Recent content by Lenus

mfb The term "not chosen at random" was the key, thanks a lot!

There is a card game where one player gets three cards and only uses two (the third one is discarded without showing to an opponenet), and the second player gets also three cards and uses only two, discarding the third card in the similar way - without showing. I am trying to enumerate all...

mfb, the result i mentioned earlier, that is known to me was achieved by the method you described, but I needed to know how to approach these problems from "this" side too )) That probability of missing some draws are easier for me, I denote them by q: q_any middle = C(40,3)/C(52,3) q_any...

mathman I realize my mistake in breakdown groups. If the first draw is the Middle card, the second could be any of the spades, that's the catch (well, for me)... I will re-run my calculation. Thank you very much

Thanks for the comment. I realized this would have been correct if I drew only 2 cards, but couldn't figure out how to apply the fact that there are 3 cards drawn. Would that be multiplying every result by C(3,2)? with regards to minor flow: S,M = 13,9 (corrected) S,Ms = 10,3 M,Ms = 9,3...

In a game of Open Face Chinese poker, I encounter this problem quite often. One draws three cards at a time and one of these cards must be discarded, so two of three drawn cards can be used. A typical problem and question in this game would be: What is the probability of drawing one spade AND...

ahh, yes of course 2/30. thanks.

May I offer my solution (I am actually checking my understanding of this type of problems) The way I would tackle this problem is: getting no ace in first four cards: q1 = C(28,4)/C(30,4). p1=1-q1 getting no ace on second four cards: q2 = C(24,4)/C(26,4). p2=1-q2 and the last draw: p3= 1/30, as...

Wow, that's interesting. Thanks again. Thats what was bothering me for weeks )) Could I have a glance at your actual calculations when you re finished. Although I am not a mathematician, I love to see the method used in action.

PeroK, This is great. Thanks a lot for your time.

Many thanks for your help. Yes I would be interested in your solution for backgammon cube (2,4,8,16,32,64), especially when to issue an initial double / redouble. In this game the cube looks like a die, so increments are just one point every time (1,2,3,4,5,6...) Consequently, you have a 25%...

You see, the introduction of doubling cube (which is of arithmetical progression, btw) complicates the strategy. The simplest examples of all are: 1. Should a first player double the stakes before first roll? 2. If he did so, and rolled 5 pts, should p2 redouble thus giving away a chance to be...

Rules are: if you roll lower or equal number than the opponent, you lose. Dice are rolled until one side loses, therefore one game can consist of several rolls. Obviously, player rolling 66 wins automatically, because ties lose, and even 66 won't save the bet. Example, p1: 7 points, p2: 8...

I tackle the following game analysis: 2 players, two 6-sided dice. Bigger sum of points win. First roller has an advantage, as he wins even if 2nd player's dice sum equals to his. As the game is played with doubling cube (potentially increasing the odds before any roll), I tried to enumerate...

I have manually double-checked the result from the formula for the deck of 8 and it is just spot on! Appreciate your help, thanks a lot!