shown above). Now when I deal, it's akin to running this "experiment" 100 times. What was the result?
13 full houses, or 13% experimental probability on this hand. Versus 8.5% theoretical probability.
Theoretical probability tells you what you can expect in the long run (thousands of hands, not 100), whereas mathematical probability tells you what you actually got. The Law of Large Numbers states that as you get more and more 2 pair deals, the percentage of full houses that you get should approach 8.5%.
I also screenshot a couple times where I was dealt "4 of a suit", since calculating the probability of being dealt a flush is pretty easy (9/47, or 19%), as well as many screenshots of being dealt 2 of a kind and seeing how often I could turn that into 3 of a kind (experimental was 54/500).
Now I have to sit down and figure out the theoretical probability of that last occurrence, making sure that my 3-of-a-kinds don't include 4-of-a-kinds.
5 comments:
I believe it would be 3*(2/47)*(45/46)*(44/45) which is .122109.
Wait ... if you want to discount 4 of a kinds, don't you also want to discount a full house? That would make it a lot trickier because the three discards have a lower probability of drawing two like cards than the nine types that were not originally dealt.
Back to the equations...
OK, let's try this:
3*(2/47)(45/46)(44/45) - [(3*(3 choose 2)) + (9*(4 choose 2)) / (45 choose 2)]
.122109 - (63/990) = .058473 probability of getting 3-of-a-kind only.
Which is way off of your 54/500 (.108) experimental result.
So, did I do the math correct this time Darren?
Corrected spelling from a comment I posted last night:
I just got home and I'm exhausted. I'll check maybe tomorrow.
Yes, you have to discount full houses, but *only* full houses wherein you get the 3rd card of the pair you were dealt but *not* those where your discards cause you to be dealt a 3-of-a-kind.
I got (3C1*45C2-3C2*45C1)/47C3, which is about 15.1%. And I haven't taken off for the full houses in which the 3-card set is made from the pair I was dealt and kept.
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