I have been trying to teach my dad about gambling (I play poker for a living) and I posed him this question "If you had a 2/3's chance of winning even money on a $100 bet but you have to make the bet 10,000 times, would you make the bet?" I explained to him it would be like rolling a 6 sided dice and 1-4 he wins, and 5/6 he loses.
He answered no because he was scared of losing money. I attempted to find how to do this math problem but I am not as good at math as I wish I was. Could someone tell me, with the formula, what the odds of him losing money would be? So what are the odds he wins 49.999999999999999999999999% with a 2/3's chance to win over 10k times.
Thank you for whoever helps or even thinks about this for a moment.
Edit: I am looking for the formula, I know it is near impossible to lose money in this sample.
If someone said to me they'd let me roll a 6-sided die and win on the 1-4, lose on the 5-6, for every penny I had, but I could only do it one time, I'd be terrified.
But if he said we could do it 10,000 times, it would be impossible for me to lose with a fair die.
Odds of being ahead after 10,000 trials are basically 100 percent.
I calculate the chances of being behind after 10k trials at 5.01237274920645×10^-3011
Start with the expected value of the wagerQuote: killamfHe answered no because he was scared of losing money.
I attempted to find how to do this math problem but I am not as good at math as I wish I was.
Could someone tell me, with the formula, what the odds of him losing money would be?
So what are the odds he wins 49.999999999999999999999999% with a 2/3's chance to win over 10k times.
Thank you for whoever helps or even thinks about this for a moment.
(2/3)*$100 + (1/3)*-100 = $33.33
EV/weighted average bet = House Edge
$33.33 / $100 = .3333...
10,000 wagers * 33.3% = $333,333.33 expected value of 10k wagers
p (prob of winning) = 2/3
Standard Deviation of 1 bet =(SQRT(p*(1-p)))*2*$100 = $94.28
of 10k bets
$94.28 * SQRT (square root) of 10,000 (100) = $9428.09
$333,333.33 / $9428.09 = 35.35 standard deviations needed to NOT show a profit
I do not know what that is or even care to know
7 SDs is 1 in 390,682,215,445
you want 5 times more.
won't happen
btw you won't convince him using logic. His reluctance is based on something emotional.
Quote: 7craps
7 SDs is 1 in 390,682,215,445
you want 5 times more.
won't happen
35 SD is way way way way way way way way more than 5 times 7 SD
Quote: sodawater35 SD is way way way way way way way way more than 5 times 7 SD
Really? 3SD is definitely 3 * 1SD. why wouldn't 35 SD be 5 * 7 SD?
Quote: rdw4potusReally? 3SD is definitely 3 * 1SD. why wouldn't 35 SD be 5 * 7 SD?
You guys are talking about 2 different measures of 'times'. 3SD is 3*1SD, but being 3SD away from the mean is not 1/3 as likely as being 1SD away.
Quote: sodawater35 SD is way way way way way way way way more than 5 times 7 SD
Yeah, this is a good estimate for it:
Pr(Z>z)≤(z*sqrt(2*PI))*e^(-z^2/2)
Using z = 35 this is: 8.665 x 10^(-265).
About the same chances as rolling 211 "Yo"s in a row.
A large enough bankroll is also still required.Quote: killamfEdit: I am looking for the formula, I know it is near impossible to lose money in this sample.
say you have $500
5 betting units for your example.
96.9697% success probability to double to $1k or be ruined.
Not a sure thing.
say you have $1000
10 betting units.
Still only a 99.9024% chance of doubling to $2k before ruin.
(you can use the Gambler's Ruin formula for this)
https://wizardofvegas.com/forum/questions-and-answers/math/1360-chance-of-winning-10-units-before-losing-10-units-with-plain-bet-at-non-even-games/#post129741
So 1,000 out of 1 million would exhaust a $1k bankroll having such a high edge in their favor
of course, They would be the one's to make the Ellen show
Playing scared to lose real money is good if it makes you have more fun
can not think of any other reason
Good Luck
"...I play poker for a living"
was the most interesting to me.
Quote: EdCollinsOut of everything that was written in this thread, this statement,
"...I play poker for a living"
was the most interesting to me.
There are a good chunk of people doing it "for a living" now. Tough road, but the best make some good bucks. If I was good enough to make $20/hr+ at NLHE (I'm not, but some are), I'd consider it.
$100 bet
win 66%
lose 33%
10,000 times
win avg of $66
lose avg of $33
the difference between the 2 is +$33
33/2 = $16.5, so on average you can expect to make $16.50 per bet. (also known as equity) - this is how poker players make money
so over 10,000 times you should net around 10,000 x 16.5 = $165,000
Quote: killamfHe answered no because he was scared of losing money. I attempted to find how to do this math problem but I am not as good at math as I wish I was.
The answer you are looking for (i.e. what are the odds of losing money after 10k bets) is not the answer to the question your dad raises.
Ask yourself, if the bet is about $1M, would you still do it ?
The question of our father is not about, how you end up after 10k play. If he is afraid of losing money, his question is "what is the probability losing it all during the course of action". That is the question commonly known as "risk of ruin", and it strongly depends on your definition of "ruin" (i.e. the amount you are forced to quit when lost).
Quote: MangoJIf he is afraid of losing money, his question is "what is the probability losing it all during the course of action".
I wouldn't be shocked he is somewhat afraid that his child may also lose money on poker, so he is not wanting to openly examine gambling from an advantage perspective. I expect many people with little knowledge of gambling mathematics to feel this way if their child is trying to make a living from poker or other forms of advantage gambling.
First you have to break the stereotype that all gamblers lose.
Now you have to explain house advantage/player advantage.
Now you have to explain that there are certain games where the player can gain an advantage over the LONG term, not the short term.
Now you have to explain risk of ruin/kelly betting in order to grow the bankroll.
If they stuck around that long they are finally educated in what you are doing. But I bet they are still skeptical if you have consecutive losing sessions (Standard for video poker players/card counters/poker players)
Even if you have winning years, they will still ask, "but how much did you lose?" - 2+2er's understand this phrase :)
Quote: tringlomane
About the same chances as rolling 211 "Yo"s in a row.
+1