Gambling probability problem

#1
A slot machine works on inserting a $1 coin. If the player wins,
the coin is returned with an additional $1 coin, otherwise the original coin is
lost. The probability of winning is 1/2 unless the previous play has resulted
in a win, in which case the probability is p < 1/2. If the cost of maintaining
the machine averages $c per play (with c < 1/3), give conditions on the value
of p that the owner of the machine must arrange in order to make a prot in
the long run.
 

Ace864

New Member
#3
Do you think that it is possible to calculate probabilities of gambling games? Of course, there are some theories and even tasks regarding calculating the dice or other gambling probabilities. But, in my opinion, it is pointless to try to figure out the chances of winning in gambling games. In real life, most of the theories do not work, and you lose despite all predictions. That’s why I prefer betting on soccer games rather than playing cards or gamble in casinos. When you bet on soccer games, you are more likely to win something if you know something about the teams you are betting on.
 
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hlsmith

Less is more. Stay pure. Stay poor.
#4
If the inputs are known, yup. Six sides to a fair die, number of cards in a deck or slots on a Roulette wheel, etc.
 

Dason

Ambassador to the humans
#5
If the inputs are known, yup. Six sides to a fair die, number of cards in a deck or slots on a Roulette wheel, etc.
Come on now - you know better. This is a 3 year old thread about a homework problem. Let's not encourage resurrecting these threads with input that doesn't really add anything.
 

hlsmith

Less is more. Stay pure. Stay poor.
#6
I couldn't remember if you couldn't post a thread until you commented say 6 time or vice versa. I am thinking it is vice versa. But I thought may be they we tryin to hit the eligibility criteria. But yeah, necromancing is all the rage!

See, I just added another non-contributing post to this gem of a relic!
 
#7
All 36 numbers are painted black and red equally (18 black and 18 red; only zero is green). You can choose whether to bet on "red" or "black. If you bet a chip on "red" and guess, the casino gives you back twice what you bet. Why double rather than triple or quadruple? Because the probability of guessing when betting on red or black is 18/36 or 1/2. And therefore, your reward will be twice the bet (i.e., equal to the fraction's denominator). According to roulette rules, you can put a chip on a specific number or the black-red and close several numbers at once with one chip. That is why the probability of winning in MGM99WIN is not tiny.
 
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#8
At school, in math lessons, we passed problems for calculating the probability of gambling, but as you know, these are fictional tasks. Yes, in life you can find the probability of what the outcome of events will be, but in any case it is luck and chance that affects the result. You cannot calculate and expect the result that you have calculated. I used to play here a lot cisdetroit.org together with his brother, he always tried to find some strategy to win, but his plan always collapsed and he did not get the desired result. No mathematician or brilliant tactician will be able to outwit the casino, because the casino in any case remains in the black.
 
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fed2

Active Member
#9
long term prob of success about pstate1inf * .5 + ( 1- pstate1inf)*p, where
pstate1inf = .5*2*q/ (2*q + 1)
according to wolfram. solve recursion pstate1(t + 1) == .5 * pstate1(t) + q * ( 1- pstate1(t) ).

C-like:
p=.2
q = 1 - p;
State = .5; #1 = fair, #0 = p
probs = c(.5,p);

f = function(i){
      y = rbinom(1,1,  State )
     
      if (y ==1){State <<- probs[2] }
      else if (y == 0){ State <<- probs[1] }
      y
    }


games = lapply(1:1000, f )
games = as.vector( unlist( games ) )


A = matrix( c(.5,p,.5,1-p), nrow=2, byrow = T)


# a recurrsion
#pstate1(t + 1) == .5 * pstate1(t) + q * ( 1- pstate1(t) )
#thanks wolfram, for solution.
state1p = function(n,q){

  (   ( .5 - q)^n + 2*q ) /  (2*q + 1)

}

pstate1inf = 2*q/ (2*q + 1)
psuc = pstate1inf * .5 + ( 1- pstate1inf)*p
better late than ever.
 
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