d*****0
发帖数: 1500 | 1 preflop hu allin
run 三次 放回和不放回
区别在哪里 | p****r
发帖数: 9164 | 2 It is the same.
This is from the blog a of Ph.d student @ CMU CS, studying AI and
game theory in poker.
" In the previous examples we assumed the cards were dealt with
replacement; but in reality they are not. Does this change anything? It isn'
t obvious at all, but it turns out that your expected payoff is the same in
both settings.
Let P be the size of the pot.
Let p be the probability P1 wins the first hand.
Let w be the probability P1 wins the second hand given he wins the first
hand.
Let L be the probability P1 wins the second hand given he loses the first
hand.
P1's expected payoff running it once is p*P
P1's expected payoff running it twice (without replacement) is
pwP + p(1-w)P/2 + (1-p)LP/2
= (P/2)(pw + p + L - pL)
Now, note that the probability that P1 wins the second hand is the same as
the probability that he wins the first hand. This is because for community
card sequences c1 and c2, the probability c1 is the first sequence and c2 is
the second sequence is the same as the probability that c2 is the first
sequence and c1 is the second sequence.
So Prob(P1 wins 2nd hand) = p.
But we know (Prob P1 wins 2nd hand) = Prob(P1 wins 2nd hand | P1 wins 1st
hand) * Prob (P1 wins 1st hand) + Prob(P1 wins 2nd hand | P1 loses 1st hand)
* Prob(P1 loses 1st hand) = wp + L(1-p)
So we have p = wp + L(1-p)
So substitution into the expression above yields:
P1's expected payoff running it twice (without replacement) is
(P/2)(2p) = p*P
So P1's expected payoff is the same running it once and twice.
"
【在 d*****0 的大作中提到】
: preflop hu allin
: run 三次 放回和不放回
: 区别在哪里
| p**********1
发帖数: 1458 | 3 I don't quite understand his argument.
This is because for community
card sequences c1 and c2, the probability c1 is the first sequence and c2 is
the second sequence is the same as the probability that c2 is the first
sequence and c1 is the second sequence.
how about the case where sequence c1=c2? for instance, P1 has AA, P2 has KK,
flop A72r. with replacement, we can have c1=c2={K K}, without replacement,
we cannot. furthermore, without replacement, if c1={K X}, then c2=/={K K}
with probability 1, but not true if replacement is allowed.
probably I misunderstood his argument or my reasoning is flawed?
edit: he assumed c1 and c2 independent?? true, with replacement, false without
replacement.
isn'
in
【在 p****r 的大作中提到】
: It is the same.
: This is from the blog a of Ph.d student @ CMU CS, studying AI and
: game theory in poker.
: " In the previous examples we assumed the cards were dealt with
: replacement; but in reality they are not. Does this change anything? It isn'
: t obvious at all, but it turns out that your expected payoff is the same in
: both settings.
: Let P be the size of the pot.
: Let p be the probability P1 wins the first hand.
: Let w be the probability P1 wins the second hand given he wins the first
| d*****0
发帖数: 1500 | 4 概率哥果然是看得非常细。
他那段的意思是，因为组合c1和组合c2实际上，相互独立，比如你的那个例子， 如果
c1 是kk 那c2就不可能是kk，但是，会是另两张牌比如7s8s，因为kk和7s8s是相互独立
的，从概率上讲，c2是kk和c1是7s8s的概率同之前的概率相同
另外，我之前纠结的是放回和不放回类似于以前概率里学的抓阄问题。
明显，如果放回的话，最后一个抓阄的同学获益。但是，后来想通了，抓阄的话，是任
何一个人抓到了，就停止了，不继续抓下去，这个与我们这个问题情况不同。
如果抓阄改个规则，放回但是每人都会轮到。那不管中间抓阄是相关概率还是独立概率
，每个人的最后的ev不变。
谢谢player | p**********1
发帖数: 1458 | 5 simple example, 1000 balls, 5 black, 995 white.
draw 100 times (i.e., run 100 times), what is the probability you get 6 or
more black balls?
without replacement, it is 0. with replacement, >0. | d*****0
发帖数: 1500 | 6 不如简化这个例子 3个球 1个黑球，抓3次，放回和不放回
ev算下来都是1
但是，如果是放回的话，有可能抓到2次或3次黑球，可是放回的ev还是1，感觉是抓不
到的概率even了抓到更多的概率 | p**********1
发帖数: 1458 | 7 I was talking about card combo, a sequence of events. if a King gives you a
win, then different story. Anyway, I am done with this topic. |
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