Ask the Wizard! (No. 244)
Jan. 7, 2010 column
For pai gow and pai gow poker, is it true that there is no table limit on bets between players when a player is banking? What happens if a player, or group of players, bets more than the banker can cover? — dk
To answer the second question first, the banker must have enough chips on the table to cover all bets. If he doesn't, the dealer will give him the choice to buy more or forfeit his turn to bank.
As for the first question, the table limit still applies when a player is banking. It would seem to be good business to allow any bet, because the casino will stand to get 5% of a larger amount. I asked about this at three different casinos. The following is what I was told, in the order I asked:
Casino 1: The Gaming Control Board needs to approve increases in the maximum bet, which they can not do on short notice.
Casino 2: The Gaming Control Board has nothing to do with it. Instead, a casino vice president needs to authorize any increase in the maximum bet, and it is generally only done for known good customers.
Casino 3: Casinos don't need Gaming Control Board approval to raise the maximum bet on a table. My source hadn't heard of a casino allowing unlimited bets in the case of player banking and added, conceptually, there isn't any exposure for the casino, so there wouldn't be a reason to preclude it.
I would add that in my many hours of playing pai gow, I have never once seen anything close to this situation come up. Usually, players don't like to bet against other players, and the maximums are sufficiently high that players rarely bump up against them, regardless of who is banking. However, if the situation happened often enough, I think casinos would indeed re-think their policy and allow unlimited bets.
This question was raised and discussed in the forum of my companion site Wizard of Vegas.
How does the house edge change in blackjack if you can only split aces? — P4u1
It would depend on the other rules, but assuming six decks and double after a split being normally allowed, then the house edge would be increased by 0.39% only. If doubling after splitting is not otherwise normally allowed, then 0.24% only. This situation actually applies in the game Triple Shot, where I get 0.33%, due to it being a single-deck game. Keep in mind, when it comes to blackjack, such figures can be off by 0.03% or so, depending on how the analysis is done.
This question was raised and discussed in the forum of my companion site Wizard of Vegas.
I saw a $1 progressive jackpot at a casino in Michigan based on the flop and player's two hole cards. It pays as follows:
Royal flush: 100% of jackpot
Straight flush: 10% of jackpot
Four of a kind: $300
Full house: $50
Flush: $40
Straight: $30
Three of a kind: $9
What would be the odds on a $105,000 jackpot? — Mike G.
The return for a jackpot of j is 0.530569 + j×0.029242. So, if j=105,000, the return would be 83.76%. For more information, see my page on Ultimate Texas Hold 'Em.
At the luggage carousel in the airport, the more bags I have to retrieve the longer I can expect to wait for all of them to come out. If I have one bag, I would have to wait until about half of the bags come out. If I take 2 bags, my wait is going to be longer and with 3, longer still. Assuming my bags are mixed up randomly among the others, what is a general formula for number of bags I'll have to wait to come out to get all my bags, in terms of my number of bags and the total number of bags? — MrPogle
Let's define some variables first, as follows:
n = number of your bags
b = total number of bags
As the number of total bags gets larger the answer will get closer to b×n/(n+1). For a large plane, that will give you a good estimate. However, if you want to be exact, the answer is
[b×combin(b,n)-(sum for i=n to b-1 of combin(i,n))]/combin(b,n)
For example, if there are 10 total bags, and four of them are yours, then the expected wait time =
[10×combin(10,4)-combin(4,4)-combin(5,4)-combin(6,4)-combin(7,4)-combin(8,4)-combin(9,4)]/combin(10,4) = 8.8 bags.
Solution:
The number of ways to pick n out of b bags is combin(b,n). So, the probability that all your bags come out within the first x bags is combin(x,n)/combin(b,n). The probability that your last bag is the xth bag to come out is (combin(x,n)-combin(x-1,n))/combin(b,n), for x>=n+1. For x=n it is 1/combin(b,n).
So, the ratio of the expected wait time to the total wait time is:
n×combin(n,n)/combin(b,n) +
(n+1)×(combin(n+1,n)-combin(n,n))/combin(b,n) +
(n+2)×(combin(n+2,n)-combin(n+1,n))/combin(b,n) +
.
.
.
+
(b-1)×(combin(b-1,n)-combin(b-2,n))/combin(b,n) +
b×(combin(b,n)-combin(b-1,n))/combin(b,n)
Taking a telescoping sum, this can be simplified to:
[b×combin(b,n)-combin(b-1,n)-combin(b-2,n)-...-combin(n,n)]/combin(b,n)
A reader later wrote in saying that the answer can be simplified to n×(b+1)/(n+1). This can be shown by induction, a legitimate method, but always leaves me emotionally unsatisfied.
This question was raised and discussed in the forum of my companion site Wizard of Vegas.
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