My Math Forum Probability and Expected Value Question regarding Blackjack

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 October 24th, 2017, 06:10 AM #1 Newbie   Joined: Oct 2017 From: United States Posts: 1 Thanks: 0 Probability and Expected Value Question regarding Blackjack Hi there, Have a homework question related to Blackjack for probabilities and expected values. For the sake of simplicity, the question gives the dealer a 50.255867% chance of winning each hand. The player has a 49.744133% chance of winning each hand. If 75 hands are dealt, and the player bets 100 on each hand, what is the probability he will win 500? What is the probability he will lose 500? Also if 100 hands are dealt, and he bets 100 on each hand, what is the total expected value given his 49.744133% chance of success on each hand?
 October 24th, 2017, 06:26 AM #2 Global Moderator   Joined: Dec 2006 Posts: 18,554 Thanks: 1479 How much does the player win if he wins a hand? If he doesn't win a hand, how much does he lose? Can he "split"?
October 24th, 2017, 01:56 PM   #3
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Quote:
 Originally Posted by Durango Hi there, Have a homework question related to Blackjack for probabilities and expected values. For the sake of simplicity, the question gives the dealer a 50.255867% chance of winning each hand. The player has a 49.744133% chance of winning each hand. If 75 hands are dealt, and the player bets 100 on each hand, what is the probability he will win 500? What is the probability he will lose 500? Also if 100 hands are dealt, and he bets 100 on each hand, what is the total expected value given his 49.744133% chance of success on each hand?
For the first question use the binomial distribution (calculation will be messy).
For the second, let p= prob. of winning. After each hand expect (2p-1)100, so after 100 hands, expect 100 times the value for one hand.

 October 26th, 2017, 07:24 PM #4 Newbie   Joined: Oct 2017 From: US Posts: 13 Thanks: 1 If the payout is 1 to 1 and each bet is 100 dollars, it needs exactly 40 wins and 35 losses to gain 500 dollars after 75 hands. It translates to a binomial probability question for exactly 40 successes out of 75 trials with a success probability of 0.49744133. I put (40/75/0.49744133) in this free web-based binomial calculator and get an exact probability of 0.075848, or 7.5848%. Similarly, it needs exactly 35 wins and 40 losses to lose 500 dollars after 75 hands. It translates to a binomial probability question for exactly 35 successes out of 75 trials with a success probability of 0.49744133. I put (35/75/0.49744133) in the same binomial calculator and get an exact probability of 0.07983, or 7.983%. Also if 100 hands are dealt, and he bets 100 on each hand, given his 49.744133% chance of success on each hand, the total expected value would be 49.744133*100-50.255867*100=-51.1734, or losing 51.1734 dollars. Last edited by James White; October 26th, 2017 at 07:27 PM.

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