Ask The Wizard #347
What are the odds of being dealt three cards towards a royal flush and then completing it twice during ten hands, all in the same suit?
For the initial royal flush, the chances of receiving three cards in any suit amounts to 4 times the combinations of 5 cards taken 3 at a time, multiplied by the combinations of 47 cards taken 2 at a time, all divided by the combinations of 52 cards taken 5 at a time, resulting in approximately 0.01663742. The likelihood of completing the royal flush with the draw is the inverse of the combinations of 47 taken 2, yielding around 0.00092507. Therefore, the cumulative probability for both occurrences is about 0.01663742 multiplied by 0.00092507, equating to 0.00001539, or 1 in 64,974 chances.
The likelihood of achieving two royal flushes in two different suits across ten hands is calculated as the combinations of 10 taken 2, multiplied by 0.00001539.2(1-0.00001539)8= 0.00000001065810. Given that the two royal flushes must be in the same suit, the probability that the second royal flush corresponds with the first is 1 in 4, therefore, dividing the earlier probability by 4 gives you about 0.00000000266453, equivalent to 1 in 375,301,378.
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Imagine a game show featuring two players who are both self-serving and highly logical. Here are how the rules play out.
- The host places a sum of $1,000,000 on the table between the contestants.
- Contestant A is asked to propose a division of the money between both players.
- Contestant B will then decide whether to accept or turn down the proposal.
- If Contestant B accepts the proposal, they split the money accordingly, and the game concludes.
- If Contestant B rejects the offer, the host deducts 10% from the total amount on the table.
- Subsequently, the host invites Contestant B to make a new suggestion, and Contestant A will again have the opportunity to accept or reject.
- If Contestant A agrees to the new proposal, they proceed to split the money as suggested. However, if A declines, another 10% is taken from the remaining balance by the host. The process then loops back to step 2 and repeats until one proposal is accepted.
The dilemma is how Contestant A should approach the initial suggestion about dividing the prize.
A should propose to take home 10/19 of the total, minus a penny, while offering Contestant B 9/19, plus a penny. Essentially:
A: $526,315.78
B: $473,684.22
The strategy revolves around getting Contestant B as close to an indifferent point as possible.
Let’s designate the ratio of the total pot to the other player as r. If B accepts A's offer, they would receive r multiplied by $1,000,000.
However, if B rejects the proposal, the host will subtract 10%. After this, B will have a strategic advantage and can propose a division retaining r for themselves while giving A 1-r.
Solving for r..
r×$1,000,000 = (1-r)×$900,000.
r×$1,900,000 = $900,000.
r = $900,000/$1,900,000 = 9/19.
A does not want B to reach total indifference, as it might lead B to choose randomly, risking the host's deduction of the prize. Therefore, A should offer B the extra penny, adjusting the offer to (9/19) multiplied by $1,000,000 plus $0.01, amounting to $473,684.22.
A: $526,315.78
B: $473,684.22
This topic is currently being discussed in my forum at Wizard of Vegas .
BetMGM occasionally provides what they describe as a 'Risk Free Bet,' although it's not completely without risk. A more accurate description could be 'a second opportunity' bet. Here are the conditions.
- Players can place a wager, capped at a maximum amount, on any event (exclusions apply for parlays, teasers, etc.).
- If the wager is successful, the player receives the payout as usual.
- If the wager fails, the player is granted a promotional bet that equals the lost amount.
- The promotional bet can also be placed on any single event.
- Should the promotional bet win, the player will receive the earnings. However, if it loses, the player receives nothing. In either scenario, the promotional bet is forfeited.
Here are my questions:
- What is the expected outcome of a $100 Risk Free Bet when placed against the spread at -110 odds?
- What strategy do you recommend?
Let’s analyze betting against the spread with -110 odds, assuming a 50% chance to win for each bet.
- There is a 50% chance that the initial bet will succeed, resulting in a profit of $90.91.
- There’s a 25% chance where the original bet fails but the second one succeeds. In this case, the player would have lost $100 but gained $90.91, leading to a net result of -$9.09.
- There’s also a 25% chance of losing both bets, which means a $100 loss.
Consequently, the expected value of this promotional betting opportunity is calculated as 0.5 multiplied by $90.91 plus 0.25 multiplied by -9.09 plus another 0.25 multiplied by -100, resulting in an expected value of $18.18.
Now, what is my recommendation? I advise placing a wager on the most significant underdog you can find. At the time of your inquiry, the best underdog I identified was this college football matchup:
Miami (FL) +575
Alabama -1000
Assuming the house edge remains consistent across both bets, the prospects of Miami winning stand at 14.01%. This leads to a house edge of 5.41% in both directions.
Let's presume that if the player suffers a loss, they’ll locate another game with the same odds to apply the second chance.
- Here are the potential outcomes:
- There’s a 14.01% probability of winning the original bet, maximizing a profit of $575.00.
- There’s also a 25% chance of losing both bets, which means a $100 loss.
There exists a 12.05% chance of losing the initial wager while winning the subsequent one. In this situation, the player loses $100 but gains $575, achieving a net profit of $475.
The anticipated value of this promotional betting situation calculates to 0.1401 multiplied by $575, plus 0.1205 multiplied by $475, plus 0.7394 multiplied by -$100, resulting in an expected value of $63.87.
This topic is currently being discussed in my forum at Wizard of Vegas .